Laplace Transform Solutions

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### Laplace Transform Formulas The Laplace Transform of a function $f(t)$, denoted as $L\{f(t)\}$ or $F(s)$, is defined as: $$L\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) dt$$ **Basic Transforms:** - $L\{1\} = \frac{1}{s}$ - $L\{t^n\} = \frac{n!}{s^{n+1}}$ for $n = 0, 1, 2, ...$ - $L\{e^{at}\} = \frac{1}{s-a}$ - $L\{\sin(at)\} = \frac{a}{s^2+a^2}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ - $L\{\sinh(at)\} = \frac{a}{s^2-a^2}$ - $L\{\cosh(at)\} = \frac{s}{s^2-a^2}$ **Properties of Laplace Transform:** - **Linearity:** $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - **First Shifting Theorem:** $L\{e^{at}f(t)\} = F(s-a)$ - **Second Shifting Theorem:** $L\{f(t-a)U(t-a)\} = e^{-as}F(s)$, where $U(t-a)$ is the Heaviside step function. - **Derivative of Transform:** $L\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$ - **Transform of Derivatives:** $L\{f'(t)\} = sF(s) - f(0)$ $L\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$ - **Transform of Integrals:** $L\{\int_0^t f(\tau) d\tau\} = \frac{F(s)}{s}$ - **Division by t:** $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$, provided $\lim_{t\to 0} \frac{f(t)}{t}$ exists. - **Convolution Theorem:** $L\{f(t) * g(t)\} = F(s)G(s)$, where $f(t) * g(t) = \int_0^t f(\tau) g(t-\tau) d\tau$. **Inverse Laplace Transform Formulas:** - $L^{-1}\{\frac{1}{s}\} = 1$ - $L^{-1}\{\frac{n!}{s^{n+1}}\} = t^n$ - $L^{-1}\{\frac{1}{s-a}\} = e^{at}$ - $L^{-1}\{\frac{a}{s^2+a^2}\} = \sin(at)$ - $L^{-1}\{\frac{s}{s^2+a^2}\} = \cos(at)$ - $L^{-1}\{\frac{a}{s^2-a^2}\} = \sinh(at)$ - $L^{-1}\{\frac{s}{s^2-a^2}\} = \cosh(at)$ ### Problems 21.1 ### Solutions to Problems 21.1 **1. Find the Laplace transforms of:** **1. $e^{2t} + 4t^2 - 3\sin 3t + 3\cos 3t$** *Formulae Used:* - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{e^{at}\} = \frac{1}{s-a}$ - $L\{t^n\} = \frac{n!}{s^{n+1}}$ - $L\{\sin(at)\} = \frac{a}{s^2+a^2}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution:* $L\{e^{2t} + 4t^2 - 3\sin 3t + 3\cos 3t\}$ $= L\{e^{2t}\} + 4L\{t^2\} - 3L\{\sin 3t\} + 3L\{\cos 3t\}$ $= \frac{1}{s-2} + 4\frac{2!}{s^{2+1}} - 3\frac{3}{s^2+3^2} + 3\frac{s}{s^2+3^2}$ $= \frac{1}{s-2} + \frac{8}{s^3} - \frac{9}{s^2+9} + \frac{3s}{s^2+9}$ **2. $1 + 2\sqrt{t} + 3/\sqrt{t}$** *Formulae Used:* - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{t^n\} = \frac{\Gamma(n+1)}{s^{n+1}}$ (for non-integer n, $\Gamma(1/2) = \sqrt{\pi}$) - $L\{t^{1/2}\} = \frac{\Gamma(3/2)}{s^{3/2}} = \frac{(1/2)\Gamma(1/2)}{s^{3/2}} = \frac{\sqrt{\pi}}{2s^{3/2}}$ - $L\{t^{-1/2}\} = \frac{\Gamma(1/2)}{s^{1/2}} = \frac{\sqrt{\pi}}{s^{1/2}}$ *Solution:* $L\{1 + 2t^{1/2} + 3t^{-1/2}\}$ $= L\{1\} + 2L\{t^{1/2}\} + 3L\{t^{-1/2}\}$ $= \frac{1}{s} + 2\frac{\sqrt{\pi}}{2s^{3/2}} + 3\frac{\sqrt{\pi}}{s^{1/2}}$ $= \frac{1}{s} + \frac{\sqrt{\pi}}{s^{3/2}} + \frac{3\sqrt{\pi}}{\sqrt{s}}$ **3. $3\cosh 5t - 4\sinh 5t$** *Formulae Used:* - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\cosh(at)\} = \frac{s}{s^2-a^2}$ - $L\{\sinh(at)\} = \frac{a}{s^2-a^2}$ *Solution:* $L\{3\cosh 5t - 4\sinh 5t\}$ $= 3L\{\cosh 5t\} - 4L\{\sinh 5t\}$ $= 3\frac{s}{s^2-5^2} - 4\frac{5}{s^2-5^2}$ $= \frac{3s}{s^2-25} - \frac{20}{s^2-25} = \frac{3s-20}{s^2-25}$ **4. $\cos(at+b)$** *Formulae Used:* - Trigonometric identity: $\cos(A+B) = \cos A \cos B - \sin A \sin B$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\cos(ct)\} = \frac{s}{s^2+c^2}$ - $L\{\sin(ct)\} = \frac{c}{s^2+c^2}$ *Solution:* $\cos(at+b) = \cos(at)\cos b - \sin(at)\sin b$ $L\{\cos(at+b)\} = L\{\cos b \cos(at) - \sin b \sin(at)\}$ $= \cos b L\{\cos(at)\} - \sin b L\{\sin(at)\}$ $= \cos b \frac{s}{s^2+a^2} - \sin b \frac{a}{s^2+a^2}$ $= \frac{s\cos b - a\sin b}{s^2+a^2}$ **5. $\sin(t-\cos t)^2$** (This problem seems to have a typo or implies a complex expression. Assuming it refers to $\sin(t) - (\cos t)^2$ or similar, or it's a non-standard function. Given the context of basic Laplace transforms, it's likely a misinterpretation of the image. If it's $\sin(t) - \cos^2(t)$, we can solve it. Let's assume it's $\sin(t) - \cos^2(t)$ for a solvable problem.) *Formulae Used:* - Trigonometric identity: $\cos^2 x = \frac{1+\cos(2x)}{2}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{\sin(at)\} = \frac{a}{s^2+a^2}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution (assuming $\sin(t) - \cos^2(t)$):* $\sin(t) - \cos^2(t) = \sin(t) - \frac{1+\cos(2t)}{2} = \sin(t) - \frac{1}{2} - \frac{1}{2}\cos(2t)$ $L\{\sin(t) - \frac{1}{2} - \frac{1}{2}\cos(2t)\}$ $= L\{\sin t\} - \frac{1}{2}L\{1\} - \frac{1}{2}L\{\cos 2t\}$ $= \frac{1}{s^2+1^2} - \frac{1}{2}\frac{1}{s} - \frac{1}{2}\frac{s}{s^2+2^2}$ $= \frac{1}{s^2+1} - \frac{1}{2s} - \frac{s}{2(s^2+4)}$ **6. $\sin 2t \cos 3t$** *Formulae Used:* - Trigonometric identity: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\sin(at)\} = \frac{a}{s^2+a^2}$ *Solution:* $\sin 2t \cos 3t = \frac{1}{2}[\sin(2t+3t) + \sin(2t-3t)]$ $= \frac{1}{2}[\sin(5t) + \sin(-t)]$ $= \frac{1}{2}[\sin(5t) - \sin(t)]$ $L\{\frac{1}{2}[\sin(5t) - \sin(t)]\}$ $= \frac{1}{2}[L\{\sin 5t\} - L\{\sin t\}]$ $= \frac{1}{2}[\frac{5}{s^2+5^2} - \frac{1}{s^2+1^2}]$ $= \frac{1}{2}[\frac{5}{s^2+25} - \frac{1}{s^2+1}]$ **7. $\sin \sqrt{t}$** *Formulae Used:* - Maclaurin series for $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ - $L\{t^n\} = \frac{\Gamma(n+1)}{s^{n+1}}$ *Solution:* This is a more advanced Laplace transform, often done using series expansion. $\sin \sqrt{t} = \sqrt{t} - \frac{(\sqrt{t})^3}{3!} + \frac{(\sqrt{t})^5}{5!} - \dots$ $= t^{1/2} - \frac{t^{3/2}}{3!} + \frac{t^{5/2}}{5!} - \dots$ $L\{\sin \sqrt{t}\} = L\{t^{1/2}\} - \frac{1}{3!}L\{t^{3/2}\} + \frac{1}{5!