If $\mu$ depends only on $x$: Calculate the expression $\frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right)$. If this expression is solely a function of $x$ (let's call it $f(x)$), then the integrating factor is $\mu(x) = e^{\int f(x)dx}$.

If $\mu$ depends only on $y$: Calculate the expression $\frac{1}{M} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)$. If this expression is solely a function of $y$ (let's call it $g(y)$), then the integrating factor is $\mu(y) = e^{\int g(y)dy}$.

Once $\mu$ is found, multiply the original equation by $\mu$ and solve the resulting exact equation using the procedure from Section 1.3.

1.5 Linear First-Order Equations

Form: A linear first-order differential equation for $y$ is of the form $y' + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are continuous functions of $x$.

Solution Procedure:

Step 1: Identify $P(x)$ and $Q(x)$. Ensure the equation is in the standard form $y' + P(x)y = Q(x)$.
Step 2: Calculate the Integrating Factor. The integrating factor is $\mu(x) = e^{\int P(x)dx}$.
Step 3: Multiply by the Integrating Factor. Multiply the entire differential equation by $\mu(x)$: $\mu(x)y' + \mu(x)P(x)y = \mu(x)Q(x)$. The left-hand side is now the derivative of a product: $(\mu(x)y)'$.
Step 4: Integrate Both Sides. Integrate both sides with respect to $x$: $\int (\mu(x)y)'dx = \int \mu(x)Q(x)dx$. This yields $\mu(x)y = \int \mu(x)Q(x)dx + C$.
Step 5: Solve for $y$. Divide by $\mu(x)$ to get the general solution: $y(x) = \frac{1}{\mu(x)} \left( \int \mu(x)Q(x)dx + C \right)$.

1.6 Bernoulli Equations

Form: Bernoulli equations are non-linear first-order differential equations of the form $y' + P(x)y = Q(x)y^n$, where $n$ is any real number except $0$ or $1$ (if $n=0$ or $n=1$, it's a linear equation).

Solution Procedure:

Step 1: Transform to Linear. Divide the entire equation by $y^n$: $y^{-n}y' + P(x)y^{1-n} = Q(x)$.
Step 2: Substitute. Let $v = y^{1-n}$. Differentiate $v$ with respect to $x$ using the chain rule: $v' = (1-n)y^{-n}y'$.
Step 3: Rewrite and Solve. Substitute $v$ and $v'$ into the transformed equation. Notice that $y^{-n}y' = \frac{1}{1-n}v'$. The equation becomes $\frac{1}{1-n}v' + P(x)v = Q(x)$, which is a linear first-order equation in $v$. Solve this linear equation for $v$ using the method described in Section 1.5.
Step 4: Substitute Back. Once $v(x)$ is found, substitute back $y = v^{1/(1-n)}$ (or $y = v^{1/(1-n)}$ if $1-n$ is negative) to get the solution for $y$.

1.7 Homogeneous Equations

Form: A first-order differential equation is homogeneous if it can be written in the form $y' = f(y/x)$. This means that the right-hand side is a function of the ratio $y/x$.

Solution Procedure:

Step 1: Substitute. Let $v = y/x$. This implies $y = vx$.
Step 2: Differentiate. Differentiate $y = vx$ with respect to $x$ using the product rule: $y' = v \cdot 1 + x \cdot v' = v + xv'$.
Step 3: Transform to Separable. Substitute $v$ and $y'$ into the original homogeneous equation: $v + xv' = f(v)$. This equation can always be rearranged into a separable form: $xv' = f(v) - v \Rightarrow \frac{dv}{f(v) - v} = \frac{dx}{x}$.
Step 4: Solve Separable Equation. Integrate both sides: $\int \frac{dv}{f(v) - v} = \int \frac{dx}{x} + C$.
Step 5: Substitute Back. Solve for $v$, then replace $v$ with $y/x$ to obtain the solution for $y$.

2. Second-Order Linear Differential Equations

Differential equations involving the second derivative of the un