Methods from First-order methods

To solve $\frac{dy}{dx}=f(x)g(y)$ on a branch where $g(y)\neq0$:

1. Write $\frac{dy}{g(y)}=f(x)\,dx$.
2. Integrate both sides: $$\int\frac{dy}{g(y)}=\int f(x)\,dx+C.$$
3. Apply initial or boundary data and state the interval or branch on which the result is valid.
4. Check separately any constant solutions arising from $g(y)=0$.

For $y'+p(x)y=q(x)$ on an interval where $p$ and $q$ are continuous:

1. Choose $$\mu(x)=e^{\int p(x)\,dx}.$$
2. Since $\mu'=p\mu$, multiply the equation by $\mu$: $$\mu y'+\mu p y=(\mu y)'=\mu q.$$
3. Integrate: $$\mu y=\int\mu q\,dx+C.$$
4. Divide by $\mu$: $$y=\frac1{\mu}\left(\int\mu q\,dx+C\right).$$

To solve an exact equation:

1. Check that $M_y=N_x$ on the domain.
2. Integrate $M$ with respect to $x$, treating $y$ as constant: $$F(x,y)=\int M(x,y)\,dx+h(y).$$
3. Differentiate this expression with respect to $y$.
4. Match $F_y$ with $N$ to determine $h(y)$.
5. Write the implicit solution as $F(x,y)=C$.

For an equation of the form

set $y=xv(x)$. Then

so

The resulting equation is separable on an interval where $x\neq0$ and the required denominators are nonzero.

For a Bernoulli equation, on a nonzero real branch where $y^{1-n}$ is defined, set

Since

multiplying the equation by $(1-n)y^{-n}$ gives the linear equation

Solve this equation with an integrating factor, then substitute back. Check $y=0$ separately because the substitution divides by a power of $y$; it is an admissible solution only when the original equation is defined there and both sides agree.