}L\{t^{5/2}\} - \dots$ $= \frac{\Gamma(3/2)}{s^{3/2}} - \frac{1}{6}\frac{\Gamma(5/2)}{s^{5/2}} + \frac{1}{120}\frac{\Gamma(7/2)}{s^{7/2}} - \dots$ $= \frac{\frac{1}{2}\sqrt{\pi}}{s^{3/2}} - \frac{1}{6}\frac{\frac{3}{2}\frac{1}{2}\sqrt{\pi}}{s^{5/2}} + \frac{1}{120}\frac{\frac{5}{2}\frac{3}{2}\frac{1}{2}\sqrt{\pi}}{s^{7/2}} - \dots$ $= \frac{\sqrt{\pi}}{2s^{3/2}} - \frac{\sqrt{\pi}}{16s^{5/2}} + \frac{\sqrt{\pi}}{128s^{7/2}} - \dots$ This form is likely expected for such a problem, or it might hint at a special function. (Alternatively, this is known to be $L\{\sin\sqrt{t}\} = \frac{\sqrt{\pi}}{2s\sqrt{s}} e^{-1/(4s)}$ for $s>0$, but this requires knowledge of special functions or contour integration, which is beyond standard introductory Laplace transforms.) **8. $\sin^5 t$** *Formulae Used:* - Trigonometric identity: $\sin^n x = \text{expand using Euler's formula or power reduction formulas}$ - $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\sin(at)\} = \frac{a}{s^2+a^2}$ *Solution:* $\sin^5 t = \left(\frac{e^{it} - e^{-it}}{2i}\right)^5 = \frac{1}{(2i)^5} (e^{it} - e^{-it})^5$ $= \frac{1}{32i^5} \sum_{k=0}^5 \binom{5}{k} (e^{it})^{5-k} (-e^{-it})^k$ $= \frac{1}{32i} [\binom{5}{0}e^{i5t} - \binom{5}{1}e^{i4t}e^{-it} + \binom{5}{2}e^{i3t}e^{-i2t} - \binom{5}{3}e^{i2t}e^{-i3t} + \binom{5}{4}e^{it}e^{-i4t} - \binom{5}{5}e^{-i5t}]$ $= \frac{1}{32i} [e^{i5t} - 5e^{i3t} + 10e^{it} - 10e^{-it} + 5e^{-i3t} - e^{-i5t}]$ $= \frac{1}{16} [\frac{e^{i5t} - e^{-i5t}}{2i} - 5\frac{e^{i3t} - e^{-i3t}}{2i} + 10\frac{e^{it} - e^{-it}}{2i}]$ $= \frac{1}{16} [\sin(5t) - 5\sin(3t) + 10\sin(t)]$ $L\{\sin^5 t\} = L\{\frac{1}{16} [\sin(5t) - 5\sin(3t) + 10\sin(t)]\}$ $= \frac{1}{16} [L\{\sin 5t\} - 5L\{\sin 3t\} + 10L\{\sin t\}]$ $= \frac{1}{16} [\frac{5}{s^2+5^2} - 5\frac{3}{s^2+3^2} + 10\frac{1}{s^2+1^2}]$ $= \frac{1}{16} [\frac{5}{s^2+25} - \frac{15}{s^2+9} + \frac{10}{s^2+1}]$ **9. $\cos^3 2t$** *Formulae Used:* - Trigonometric identity: $\cos 3x = 4\cos^3 x - 3\cos x \implies \cos^3 x = \frac{1}{4}(\cos 3x + 3\cos x)$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution:* Using $\cos^3 x = \frac{1}{4}(\cos 3x + 3\cos x)$, with $x=2t$: $\cos^3 2t = \frac{1}{4}(\cos(3 \cdot 2t) + 3\cos(2t))$ $= \frac{1}{4}(\cos(6t) + 3\cos(2t))$ $L\{\cos^3 2t\} = L\{\frac{1}{4}(\cos(6t) + 3\cos(2t))\}$ $= \frac{1}{4}[L\{\cos 6t\} + 3L\{\cos 2t\}]$ $= \frac{1}{4}[\frac{s}{s^2+6^2} + 3\frac{s}{s^2+2^2}]$ $= \frac{1}{4}[\frac{s}{s^2+36} + \frac{3s}{s^2+4}]$ **10. $e^{-2t} \sinh bt$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$, where $F(s) = L\{f(t)\}$ - $L\{\sinh(bt)\} = \frac{b}{s^2-b^2}$ *Solution:* Let $f(t) = \sinh bt$. Then $F(s) = \frac{b}{s^2-b^2}$. Using the First Shifting Theorem with $a=-2$: $L\{e^{-2t} \sinh bt\} = F(s-(-2)) = F(s+2)$ $= \frac{b}{(s+2)^2-b^2}$ **11. $e^{2t}(3^t - 4t)$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$, where $F(s) = L\{f(t)\}$ - $3^t = e^{t \ln 3}$ - $L\{e^{ct}\} = \frac{1}{s-c}$ - $L\{t^n\} = \frac{n!}{s^{n+1}}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ *Solution:* Let $f(t) = 3^t - 4t = e^{t \ln 3} - 4t$. $F(s) = L\{e^{t \ln 3} - 4t\} = L\{e^{t \ln 3}\} - 4L\{t\}$ $= \frac{1}{s - \ln 3} - 4\frac{1!}{s^{1+1}}$ $= \frac{1}{s - \ln 3} - \frac{4}{s^2}$ Using the First Shifting Theorem with $a=2$: $L\{e^{2t}(3^t - 4t)\} = F(s-2)$ $= \frac{1}{(s-2) - \ln 3} - \frac{4}{(s-2)^2}$ **12. $e^{3t} \sin 5t \sin 3t$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$, where $F(s) = L\{f(t)\}$ - Trigonometric identity: $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = \sin 5t \sin 3t$. $\sin 5t \sin 3t = \frac{1}{2}[\cos(5t-3t) - \cos(5t+3t)]$ $= \frac{1}{2}[\cos(2t) - \cos(8t)]$ $F(s) = L\{\frac{1}{2}[\cos(2t) - \cos(8t)]\}$ $= \frac{1}{2}[L\{\cos 2t\} - L\{\cos 8t\}]$ $= \frac{1}{2}[\frac{s}{s^2+2^2} - \frac{s}{s^2+8^2}]$ $= \frac{1}{2}[\frac{s}{s^2+4} - \frac{s}{s^2+64}]$ Using the First Shifting Theorem with $a=3$: $L\{e^{3t} \sin 5t \sin 3t\} = F(s-3)$ $= \frac{1}{2}[\frac{s-3}{(s-3)^2+4} - \frac{s-3}{(s-3)^2+64}]$ **13. $e^{-t} \sin^2 t$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$, where $F(s) = L\{f(t)\}$ - Trigonometric identity: $\sin^2 x = \frac{1-\cos(2x)}{2}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = \sin^2 t$. $\sin^2 t = \frac{1-\cos(2t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2t)$ $F(s) = L\{\frac{1}{2} - \frac{1}{2}\cos(2t)\}$ $= \frac{1}{2}L\{1\} - \frac{1}{2}L\{\cos 2t\}$ $= \frac{1}{2}\frac{1}{s} - \frac{1}{2}\frac{s}{s^2+2^2}$ $= \frac{1}{2s} - \frac{s}{2(s^2+4)}$ Using the First Shifting Theorem with $a=-1$: $L\{e^{-t} \sin^2 t\} = F(s-(-1)) = F(s+1)$ $= \frac{1}{2(s+1)} - \frac{s+1}{2((s+1)^2+4)}$ **14. $e^{2t} \sin^4 t$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$, where $F(s) = L\{f(t)\}$ - Trigonometric identity: $\sin^2 x = \frac{1-\cos(2x)}{2}$, so $\sin^4 x = (\frac{1-\cos(2x)}{2})^2 = \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x))$ - $\cos^2 x = \frac{1+\cos(2x)}{2}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = \sin^4 t$. $\sin^4 t = \frac{1}{4}(1 - 2\cos(2t) + \cos^2(2t))$ $= \frac{1}{4}(1 - 2\cos(2t) + \frac{1+\cos(4t)}{2})$ $= \frac{1}{4}(1 - 2\cos(2t) + \frac{1}{2} + \frac{1}{2}\cos(4t))$ $= \frac{1}{4}(\frac{3}{2} - 2\cos(2t) + \frac{1}{2}\cos(4t))$ $= \frac{3}{8} - \frac{1}{2}\cos(2t) + \frac{1}{8}\cos(4t)$ $F(s) = L\{\frac{3}{8} - \frac{1}{2}\cos(2t) + \frac{1}{8}\cos(4t)\}$ $= \frac{3}{8}L\{1\} - \frac{1}{2}L\{\cos 2t\} + \frac{1}{8}L\{\cos 4t\}$ $= \frac{3}{8s} - \frac{1}{2}\frac{s}{s^2+2^2} + \frac{1}{8}\frac{s}{s^2+4^2}$ $= \frac{3}{8s} - \frac{s}{2(s^2+4)} + \frac{s}{8(s^2+16)}$ Using the First Shifting Theorem with $a=2$: $L\{e^{2t} \sin^4 t\} = F(s-2)$ $= \frac{3}{8(s-2)} - \frac{s-2}{2((s-2)^2+4)} + \frac{s-2}{8((s-2)^2+16)}$ **15. $\cosh at \sin at$** *Formulae Used:* - $\cosh x = \frac{e^x + e^{-x}}{2}$ - Trigonometric identity: $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{e^{at}\sin(bt)\} = \frac{b}{(s-a)^2+b^2}$ (from First Shifting Theorem) *Solution (Method 1: Using Euler's form for cosh):* $\cosh at \sin at = \frac{e^{at} + e^{-at}}{2} \sin at$ $= \frac{1}{2} (e^{at}\sin at + e^{-at}\sin at)$ $L\{\cosh at \sin at\} = \frac{1}{2} (L\{e^{at}\sin at\} + L\{e^{-at}\sin at\})$ Using $L\{e^{ct}\sin(bt)\} = \frac{b}{(s-c)^2+b^2}$: For $e^{at}\sin at$: $c=a, b=a \implies \frac{a}{(s-a)^2+a^2}$ For $e^{-at}\sin at$: $c=-a, b=a \implies \frac{a}{(s-(-a))^2+a^2} = \frac{a}{(s+a)^2+a^2}$ So, $L\{\cosh at \sin at\} = \frac{1}{2} \left[ \frac{a}{(s-a)^2+a^2} + \frac{a}{(s+a)^2+a^2} \right]$ $= \frac{a}{2} \left[ \frac{(s+a)^2+a^2 + (s-a)^2+a^2}{((s-a)^2+a^2)((s+a)^2+a^2)} \right]$ Numerator: $s^2+2as+a^2+a^2 + s^2-2as+a^2+a^2 = 2s^2+4a^2$ Denominator: $(s^2-2as+2a^2)(s^2+2as+2a^2) = ((s^2+2a^2)-2as)((s^2+2a^2)+2as)$ $= (s^2+2a^2)^2 - (2as)^2 = s^4+4a^2s^2+4a^4 - 4a^2s^2 = s^4+4a^4$ So, $L\{\cosh at \sin at\} = \frac{a}{2} \frac{2s^2+4a^2}{s^4+4a^4} = \frac{a(s^2+2a^2)}{s^4+4a^4}$ **16. $\sinh 3t \cos^2 t$** *Formulae Used:* - $\sinh x = \frac{e^x - e^{-x}}{2}$ - Trigonometric identity: $\cos^2 x = \frac{1+\cos(2x)}{2}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - First Shifting Theorem: $L\{e^{ct}f(t)\} = F(s-c)$ - $L\{1\} = \frac{1}{s}$ - $L\{\cos(bt)\} = \frac{s}{s^2+b^2}$ *Solution:* $\sinh 3t \cos^2 t = \frac{e^{3t} - e^{-3t}}{2} \cdot \frac{1+\cos(2t)}{2}$ $= \frac{1}{4} (e^{3t} - e^{-3t})(1+\cos(2t))$ $= \frac{1}{4} (e^{3t} + e^{3t}\cos(2t) - e^{-3t} - e^{-3t}\cos(2t))$ $L\{\sinh 3t \cos^2 t\} = \frac{1}{4} [L\{e^{3t}\} + L\{e^{3t}\cos(2t)\} - L\{e^{-3t}\} - L\{e^{-3t}\cos(2t)\}]$ Using $L\{e^{ct}\} = \frac{1}{s-c}$ and $L\{e^{ct}\cos(bt)\} = \frac{s-c}{(s-c)^2+b^2}$: $L\{e^{3t}\} = \frac{1}{s-3}$ $L\{e^{3t}\cos(2t)\} = \frac{s-3}{(s-3)^2+2^2} = \frac{s-3}{(s-3)^2+4}$ $L\{e^{-3t}\} = \frac{1}{s-(-3)} = \frac{1}{s+3}$ $L\{e^{-3t}\cos(2t)\} = \frac{s-(-3)}{(s-(-3))^2+2^2} = \frac{s+3}{(s+3)^2+4}$ So, $L\{\sinh 3t \cos^2 t\} = \frac{1}{4} \left[ \frac{1}{s-3} + \frac{s-3}{(s-3)^2+4} - \frac{1}{s+3} - \frac{s+3}{(s+3)^2+4} \right]$ **17. $t^2 e^{2t}$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$, where $F(s) = L\{f(t)\}$ - $L\{t^n\} = \frac{n!}{s^{n+1}}$ *Solution:* Let $f(t) = t^2$. Then $F(s) = L\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}$. Using the First Shifting Theorem with $a=2$: $L\{t^2 e^{2t}\} = F(s-2) = \frac{2}{(s-2)^3}$ **18. $(1+e^{-t})^3$** *Formulae Used:* - Binomial Expansion: $(A+B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{e^{at}\} = \frac{1}{s-a}$ *Solution:* $(1+e^{-t})^3 = 1^3 + 3(1)^2(e^{-t}) + 3(1)(e^{-t})^2 + (e^{-t})^3$ $= 1 + 3e^{-t} + 3e^{-2t} + e^{-3t}$ $L\{(1+e^{-t})^3\} = L\{1 + 3e^{-t} + 3e^{-2t} + e^{-3t}\}$ $= L\{1\} + 3L\{e^{-t}\} + 3L\{e^{-2t}\} + L\{e^{-3t}\}$ $= \frac{1}{s} + 3\frac{1}{s-(-1)} + 3\frac{1}{s-(-2)} + \frac{1}{s-(-3)}$ $= \frac{1}{s} + \frac{3}{s+1} + \frac{3}{s+2} + \frac{1}{s+3}$ **19. $t(1+\sin t)$** *Formulae Used:* - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$ - $L\{t\} = \frac{1}{s^2}$ - $L\{\sin t\} = \frac{1}{s^2+1}$ *Solution:* $t(1+\sin t) = t + t\sin t$ $L\{t(1+\sin t)\} = L\{t\} + L\{t\sin t\}$ $L\{t\} = \frac{1}{s^2}$ For $L\{t\sin t\}$, let $f(t) = \sin t$. Then $F(s) = \frac{1}{s^2+1}$. Using $L\{t f(t)\} = -\frac{d}{ds} F(s)$: $L\{t\sin t\} = -\frac{d}{ds}\left(\frac{1}{s^2+1}\right)$ $= - \frac{d}{ds}(s^2+1)^{-1} = - (-1)(s^2+1)^{-2} (2s)$ $= \frac{2s}{(s^2+1)^2}$ So, $L\{t(1+\sin t)\} = \frac{1}{s^2} + \frac{2s}{(s^2+1)^2}$ **20. $f(t) = \begin{cases} 4 & 0 \le t 0$, $\lim_{T\to\infty} e^{-sT} = 0$: $= 4 \left( -\frac{e^{-s}}{s} + \frac{1}{s} \right) + 3 \left( 0 + \frac{e^{-s}}{s} \right)$ $= \frac{4}{s} - \frac{4e^{-s}}{s} + \frac{3e^{-s}}{s}$ $= \frac{4}{s} - \frac{e^{-s}}{s} = \frac{4-e^{-s}}{s}$ *Solution (Method 2: Using Heaviside function):* $f(t) = 4U(t) - 4U(t-1) + 3U(t-1)$ $f(t) = 4U(t) - U(t-1)$ (since $U(t)$ is implied for $t \ge 0$) $L\{f(t)\} = 4L\{U(t)\} - L\{U(t-1)\}$ $L\{U(t)\} = L\{1\} = \frac{1}{s}$ For $L\{U(t-1)\}$, use $L\{f(t-a)U(t-a)\} = e^{-as}F(s)$. Here $f(t)=1$, so $F(s)=1/s$. $a=1$. $L\{U(t-1)\} = e^{-1s} \frac{1}{s} = \frac{e^{-s}}{s}$ So, $L\{f(t)\} = 4\frac{1}{s} - \frac{e^{-s}}{s} = \frac{4-e^{-s}}{s}$ **21. $f(t) = \begin{cases} \sin t & 0 \pi \end{cases}$** *Formulae Used:* - Definition of Laplace Transform: $L\{f(t)\} = \int_0^\infty e^{-st} f(t) dt$ - Heaviside step function: $U(t-a)$ - $f(t) = g(t)[U(t)-U(t-a)]$ for a function defined on $[0,a)$ and 0 otherwise. - $L\{\sin at\} = \frac{a}{s^2+a^2}$ - Second Shifting Theorem: $L\{f(t-a)U(t-a)\} = e^{-as}F(s)$ *Solution (Method 1: Direct Integration):* $L\{f(t)\} = \int_0^\pi e^{-st}\sin t dt + \int_\pi^\infty e^{-st}(0)dt$ $= \int_0^\pi e^{-st}\sin t dt$ Using the formula $\int e^{ax}\sin(bx)dx = \frac{e^{ax}}{a^2+b^2}(a\sin(bx) - b\cos(bx))$: Here $a=-s, b=1$. $= \left[ \frac{e^{-st}}{(-s)^2+1^2}(-s\sin t - 1\cos t) \right]_0^\pi$ $= \left[ \frac{e^{-st}}{s^2+1}(-s\sin t - \cos t) \right]_0^\pi$ $= \frac{e^{-s\pi}}{s^2+1}(-s\sin\pi - \cos\pi) - \frac{e^0}{s^2+1}(-s\sin 0 - \cos 0)$ $= \frac{e^{-s\pi}}{s^2+1}(-s(0) - (-1)) - \frac{1}{s^2+1}(-s(0) - 1)$ $= \frac{e^{-s\pi}}{s^2+1}(1) - \frac{1}{s^2+1}(-1)$ $= \frac{e^{-s\pi}+1}{s^2+1}$ *Solution (Method 2: Using Heaviside function):* $f(t) = \sin t [U(t) - U(t-\pi)]$ $= \sin t U(t) - \sin t U(t-\pi)$ We know $L\{\sin t\} = \frac{1}{s^2+1}$. For the second term: $\sin t U(t-\pi) = \sin(t-\pi+\pi) U(t-\pi) = (\sin(t-\pi)\cos\pi + \cos(t-\pi)\sin\pi) U(t-\pi)$ $= (\sin(t-\pi)(-1) + \cos(t-\pi)(0)) U(t-\pi)$ $= -\sin(t-\pi) U(t-\pi)$ Let $g(t) = \sin t$, then $G(s) = \frac{1}{s^2+1}$. $L\{-\sin(t-\pi) U(t-\pi)\} = -e^{-\pi s} L\{\sin t\} = -e^{-\pi s} \frac{1}{s^2+1}$ So, $L\{f(t)\} = \frac{1}{s^2+1} - \frac{e^{-\pi s}}{s^2+1} = \frac{1-e^{-\pi s}}{s^2+1}$ (Note: The problem statement says $0 \pi$. So $f(t) = \sin t \cdot (U(t) - U(t-\pi))$ for $t>0$. Since $\sin(t-\pi) = -\sin t$, then $\sin t = -\sin(t-\pi)$. $L\{f(t)\} = L\{\sin t U(t) - \sin t U(t-\pi)\}$ $= L\{\sin t\} - L\{(-\sin(t-\pi)) U(t-\pi)\}$ $= \frac{1}{s^2+1} - (-e^{-\pi s}L\{\sin t\})$ $= \frac{1}{s^2+1} + e^{-\pi s}\frac{1}{s^2+1} = \frac{1+e^{-\pi s}}{s^2+1}$. This matches Method 1. **22. $f(t) = \begin{cases} \sin(x-\pi/3) & x > \pi/3 \\ 0 & x 2\pi/3 \\ 0 & t 3 \end{cases}$** *Formulae Used:* - Definition of Laplace Transform: $L\{f(t)\} = \int_0^\infty e^{-st} f(t) dt$ - Heaviside step function: $U(t-a)$ - $f(t) = f_1(t)U(t) + (f_2(t)-f_1(t))U(t-a) + (f_3(t)-f_2(t))U(t-b) + \dots$ - Second Shifting Theorem: $L\{g(t-a)U(t-a)\} = e^{-as}G(s)$ - $L\{t^n\} = \frac{n!}{s^{n+1}}$ - $L\{1\} = \frac{1}{s}$ *Solution (Using Heaviside function):* $f(t) = t^2 U(t) + ((7-t)-t^2)U(t-2) + (7-(7-t))U(t-3)$ $f(t) = t^2 U(t) + (7-t-t^2)U(t-2) + t U(t-3)$ Let's find the Laplace transform of each term using the Second Shifting Theorem. For $L\{t^2 U(t)\}$: $L\{t^2\} = \frac{2}{s^3}$. For $L\{(7-t-t^2)U(t-2)\}$: Let $g(t) = 7-t-t^2$. We need to express $g(t)$ in terms of $(t-2)$. Let $\tau = t-2$, so $t = \tau+2$. $g(\tau+2) = 7 - (\tau+2) - (\tau+2)^2$ $= 7 - \tau - 2 - (\tau^2 + 4\tau + 4)$ $= 7 - \tau - 2 - \tau^2 - 4\tau - 4$ $= 1 - 5\tau - \tau^2$ So, $L\{(7-t-t^2)U(t-2)\} = e^{-2s} L\{1 - 5t - t^2\}$ $= e^{-2s} [L\{1\} - 5L\{t\} - L\{t^2\}]$ $= e^{-2s} [\frac{1}{s} - 5\frac{1}{s^2} - \frac{2}{s^3}]$ For $L\{t U(t-3)\}$: Let $h(t) = t$. We need to express $h(t)$ in terms of $(t-3)$. Let $\tau = t-3$, so $t = \tau+3$. $h(\tau+3) = \tau+3$ So, $L\{t U(t-3)\} = e^{-3s} L\{t+3\}$ $= e^{-3s} [L\{t\} + 3L\{1\}]$ $= e^{-3s} [\frac{1}{s^2} + \frac{3}{s}]$ Combining these: $L\{f(t)\} = \frac{2}{s^3} + e^{-2s} [\frac{1}{s} - \frac{5}{s^2} - \frac{2}{s^3}] + e^{-3s} [\frac{1}{s^2} + \frac{3}{s}]$ **25. If $L\{F(t)\} = \frac{1}{s(s^2+1)}$, find $L^{-1}\{F(s)(2s)\}$** *Formulae Used:* - $L\{f'(t)\} = sF(s) - f(0)$ - $L\{f(t)\} = F(s)$ - Inverse Laplace Transform *Solution:* Given $L\{F(t)\} = \frac{1}{s(s^2+1)}$. We want to find $L^{-1}\{F(s)(2s)\}$. $L^{-1}\{2s F(s)\} = 2L^{-1}\{s F(s)\}$ We know that $L\{f'(t)\} = sF(s) - f(0)$. So $L^{-1}\{sF(s)\} = f'(t) + f(0)\delta(t)$ (this is for general $f(t)$, for functions piecewise continuous with jump discontinuities, the derivative involves delta functions). However, usually in these types of problems, $f(0)$ is assumed to be 0 or can be found. Let's find $F(t) = L^{-1}\{\frac{1}{s(s^2+1)}\}$ first using partial fractions. $\frac{1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1}$ $1 = A(s^2+1) + (Bs+C)s$ If $s=0$, $1 = A(1) \implies A=1$. $1 = 1(s^2+1) + Bs^2+Cs$ $1 = s^2+1+Bs^2+Cs$ $0s^2+0s+1 = (1+B)s^2 + Cs + 1$ Comparing coefficients: $1+B=0 \implies B=-1$ $C=0$ So, $F(s) = \frac{1}{s} - \frac{s}{s^2+1}$ $F(t) = L^{-1}\{\frac{1}{s} - \frac{s}{s^2+1}\} = L^{-1}\{\frac{1}{s}\} - L^{-1}\{\frac{s}{s^2+1}\}$ $F(t) = 1 - \cos t$. Now we need to find $L^{-1}\{2sF(s)\}$. $2sF(s) = 2s \left( \frac{1}{s} - \frac{s}{s^2+1} \right) = 2 - \frac{2s^2}{s^2+1}$ $2sF(s) = 2 - \frac{2(s^2+1-1)}{s^2+1} = 2 - 2\left(1 - \frac{1}{s^2+1}\right)$ $= 2 - 2 + \frac{2}{s^2+1} = \frac{2}{s^2+1}$ $L^{-1}\{2sF(s)\} = L^{-1}\{\frac{2}{s^2+1}\} = 2L^{-1}\{\frac{1}{s^2+1}\} = 2\sin t$. This method seems more straightforward and avoids issues with $f(0)$ for $f'(t)$. Let's verify $f(0)$ for $F(t) = 1-\cos t$. $F(0) = 1-\cos 0 = 1-1=0$. So $L^{-1}\{sF(s)\} = F'(t)$ when $F(0)=0$. $F'(t) = \frac{d}{dt}(1-\cos t) = \sin t$. So $L^{-1}\{2sF(s)\} = 2F'(t) = 2\sin t$. Both methods yield the same result. ### Problems 21.2 ### Solutions to Problems 21.2 **1. Find the Laplace transform of the saw-toothed wave of period $T$, given $f(t) = t/T$ for $0 ### Problems 21.3 ### Solutions to Problems 21.3 **1. Find $L\{\int_0^t e^{-\tau} \cos \tau d\tau\}$** *Formulae Used:* - Transform of Integrals: $L\{\int_0^t f(\tau) d\tau\} = \frac{F(s)}{s}$, where $F(s) = L\{f(t)\}$ - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$ - $L\{\cos(bt)\} = \frac{s}{s^2+b^2}$ *Solution:* Let $f(\tau) = e^{-\tau} \cos \tau$. First, find $L\{e^{-\tau} \cos \tau\}$. Let $g(\tau) = \cos \tau$. Then $G(s) = L\{\cos \tau\} = \frac{s}{s^2+1^2} = \frac{s}{s^2+1}$. Using the First Shifting Theorem with $a=-1$: $F(s) = L\{e^{-\tau}\cos \tau\} = G(s-(-1)) = G(s+1) = \frac{s+1}{(s+1)^2+1}$ Now, apply the transform of integrals: $L\{\int_0^t e^{-\tau} \cos \tau d\tau\} = \frac{F(s)}{s} = \frac{1}{s} \frac{s+1}{(s+1)^2+1}$ $= \frac{s+1}{s((s+1)^2+1)} = \frac{s+1}{s(s^2+2s+1+1)} = \frac{s+1}{s(s^2+2s+2)}$ **2. Given $L\{ \frac{1}{\sqrt{\pi t}} \} = \frac{1}{\sqrt{s}}$, show that $L\{ \frac{1}{\sqrt{\pi (t-1)}} H(t-1) \} = \frac{e^{-s}}{\sqrt{s}}$.** *Formulae Used:* - Second Shifting Theorem: $L\{f(t-a)U(t-a)\} = e^{-as}F(s)$ (where $H(t-1)$ is the Heaviside step function $U(t-1)$) *Solution:* Let $f(t) = \frac{1}{\sqrt{\pi t}}$. We are given $L\{f(t)\} = F(s) = \frac{1}{\sqrt{s}}$. We need to find $L\{ \frac{1}{\sqrt{\pi (t-1)}} H(t-1) \}$. This is of the form $L\{f(t-a)U(t-a)\}$ with $a=1$. Replacing $t$ with $(t-1)$ in $f(t)$ gives $f(t-1) = \frac{1}{\sqrt{\pi (t-1)}}$. So, using the Second Shifting Theorem with $a=1$: $L\{ f(t-1) H(t-1) \} = e^{-1s} F(s)$ $= e^{-s} \frac{1}{\sqrt{s}} = \frac{e^{-s}}{\sqrt{s}}$ This matches the required result. **3. Given $L\{\sin(\sqrt{t})/\sqrt{t}\} = \sqrt{\pi/(s(s+1))}$, prove that $L\{\cos(\sqrt{t})/\sqrt{t}\} = \sqrt{\pi/s} e^{-1/(4s)}$.** *Formulae Used:* - $L\{f'(t)\} = sF(s) - f(0)$ - Derivative of $\sin(\sqrt{t})/\sqrt{t}$ - Integration by parts - Differentiation of $L\{f(t)\}$ with respect to $s$ (not directly used here, but for derivation) *Solution:* Let $g(t) = \frac{\sin(\sqrt{t})}{\sqrt{t}}$. We are given $G(s) = L\{g(t)\} = \sqrt{\frac{\pi}{s(s+1)}}$. We need to find $L\{\frac{\cos(\sqrt{t})}{\sqrt{t}}\}$. Consider the derivative of $g(t)$: $g'(t) = \frac{d}{dt} (t^{-1/2} \sin(t^{1/2}))$ Using product rule and chain rule: $g'(t) = (-\frac{1}{2}t^{-3/2})\sin(t^{1/2}) + t^{-1/2}(\cos(t^{1/2}) \cdot \frac{1}{2}t^{-1/2})$ $= -\frac{1}{2}\frac{\sin(\sqrt{t})}{t^{3/2}} + \frac{1}{2}\frac{\cos(\sqrt{t})}{t}$ This doesn't seem to lead directly to the desired term. Let's consider the relationship between $L\{f(t)\}$ and $L\{f'(t)\}$. If $f(t) = \frac{\sin(\sqrt{t})}{\sqrt{t}}$, then $f(0) = \lim_{t\to 0} \frac{\sin(\sqrt{t})}{\sqrt{t}} = 1$. $L\{f'(t)\} = sG(s) - f(0) = s\sqrt{\frac{\pi}{s(s+1)}} - 1 = \sqrt{\frac{\pi s}{s+1}} - 1$. This also doesn't directly simplify. This problem requires knowledge of special functions or a specific identity. The standard approach for $L\{\cos(\sqrt{t})/\sqrt{t}\}$ is not straightforward from $L\{\sin(\sqrt{t})/\sqrt{t}\}$ using basic properties. However, if we assume $L\{\frac{\cos(\sqrt{t})}{\sqrt{t}}\}$ is a known transform: We know $L\{t^a\} = \frac{\Gamma(a+1)}{s^{a+1}}$. $L\{ \frac{1}{\sqrt{t}} \} = \frac{\Gamma(1/2)}{s^{1/2}} = \sqrt{\frac{\pi}{s}}$. The form $e^{-1/(4s)}$ suggests a relation to $L\{ e^{-a^2/t} \}$. The transform $L\{ \frac{\cos(a\sqrt{t})}{\sqrt{t}} \} = \sqrt{\frac{\pi}{s}} e^{-a^2/(4s)}$. In our case, $a=1$. So $L\{ \frac{\cos(\sqrt{t})}{\sqrt{t}} \} = \sqrt{\frac{\pi}{s}} e^{-1/(4s)}$. This is a standard result, but proving it from the given $\sin(\sqrt{t})/\sqrt{t}$ transform is complex. It typically involves using the Convolution Theorem or advanced integral transforms. Given the context, it's likely meant to be a known result or a verification. **Find the Laplace transforms of the following functions:** **4. $t \sin^2 t$** *Formulae Used:* - $L\{t f(t)\} = -\frac{d}{ds} F(s)$ - Trigonometric identity: $\sin^2 x = \frac{1-\cos(2x)}{2}$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = \sin^2 t = \frac{1-\cos(2t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2t)$. $F(s) = L\{f(t)\} = L\{\frac{1}{2} - \frac{1}{2}\cos(2t)\}$ $= \frac{1}{2}L\{1\} - \frac{1}{2}L\{\cos(2t)\}$ $= \frac{1}{2s} - \frac{1}{2}\frac{s}{s^2+2^2} = \frac{1}{2s} - \frac{s}{2(s^2+4)}$ Now apply $L\{t f(t)\} = -\frac{d}{ds} F(s)$: $L\{t \sin^2 t\} = -\frac{d}{ds} \left( \frac{1}{2s} - \frac{s}{2(s^2+4)} \right)$ $= -\left( -\frac{1}{2s^2} - \frac{1}{2}\frac{d}{ds}\left(\frac{s}{s^2+4}\right) \right)$ Using quotient rule for $\frac{s}{s^2+4}$: $\frac{1(s^2+4) - s(2s)}{(s^2+4)^2} = \frac{s^2+4-2s^2}{(s^2+4)^2} = \frac{4-s^2}{(s^2+4)^2}$ $= \frac{1}{2s^2} + \frac{1}{2}\frac{4-s^2}{(s^2+4)^2}$ $= \frac{1}{2s^2} + \frac{4-s^2}{2(s^2+4)^2}$ **5. $\sin 2t - 2t \cos 2t$** *Formulae Used:* - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\sin(at)\} = \frac{a}{s^2+a^2}$ - $L\{\cos(at)\} = \frac{s}{s^2+a^2}$ - $L\{t f(t)\} = -\frac{d}{ds} F(s)$ *Solution:* $L\{\sin 2t - 2t \cos 2t\} = L\{\sin 2t\} - 2L\{t \cos 2t\}$ $L\{\sin 2t\} = \frac{2}{s^2+2^2} = \frac{2}{s^2+4}$ For $L\{t \cos 2t\}$, let $f(t) = \cos 2t$. Then $F(s) = \frac{s}{s^2+4}$. $L\{t \cos 2t\} = -\frac{d}{ds} \left( \frac{s}{s^2+4} \right)$ Using quotient rule: $-\frac{1(s^2+4) - s(2s)}{(s^2+4)^2} = -\frac{s^2+4-2s^2}{(s^2+4)^2} = -\frac{4-s^2}{(s^2+4)^2} = \frac{s^2-4}{(s^2+4)^2}$ So, $L\{\sin 2t - 2t \cos 2t\} = \frac{2}{s^2+4} - 2\frac{s^2-4}{(s^2+4)^2}$ $= \frac{2(s^2+4) - 2(s^2-4)}{(s^2+4)^2}$ $= \frac{2s^2+8 - 2s^2+8}{(s^2+4)^2} = \frac{16}{(s^2+4)^2}$ **6. $t^2 \cos at$** *Formulae Used:* - $L\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s)$ - $L\{\cos at\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = \cos at$. Then $F(s) = \frac{s}{s^2+a^2}$. $L\{t^2 \cos at\} = (-1)^2 \frac{d^2}{ds^2} F(s) = \frac{d^2}{ds^2} \left( \frac{s}{s^2+a^2} \right)$ First derivative: $\frac{d}{ds} \left( \frac{s}{s^2+a^2} \right) = \frac{1(s^2+a^2) - s(2s)}{(s^2+a^2)^2} = \frac{a^2-s^2}{(s^2+a^2)^2}$ Second derivative: $\frac{d}{ds} \left( \frac{a^2-s^2}{(s^2+a^2)^2} \right)$ Using quotient rule: $\frac{(-2s)(s^2+a^2)^2 - (a^2-s^2) \cdot 2(s^2+a^2)(2s)}{((s^2+a^2)^2)^2}$ $= \frac{-2s(s^2+a^2) - 4s(a^2-s^2)}{(s^2+a^2)^3}$ $= \frac{-2s^3-2a^2s - 4a^2s+4s^3}{(s^2+a^2)^3}$ $= \frac{2s^3 - 6a^2s}{(s^2+a^2)^3} = \frac{2s(s^2-3a^2)}{(s^2+a^2)^3}$ **7. $t \sin at$** *Formulae Used:* - $L\{t f(t)\} = -\frac{d}{ds} F(s)$ - $L\{\sin at\} = \frac{a}{s^2+a^2}$ *Solution:* Let $f(t) = \sin at$. Then $F(s) = \frac{a}{s^2+a^2}$. $L\{t \sin at\} = -\frac{d}{ds} \left( \frac{a}{s^2+a^2} \right)$ $= -a \frac{d}{ds} (s^2+a^2)^{-1}$ $= -a (-1)(s^2+a^2)^{-2} (2s)$ $= \frac{2as}{(s^2+a^2)^2}$ **8. $e^{at} \sin 3t$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$ - $L\{\sin bt\} = \frac{b}{s^2+b^2}$ *Solution:* Let $f(t) = \sin 3t$. Then $F(s) = \frac{3}{s^2+3^2} = \frac{3}{s^2+9}$. Using the First Shifting Theorem: $L\{e^{at} \sin 3t\} = F(s-a) = \frac{3}{(s-a)^2+9}$ **9. $t e^{-2t} \sin 4t$** *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$ - $L\{t g(t)\} = -\frac{d}{ds} G(s)$ - $L\{\sin bt\} = \frac{b}{s^2+b^2}$ *Solution:* Let $g(t) = \sin 4t$. Then $G(s) = \frac{4}{s^2+4^2} = \frac{4}{s^2+16}$. Now consider $L\{t \sin 4t\}$: $L\{t \sin 4t\} = -\frac{d}{ds} G(s) = -\frac{d}{ds} \left( \frac{4}{s^2+16} \right)$ $= -4 \frac{d}{ds} (s^2+16)^{-1} = -4 (-1)(s^2+16)^{-2} (2s)$ $= \frac{8s}{(s^2+16)^2}$. Call this $H(s)$. Now apply the First Shifting Theorem for $e^{-2t}$: $L\{e^{-2t} (t \sin 4t)\} = H(s-(-2)) = H(s+2)$ $= \frac{8(s+2)}{((s+2)^2+16)^2}$ **10. $e^{-t} (e^{-t} - e^{t})^2$** *Formulae Used:* - Algebraic expansion - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{e^{at}\} = \frac{1}{s-a}$ *Solution:* $e^{-t} (e^{-t} - e^{t})^2 = e^{-t} ( (e^{-t})^2 - 2e^{-t}e^{t} + (e^{t})^2 )$ $= e^{-t} ( e^{-2t} - 2e^0 + e^{2t} )$ $= e^{-t} ( e^{-2t} - 2 + e^{2t} )$ $= e^{-3t} - 2e^{-t} + e^{t}$ $L\{e^{-3t} - 2e^{-t} + e^{t}\}$ $= L\{e^{-3t}\} - 2L\{e^{-t}\} + L\{e^{t}\}$ $= \frac{1}{s-(-3)} - 2\frac{1}{s-(-1)} + \frac{1}{s-1}$ $= \frac{1}{s+3} - \frac{2}{s+1} + \frac{1}{s-1}$ **11. $(e^{4t} - e^{-4t})^2 / t$** *Formulae Used:* - Algebraic expansion: $(A-B)^2 = A^2 - 2AB + B^2$ - Division by t: $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - $L\{e^{at}\} = \frac{1}{s-a}$ - $L\{1\} = \frac{1}{s}$ *Solution:* First, simplify the numerator: $(e^{4t} - e^{-4t})^2 = (e^{4t})^2 - 2e^{4t}e^{-4t} + (e^{-4t})^2$ $= e^{8t} - 2e^0 + e^{-8t} = e^{8t} - 2 + e^{-8t}$ So, $f(t) = \frac{e^{8t} - 2 + e^{-8t}}{t}$. Let $g(t) = e^{8t} - 2 + e^{-8t}$. $G(s) = L\{g(t)\} = L\{e^{8t}\} - 2L\{1\} + L\{e^{-8t}\}$ $= \frac{1}{s-8} - \frac{2}{s} + \frac{1}{s-(-8)}$ $= \frac{1}{s-8} - \frac{2}{s} + \frac{1}{s+8}$ Now apply the division by t property: $L\{\frac{g(t)}{t}\} = \int_s^\infty G(u) du = \int_s^\infty \left( \frac{1}{u-8} - \frac{2}{u} + \frac{1}{u+8} \right) du$ $= \left[ \ln|u-8| - 2\ln|u| + \ln|u+8| \right]_s^\infty$ $= \left[ \ln\left|\frac{(u-8)(u+8)}{u^2}\right| \right]_s^\infty = \left[ \ln\left|\frac{u^2-64}{u^2}\right| \right]_s^\infty$ $= \left[ \ln\left|1-\frac{64}{u^2}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \ln\left|1-\frac{64}{u^2}\right| - \ln\left|1-\frac{64}{s^2}\right|$ $= \ln(1) - \ln\left|\frac{s^2-64}{s^2}\right|$ $= 0 - \ln\left|\frac{s^2-64}{s^2}\right| = \ln\left|\frac{s^2}{s^2-64}\right|$ For the transform to exist, $\lim_{t\to 0} \frac{g(t)}{t}$ must exist. $\lim_{t\to 0} \frac{e^{8t} - 2 + e^{-8t}}{t} = \lim_{t\to 0} \frac{(1+8t+\dots) - 2 + (1-8t+\dots)}{t} = \lim_{t\to 0} \frac{0 + \dots}{t} = 0$. So the limit exists. **12. $(\sin t - \sin 5t)/t$** *Formulae Used:* - Division by t: $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\sin at\} = \frac{a}{s^2+a^2}$ *Solution:* Let $g(t) = \sin t - \sin 5t$. $G(s) = L\{g(t)\} = L\{\sin t\} - L\{\sin 5t\}$ $= \frac{1}{s^2+1^2} - \frac{5}{s^2+5^2} = \frac{1}{s^2+1} - \frac{5}{s^2+25}$ Now apply the division by t property: $L\{\frac{\sin t - \sin 5t}{t}\} = \int_s^\infty G(u) du = \int_s^\infty \left( \frac{1}{u^2+1} - \frac{5}{u^2+25} \right) du$ $= \left[ \arctan(u) - 5 \cdot \frac{1}{5}\arctan\left(\frac{u}{5}\right) \right]_s^\infty$ $= \left[ \arctan(u) - \arctan\left(\frac{u}{5}\right) \right]_s^\infty$ $= \lim_{u\to\infty} (\arctan(u) - \arctan(u/5)) - (\arctan(s) - \arctan(s/5))$ As $u\to\infty$, $\arctan(u) \to \pi/2$ and $\arctan(u/5) \to \pi/2$. So the first term is $\pi/2 - \pi/2 = 0$. $= 0 - (\arctan(s) - \arctan(s/5))$ $= \arctan(s/5) - \arctan(s)$ For the transform to exist, $\lim_{t\to 0} \frac{\sin t - \sin 5t}{t}$ must exist. $\lim_{t\to 0} \frac{\sin t - \sin 5t}{t} = \lim_{t\to 0} (\frac{\sin t}{t} - 5\frac{\sin 5t}{5t}) = 1 - 5(1) = -4$. So the limit exists. **13. $(\frac{1-\cos 5t}{t})$** *Formulae Used:* - Division by t: $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{\cos at\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = 1-\cos 5t$. $F(s) = L\{1-\cos 5t\} = L\{1\} - L\{\cos 5t\}$ $= \frac{1}{s} - \frac{s}{s^2+5^2} = \frac{1}{s} - \frac{s}{s^2+25}$ Now apply the division by t property: $L\{\frac{1-\cos 5t}{t}\} = \int_s^\infty F(u) du = \int_s^\infty \left( \frac{1}{u} - \frac{u}{u^2+25} \right) du$ $= \left[ \ln|u| - \frac{1}{2}\ln|u^2+25| \right]_s^\infty$ $= \left[ \ln|u| - \ln\sqrt{u^2+25} \right]_s^\infty = \left[ \ln\left|\frac{u}{\sqrt{u^2+25}}\right| \right]_s^\infty$ $= \left[ \ln\left|\frac{1}{\sqrt{1+25/u^2}}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \ln\left|\frac{1}{\sqrt{1+25/u^2}}\right| - \ln\left|\frac{s}{\sqrt{s^2+25}}\right|$ $= \ln(1) - \ln\left|\frac{s}{\sqrt{s^2+25}}\right|$ $= 0 - \ln\left|\frac{s}{\sqrt{s^2+25}}\right| = \ln\left|\frac{\sqrt{s^2+25}}{s}\right|$ For the transform to exist, $\lim_{t\to 0} \frac{1-\cos 5t}{t}$ must exist. $\lim_{t\to 0} \frac{1-\cos 5t}{t} = \lim_{t\to 0} \frac{1-(1-\frac{(5t)^2}{2!}+\dots)}{t} = \lim_{t\to 0} \frac{\frac{25t^2}{2}+\dots}{t} = 0$. So the limit exists. **14. $(e^{-at} - \cos bt)/t$** *Formulae Used:* - Division by t: $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{e^{at}\} = \frac{1}{s-a}$ - $L\{\cos bt\} = \frac{s}{s^2+b^2}$ *Solution:* Let $f(t) = e^{-at} - \cos bt$. $F(s) = L\{e^{-at}\} - L\{\cos bt\}$ $= \frac{1}{s-(-a)} - \frac{s}{s^2+b^2} = \frac{1}{s+a} - \frac{s}{s^2+b^2}$ Now apply the division by t property: $L\{\frac{e^{-at} - \cos bt}{t}\} = \int_s^\infty F(u) du = \int_s^\infty \left( \frac{1}{u+a} - \frac{u}{u^2+b^2} \right) du$ $= \left[ \ln|u+a| - \frac{1}{2}\ln|u^2+b^2| \right]_s^\infty$ $= \left[ \ln|u+a| - \ln\sqrt{u^2+b^2} \right]_s^\infty = \left[ \ln\left|\frac{u+a}{\sqrt{u^2+b^2}}\right| \right]_s^\infty$ $= \left[ \ln\left|\frac{1+a/u}{\sqrt{1+b^2/u^2}}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \ln\left|\frac{1+a/u}{\sqrt{1+b^2/u^2}}\right| - \ln\left|\frac{s+a}{\sqrt{s^2+b^2}}\right|$ $= \ln(1) - \ln\left|\frac{s+a}{\sqrt{s^2+b^2}}\right|$ $= 0 - \ln\left|\frac{s+a}{\sqrt{s^2+b^2}}\right| = \ln\left|\frac{\sqrt{s^2+b^2}}{s+a}\right|$ For the transform to exist, $\lim_{t\to 0} \frac{e^{-at} - \cos bt}{t}$ must exist. $\lim_{t\to 0} \frac{(1-at+\dots) - (1-\frac{(bt)^2}{2!}+\dots)}{t} = \lim_{t\to 0} \frac{-at + \dots}{t} = -a$. So the limit exists. **15. $(e^{-t} - e^{-3t})/t$** *Formulae Used:* - Division by t: $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{e^{at}\} = \frac{1}{s-a}$ *Solution:* Let $f(t) = e^{-t} - e^{-3t}$. $F(s) = L\{e^{-t}\} - L\{e^{-3t}\}$ $= \frac{1}{s-(-1)} - \frac{1}{s-(-3)} = \frac{1}{s+1} - \frac{1}{s+3}$ Now apply the division by t property: $L\{\frac{e^{-t} - e^{-3t}}{t}\} = \int_s^\infty F(u) du = \int_s^\infty \left( \frac{1}{u+1} - \frac{1}{u+3} \right) du$ $= \left[ \ln|u+1| - \ln|u+3| \right]_s^\infty$ $= \left[ \ln\left|\frac{u+1}{u+3}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \ln\left|\frac{u+1}{u+3}\right| - \ln\left|\frac{s+1}{s+3}\right|$ $= \lim_{u\to\infty} \ln\left|\frac{1+1/u}{1+3/u}\right| - \ln\left|\frac{s+1}{s+3}\right|$ $= \ln(1) - \ln\left|\frac{s+1}{s+3}\right|$ $= 0 - \ln\left|\frac{s+1}{s+3}\right| = \ln\left|\frac{s+3}{s+1}\right|$ For the transform to exist, $\lim_{t\to 0} \frac{e^{-t} - e^{-3t}}{t}$ must exist. $\lim_{t\to 0} \frac{(1-t+\dots) - (1-3t+\dots)}{t} = \lim_{t\to 0} \frac{2t + \dots}{t} = 2$. So the limit exists. **16. $(1-\cos t)/t^2$** *Formulae Used:* - $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ (applied twice) - $L\{1\} = \frac{1}{s}$ - $L\{\cos t\} = \frac{s}{s^2+1}$ *Solution:* Let $f(t) = 1-\cos t$. $F(s) = L\{1-\cos t\} = \frac{1}{s} - \frac{s}{s^2+1}$. Now, first division by $t$: $L\{\frac{1-\cos t}{t}\} = \int_s^\infty \left( \frac{1}{u} - \frac{u}{u^2+1} \right) du$ $= \left[ \ln|u| - \frac{1}{2}\ln|u^2+1| \right]_s^\infty = \left[ \ln\left|\frac{u}{\sqrt{u^2+1}}\right| \right]_s^\infty$ $= \left[ \ln\left|\frac{1}{\sqrt{1+1/u^2}}\right| \right]_s^\infty = 0 - \ln\left|\frac{s}{\sqrt{s^2+1}}\right| = \ln\left|\frac{\sqrt{s^2+1}}{s}\right|$. Call this $G_1(s)$. For the second division by $t$: $L\{\frac{1-\cos t}{t^2}\} = \int_s^\infty G_1(u) du = \int_s^\infty \ln\left(\frac{\sqrt{u^2+1}}{u}\right) du$ $= \int_s^\infty (\frac{1}{2}\ln(u^2+1) - \ln u) du$ This integral is more complex and typically requires integration by parts with $\ln(x)$ functions which results in $x\ln(x)-x$. This specific transform is not easily solvable via elementary integral. A known identity: $L\{\frac{1-\cos at}{t^2}\} = \arctan(\frac{s}{a}) - \frac{as}{s^2+a^2}$ No, that's not correct. $L\{\frac{1-\cos at}{t}\} = \frac{1}{2}\ln(\frac{s^2+a^2}{s^2})$. So $L\{\frac{1-\cos at}{t^2}\} = \int_s^\infty \frac{1}{2}\ln(\frac{u^2+a^2}{u^2}) du = \frac{1}{2} \int_s^\infty (\ln(u^2+a^2) - 2\ln u) du$. Using $\int \ln x dx = x\ln x - x$: $\int \ln(u^2+a^2) du = u\ln(u^2+a^2) - \int u \frac{2u}{u^2+a^2} du = u\ln(u^2+a^2) - \int (2 - \frac{2a^2}{u^2+a^2}) du$ $= u\ln(u^2+a^2) - 2u + 2a \arctan(u/a)$. $\int 2\ln u du = 2u\ln u - 2u$. So, $\frac{1}{2} \left[ u\ln(u^2+a^2) - 2u + 2a \arctan(u/a) - (2u\ln u - 2u) \right]_s^\infty$ $= \frac{1}{2} \left[ u\ln(u^2+a^2) - 2u\ln u + 2a \arctan(u/a) \right]_s^\infty$ $= \left[ \frac{u}{2}\ln(\frac{u^2+a^2}{u^2}) + a \arctan(u/a) \right]_s^\infty$ $= \left[ \frac{u}{2}\ln(1+\frac{a^2}{u^2}) + a \arctan(u/a) \right]_s^\infty$ For $a=1$: $= \left[ \frac{u}{2}\ln(1+\frac{1}{u^2}) + \arctan(u) \right]_s^\infty$ As $u\to\infty$, $\ln(1+1/u^2) \approx 1/u^2$. So $\frac{u}{2}\ln(1+1/u^2) \approx \frac{u}{2} \frac{1}{u^2} = \frac{1}{2u} \to 0$. $\arctan(u) \to \pi/2$. So, the upper limit is $\pi/2$. $= \frac{\pi}{2} - \left( \frac{s}{2}\ln(1+\frac{1}{s^2}) + \arctan(s) \right)$ $= \frac{\pi}{2} - \frac{s}{2}\ln(\frac{s^2+1}{s^2}) - \arctan(s)$ $= \frac{\pi}{2} - \arctan(s) - \frac{s}{2}(\ln(s^2+1) - 2\ln s)$ This is the result. For the transform to exist, $\lim_{t\to 0} \frac{1-\cos t}{t^2}$ must exist. $\lim_{t\to 0} \frac{1-(1-t^2/2!+t^4/4!-\dots)}{t^2} = \lim_{t\to 0} \frac{t^2/2 - t^4/24 + \dots}{t^2} = \frac{1}{2}$. So the limit exists. **17. $(1-\cos t)/t$** (This is problem 13 with $a=1$, so the solution is $\ln\left|\frac{\sqrt{s^2+1}}{s}\right|$). *Formulae Used:* - Division by t: $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{1\} = \frac{1}{s}$ - $L\{\cos at\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = 1-\cos t$. $F(s) = L\{1-\cos t\} = L\{1\} - L\{\cos t\}$ $= \frac{1}{s} - \frac{s}{s^2+1}$ Now apply the division by t property: $L\{\frac{1-\cos t}{t}\} = \int_s^\infty F(u) du = \int_s^\infty \left( \frac{1}{u} - \frac{u}{u^2+1} \right) du$ $= \left[ \ln|u| - \frac{1}{2}\ln|u^2+1| \right]_s^\infty$ $= \left[ \ln\left|\frac{u}{\sqrt{u^2+1}}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \ln\left|\frac{u}{\sqrt{u^2+1}}\right| - \ln\left|\frac{s}{\sqrt{s^2+1}}\right|$ $= \lim_{u\to\infty} \ln\left|\frac{1}{\sqrt{1+1/u^2}}\right| - \ln\left|\frac{s}{\sqrt{s^2+1}}\right|$ $= \ln(1) - \ln\left|\frac{s}{\sqrt{s^2+1}}\right|$ $= 0 - \ln\left|\frac{s}{\sqrt{s^2+1}}\right| = \ln\left|\frac{\sqrt{s^2+1}}{s}\right|$ **18. $2t^2 + \cos 2t - 3t \sin t$** *Formulae Used:* - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{t^n\} = \frac{n!}{s^{n+1}}$ - $L\{\cos at\} = \frac{s}{s^2+a^2}$ - $L\{t \sin at\} = \frac{2as}{(s^2+a^2)^2}$ (from problem 7) *Solution:* $L\{2t^2 + \cos 2t - 3t \sin t\}$ $= 2L\{t^2\} + L\{\cos 2t\} - 3L\{t \sin t\}$ $L\{t^2\} = \frac{2!}{s^3} = \frac{2}{s^3}$ $L\{\cos 2t\} = \frac{s}{s^2+2^2} = \frac{s}{s^2+4}$ $L\{t \sin t\} = \frac{2(1)s}{(s^2+1^2)^2} = \frac{2s}{(s^2+1)^2}$ (using result from problem 7 with $a=1$) So, $L\{2t^2 + \cos 2t - 3t \sin t\} = 2\frac{2}{s^3} + \frac{s}{s^2+4} - 3\frac{2s}{(s^2+1)^2}$ $= \frac{4}{s^3} + \frac{s}{s^2+4} - \frac{6s}{(s^2+1)^2}$ **19. Evaluate $L\{\int_0^t \frac{\cos 6\tau - \cos 4\tau}{\tau} d\tau\}$** *Formulae Used:* - Transform of Integrals: $L\{\int_0^t f(\tau) d\tau\} = \frac{F(s)}{s}$, where $F(s) = L\{f(t)\}$ - Division by t: $L\{\frac{g(t)}{t}\} = \int_s^\infty G(u) du$ - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$ - $L\{\cos at\} = \frac{s}{s^2+a^2}$ *Solution:* Let $f(t) = \frac{\cos 6t - \cos 4t}{t}$. First, find $L\{f(t)\}$. Let $g(t) = \cos 6t - \cos 4t$. $G(s) = L\{\cos 6t\} - L\{\cos 4t\} = \frac{s}{s^2+6^2} - \frac{s}{s^2+4^2} = \frac{s}{s^2+36} - \frac{s}{s^2+16}$. Now apply the division by t property to find $F(s) = L\{f(t)\}$: $F(s) = \int_s^\infty G(u) du = \int_s^\infty \left( \frac{u}{u^2+36} - \frac{u}{u^2+16} \right) du$ $= \left[ \frac{1}{2}\ln|u^2+36| - \frac{1}{2}\ln|u^2+16| \right]_s^\infty$ $= \left[ \frac{1}{2}\ln\left|\frac{u^2+36}{u^2+16}\right| \right]_s^\infty$ $= \left[ \frac{1}{2}\ln\left|\frac{1+36/u^2}{1+16/u^2}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \frac{1}{2}\ln\left|\frac{1+36/u^2}{1+16/u^2}\right| - \frac{1}{2}\ln\left|\frac{s^2+36}{s^2+16}\right|$ $= \frac{1}{2}\ln(1) - \frac{1}{2}\ln\left|\frac{s^2+36}{s^2+16}\right|$ $= 0 - \frac{1}{2}\ln\left|\frac{s^2+36}{s^2+16}\right| = \frac{1}{2}\ln\left|\frac{s^2+16}{s^2+36}\right|$. For the transform to exist, $\lim_{t\to 0} \frac{\cos 6t - \cos 4t}{t}$ must exist. $\lim_{t\to 0} \frac{(1-\frac{(6t)^2}{2!}+\dots) - (1-\frac{(4t)^2}{2!}+\dots)}{t} = \lim_{t\to 0} \frac{-\frac{36t^2}{2} + \frac{16t^2}{2} + \dots}{t} = \lim_{t\to 0} \frac{-10t^2+\dots}{t} = 0$. So the limit exists. Now, apply the transform of integrals: $L\{\int_0^t f(\tau) d\tau\} = \frac{F(s)}{s}$ $= \frac{1}{s} \cdot \frac{1}{2}\ln\left|\frac{s^2+16}{s^2+36}\right|$ $= \frac{1}{2s}\ln\left(\frac{s^2+16}{s^2+36}\right)$ **20. Prove that:** **(i) $\int_0^\infty \frac{e^{-at} - e^{-bt}}{t} dt = \log(b/a)$** *Formulae Used:* - $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Property of Laplace Transform: $\int_0^\infty e^{-st} f(t) dt = F(s)$. So, $\int_0^\infty f(t) dt = F(0)$ (if $s=0$ is valid for the integral). However, $\int_0^\infty \frac{e^{-at} - e^{-bt}}{t} dt$ is $L\{\frac{e^{-at} - e^{-bt}}{t}\}|_{s=0}$. *Solution:* Let $f(t) = e^{-at} - e^{-bt}$. $F(s) = L\{f(t)\} = L\{e^{-at}\} - L\{e^{-bt}\} = \frac{1}{s+a} - \frac{1}{s+b}$. Using the division by t property: $L\{\frac{e^{-at} - e^{-bt}}{t}\} = \int_s^\infty \left( \frac{1}{u+a} - \frac{1}{u+b} \right) du$ $= \left[ \ln|u+a| - \ln|u+b| \right]_s^\infty = \left[ \ln\left|\frac{u+a}{u+b}\right| \right]_s^\infty$ $= \lim_{u\to\infty} \ln\left|\frac{u+a}{u+b}\right| - \ln\left|\frac{s+a}{s+b}\right|$ $= \lim_{u\to\infty} \ln\left|\frac{1+a/u}{1+b/u}\right| - \ln\left|\frac{s+a}{s+b}\right|$ $= \ln(1) - \ln\left|\frac{s+a}{s+b}\right|$ $= 0 - \ln\left|\frac{s+a}{s+b}\right| = \ln\left|\frac{s+b}{s+a}\right|$. Now, we need to evaluate this at $s=0$: $\int_0^\infty \frac{e^{-at} - e^{-bt}}{t} dt = \lim_{s\to 0} L\{\frac{e^{-at} - e^{-bt}}{t}\}$ $= \ln\left|\frac{0+b}{0+a}\right| = \ln\left(\frac{b}{a}\right)$. (Assuming $a, b > 0$) This matches the required result. **(ii) $\int_0^\infty e^{-2t} \frac{\sinh t}{t} dt = \frac{1}{2}\log 3$** *Formulae Used:* - $\sinh t = \frac{e^t - e^{-t}}{2}$ - $L\{\frac{f(t)}{t}\} = \int_s^\infty F(u) du$ - Property of Laplace Transform: $\int_0^\infty e^{-st} f(t) dt = F(s)$. *Solution:* The integral is $\int_0^\infty e^{-2t} \frac{\sinh t}{t} dt$, which is the Laplace transform of $\frac{\sinh t}{t}$ evaluated at $s=2$. First, find $L\{\frac{\sinh t}{t}\}$. Let $f(t) = \sinh t = \frac{e^t - e^{-t}}{2}$. $F(s) = L\{\sinh t\} = \frac{1}{2}(L\{e^t\} - L\{e^{-t}\})$ $= \frac{1}{2}(\frac{1}{s-1} - \frac{1}{s+1})$ Now apply the division by t property: $L\{\frac{\sinh t}{t}\} = \int_s^\infty F(u) du = \int_s^\infty \frac{1}{2}\left( \frac{1}{u-1} - \frac{1}{u+1} \right) du$ $= \frac{1}{2} \left[ \ln|u-1| - \ln|u+1| \right]_s^\infty = \frac{1}{2} \left[ \ln\left|\frac{u-1}{u+1}\right| \right]_s^\infty$ $= \frac{1}{2} \left( \lim_{u\to\infty} \ln\left|\frac{u-1}{u+1}\right| - \ln\left|\frac{s-1}{s+1}\right| \right)$ $= \frac{1}{2} \left( \ln(1) - \ln\left|\frac{s-1}{s+1}\right| \right)$ $= -\frac{1}{2}\ln\left|\frac{s-1}{s+1}\right| = \frac{1}{2}\ln\left|\frac{s+1}{s-1}\right|$. Now evaluate this at $s=2$: $\int_0^\infty e^{-2t} \frac{\sinh t}{t} dt = \frac{1}{2}\ln\left|\frac{2+1}{2-1}\right| = \frac{1}{2}\ln\left|\frac{3}{1}\right| = \frac{1}{2}\ln 3$. This matches the required result. **(iii) $\int_0^\infty e^{-2t} \sin 3t dt = 3/13$** *Formulae Used:* - Property of Laplace Transform: $\int_0^\infty e^{-st} f(t) dt = F(s)$. - $L\{\sin at\} = \frac{a}{s^2+a^2}$. *Solution:* The integral is $\int_0^\infty e^{-2t} \sin 3t dt$, which is the Laplace transform of $\sin 3t$ evaluated at $s=2$. Let $f(t) = \sin 3t$. $F(s) = L\{\sin 3t\} = \frac{3}{s^2+3^2} = \frac{3}{s^2+9}$. Now evaluate $F(s)$ at $s=2$: $F(2) = \frac{3}{2^2+9} = \frac{3}{4+9} = \frac{3}{13}$. This matches the required result. **(iv) $\int_0^\infty e^{-t} \sin^2 t dt = 1/4$** *Formulae Used:* - Property of Laplace Transform: $\int_0^\infty e^{-st} f(t) dt = F(s)$. - Trigonometric identity: $\sin^2 x = \frac{1-\cos(2x)}{2}$. - Linearity: $L\{af(t) + bg(t)\} = aL\{f(t)\} + bL\{g(t)\}$. - $L\{1\} = \frac{1}{s}$. - $L\{\cos at\} = \frac{s}{s^2+a^2}$. *Solution:* The integral is $\int_0^\infty e^{-t} \sin^2 t dt$, which is the Laplace transform of $\sin^2 t$ evaluated at $s=1$. Let $f(t) = \sin^2 t$. $\sin^2 t = \frac{1-\cos(2t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2t)$. $F(s) = L\{\frac{1}{2} - \frac{1}{2}\cos(2t)\}$ $= \frac{1}{2}L\{1\} - \frac{1}{2}L\{\cos 2t\}$ $= \frac{1}{2s} - \frac{1}{2}\frac{s}{s^2+2^2} = \frac{1}{2s} - \frac{s}{2(s^2+4)}$. Now evaluate $F(s)$ at $s=1$: $F(1) = \frac{1}{2(1)} - \frac{1}{2(1^2+4)} = \frac{1}{2} - \frac{1}{2(5)} = \frac{1}{2} - \frac{1}{10}$ $= \frac{5}{10} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5}$. Wait, the required result is $1/4$. Let me recheck. Ah, there was a problem number 13 in 21.1 $e^{-t} \sin^2 t$, the transform was $\frac{1}{2(s+1)} - \frac{s+1}{2((s+1)^2+4)}$. The current problem is $\int_0^\infty e^{-t} \sin^2 t dt$. This is $L\{\sin^2 t\}|_{s=1}$. My $F(s)$ for $\sin^2 t$ was $\frac{1}{2s} - \frac{s}{2(s^2+4)}$. $F(1) = \frac{1}{2(1)} - \frac{1}{2(1^2+4)} = \frac{1}{2} - \frac{1}{10} = \frac{4}{10} = \frac{2}{5}$. The given answer is $1/4$. There might be an issue in the problem statement or a common mistake. Let's carefully re-evaluate. $L\{\sin^2 t\} = \frac{1}{2s} - \frac{s}{2(s^2+4)}$. At $s=1$: $\frac{1}{2} - \frac{1}{2(1+4)} = \frac{1}{2} - \frac{1}{10} = \frac{5-1}{10} = \frac{4}{10} = \frac{2}{5}$. The result $1/4$ is incorrect for this integral. Or perhaps it's related to a different function. If the integral was $\int_0^\infty e^{-2t} \sin^2 t dt$, then $s=2$. $F(2) = \frac{1}{2(2)} - \frac{2}{2(2^2+4)} = \frac{1}{4} - \frac{2}{2(8)} = \frac{1}{4} - \frac{1}{8} = \frac{2-1}{8} = \frac{1}{8}$. If the integral was $\int_0^\infty e^{-t} \sin^2(t/2) dt$. Then $L\{\sin^2(t/2)\}|_{s=1}$. $\sin^2(t/2) = \frac{1-\cos t}{2}$. $L\{\frac{1-\cos t}{2}\} = \frac{1}{2}(\frac{1}{s} - \frac{s}{s^2+1})$. At $s=1$: $\frac{1}{2}(\frac{1}{1} - \frac{1}{1^2+1}) = \frac{1}{2}(1 - \frac{1}{2}) = \frac{1}{2}(\frac{1}{2}) = \frac{1}{4}$. So the problem might have intended $\sin^2(t/2)$ or had a typo in the $e^{-t}$ term. Assuming the problem statement is exact as printed, my calculation gives $2/5$. If the answer $1/4$ is correct, then it implies the function was $\sin^2(t/2)$. I will proceed with the calculation for $\sin^2 t$ as written. **21. Evaluate:** **(i) $L\{\int_0^t \frac{\sin \tau}{\tau} d\tau\}$** *Formulae Used:* - Transform of Integrals: $L\{\int_0^t f(\tau) d\tau\} = \frac{F(s)}{s}$, where $F(s) = L\{f(t)\}$ - Division by t: $L\{\frac{g(t)}{t}\} = \int_s^\infty G(u) du$ - $L\{\sin at\} = \frac{a}{s^2+a^2}$ *Solution:* Let $f(t) = \frac{\sin t}{t}$. First, find $L\{f(t)\}$. Let $g(t) = \sin t$. $G(s) = L\{\sin t\} = \frac{1}{s^2+1}$. Now apply the division by t property to find $F(s) = L\{f(t)\}$: $F(s) = \int_s^\infty G(u) du = \int_s^\infty \frac{1}{u^2+1} du$ $= \left[ \arctan(u) \right]_s^\infty$ $= \lim_{u\to\infty} \arctan(u) - \arctan(s)$ $= \frac{\pi}{2} - \arctan(s)$. For the transform to exist, $\lim_{t\to 0} \frac{\sin t}{t} = 1$, which exists. Now, apply the transform of integrals: $L\{\int_0^t \frac{\sin \tau}{\tau} d\tau\} = \frac{F(s)}{s} = \frac{1}{s} \left( \frac{\pi}{2} - \arctan(s) \right)$. **(ii) $L\{\int_0^t e^{-\tau} \cos t d\tau\}$** (This looks like a typo, perhaps $e^{-\tau} \cos \tau d\tau$ or $e^{-t} \cos \tau d\tau$. Assuming $e^{-\tau} \cos \tau d\tau$ for a solvable problem, which is problem 1 again). If it is $L\{\int_0^t e^{-\tau} \cos t d\tau\}$, then $\cos t$ is constant w.r.t. $\tau$. $L\{\cos t \int_0^t e^{-\tau} d\tau\} = L\{\cos t [-e^{-\tau}]_0^t\}$ $= L\{\cos t (-e^{-t} - (-e^0))\} = L\{\cos t (1 - e^{-t})\}$ $= L\{\cos t - e^{-t}\cos t\}$ $= L\{\cos t\} - L\{e^{-t}\cos t\}$ $L\{\cos t\} = \frac{s}{s^2+1}$. $L\{e^{-t}\cos t\}$ (using First Shifting Theorem on $\cos t$ with $a=-1$) $= \frac{s-(-1)}{(s-(-1))^2+1^2} = \frac{s+1}{(s+1)^2+1} = \frac{s+1}{s^2+2s+2}$. So, $L\{\int_0^t e^{-\tau} \cos t d\tau\} = \frac{s}{s^2+1} - \frac{s+1}{s^2+2s+2}$. **(iii) $L\{\int_0^t e^{-t} \sin t dt\}$** (This looks like a typo, if it's $\int_0^t e^{-\tau} \sin \tau d\tau$, then it's similar to problem 1. If it's $e^{-t} \int_0^t \sin \tau d\tau$, it's different. Assuming $e^{-t}$ is outside the integral or part of the function to be transformed, so $L\{e^{-t} \cdot \text{integral}\}$). Let's assume the question meant $L\{e^{-t} \int_0^t \sin \tau d\tau\}$. *Formulae Used:* - First Shifting Theorem: $L\{e^{at}f(t)\} = F(s-a)$ - Transform of Integrals: $L\{\int_0^t g(\tau) d\tau\} = \frac{G(s)}{s}$ - $L\{\sin at\} = \frac{a}{s^2+a^2}$ *Solution:* Let $g(\tau) = \sin \tau$. Then $G(s) = L\{\sin \tau\} = \frac{1}{s^2+1}$. Let $h(t) = \int_0^t \sin \tau d\tau$. $L\{h(t)\} = \frac{G(s)}{s} = \frac{1}{s(s^2+1)}$. Call this $H(s)$. We need to find $L\{e^{-t} h(t)\}$. Using the First Shifting Theorem with $a=-1$: $L\{e^{-t} h(t)\} = H(s-(-1)) = H(s+1)$ $= \frac{1}{(s+1)((s+1)^2+1)} = \frac{1}{(s+1)(s^2+2s+2)}$. **22. Show that:** **(i) $L\{t J_0(at)\} = \frac{s}{(s^2+a^2)^{3/2}}$** *Formulae Used:* - $L\{t f(t)\} = -\frac{d}{ds} F(s)$ - $L\{J_0(at)\} = \frac{1}{\sqrt{s^2+a^2}}$ (standard result for Bessel function of first kind, order 0) *Solution:* Let $f(t) = J_0(at)$. Then $F(s) = L\{J_0(at)\} = \frac{1}{\sqrt{s^2+a^2}}$. Using the property $L\{t f(t)\} = -\frac{d}{ds} F(s)$: $L\{t J_0(at)\} = -\frac{d}{ds} \left( \frac{1}{\sqrt{s^2+a^2}} \right)$ $= -\frac{d}{ds} (s^2+a^2)^{-1/2}$ $= - (-\frac{1}{2})(s^2+a^2)^{-3/2} (2s)$ $= s(s^2+a^2)^{-3/2} = \frac{s}{(s^2+a^2)^{3/2}}$. This matches the required result. **(ii) $\int_0^\infty t e^{-3t} J_0(4t) dt = 3/125$** *Formulae Used:* - Property of Laplace Transform: $\int_0^\infty e^{-st} f(t) dt = F(s)$. - Result from (i): $L\{t J_0(at)\} = \frac{s}{(s^2+a^2)^{3/2}}$. *Solution:* The integral is $\int_0^\infty t e^{-3t} J_0(4t) dt$, which is the Laplace transform of $t J_0(4t)$ evaluated at $s=3$. From part (i), we have $L\{t J_0(at)\} = \frac{s}{(s^2+a^2)^{3/2}}$. Here, $a=4$. So, $L\{t J_0(4t)\} = \frac{s}{(s^2+4^2)^{3/2}} = \frac{s}{(s^2+16)^{3/2}}$. Now evaluate this at $s=3$: $\int_0^\infty t e^{-3t} J_0(4t) dt = \frac{3}{(3^2+16)^{3/2}} = \frac{3}{(9+16)^{3/2}}$ $= \frac{3}{(25)^{3/2}} = \frac{3}{(\sqrt{25})^3} = \frac{3}{5^3} = \frac{3}{125}$. This matches the required result.