First Order Differential Equations
Separable Equations
$\frac{dy}{dx} = g(x)h(y)$
To solve:

Rearrange: \(\frac{1}{h(y)}dy = g(x)dx\)

Integrate both sides: \(\int \frac{1}{h(y)}dy = \int g(x)dx\)

Solve for $y$

Lump all constants together into $c_1$
Linear Equations
$a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)$
If $g(x) = 0$, then the equation is said to be homogeneous, otherwise its nonhomogeneous.
To solve:

Put into standard form: \(\frac{dy}{dx} + \frac{a_0(x)}{a_1(x)}y = \frac{g(x)}{a_1(x)}\)

$P(x) = \frac{a_0(x)}{a_1(x)}$ and $f(x) = \frac{g(x)}{a_1(x)}$

Determine the integrating factor: $\mu(x) = e^{\int P(x)dx}$
Note: Ignore the constant when integrating $P(x)$. 
Multiply both sides of the equation by the integrating factor: \(\mu(x)[\frac{dy}{dx} + P(x)y] = \mu(x)f(x)\)

Notice that the left side of the equation will always become $(\mu(x)y)’$: \((\mu(x)y)' = \mu(x)f(x)\)

Integrate both sides: \(\mu(x)y = \int \mu(x)f(x)dx\)

Solve for $y$

Lump all constants together into $c_1$
Exact Equations
$M(x, y)dx + N(x, y)dy = 0$
If $f(x, y)$ exists such that $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = M(x, y)dx + N(x, y)dy$, then $df = 0$ and $f(x, y) = c$
To solve:

Verify that $M_y = N_x$ or $M_x = N_y$ (whichever is easier): \(\frac{\partial}{\partial y}M(x, y) = \frac{\partial}{\partial x}N(x, y)\)

Integrate $M(x, y)$. The constant of integration will be replaced by a function that only depends on $y$: \(\int M(x, y)dx = \mathbf{M}(x, y) + h(y)\)

Integrate $N(x, y)$. The constant of integration will be replaced by a function that only depends on $x$: \(\int N(x, y)dy = \mathbf{N}(x, y) + g(x)\)

$h(y)$ is equal to the terms in $\mathbf{N}(x, y)$ that aren’t also in $\mathbf{M}(x, y)$.

$g(y)$ is equal to the terms in $\mathbf{M}(x, y)$ that aren’t also in $\mathbf{N}(x, y)$.

Determine $f(x, y)$. Terms that exist in both $\mathbf{M}(x, y)$ and $\mathbf{N}(x, y)$ exist only once in $f(x, y)$. Also include $h(y)$ and $g(y)$ in $f(x, y)$.

Rewrite as $f(x, y) = c$ to get an implicit solution.
Bernoulli’s Equation
$a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)y^n$
If $n = 0$ or $1$, then the equation becomes linear and can be solved using the method above.
To solve:

Rewrite in standard form: \(\frac{dy}{dx} + \frac{a_0(x)}{a_1(x)}y = \frac{g(x)}{a_1(x)}y^n\)

$P(x) = \frac{a_0(x)}{a_1(x)}$ and $f(x) = \frac{g(x)}{a_1(x)}$

Compute $u = y^{n  1}$. Solve in terms of $y$, then take the derivative to get $\frac{dy}{dx}$: \(u = y^{n  1} \Rightarrow y = u^{(1  n)}\) \(\frac{dy}{dx} = (1  n)u^{n}\)

Substitute in $y$ and $y’$: \((1  n)u^{n} + P(x)u^{(1  n)} = f(x)(u^{(1  n)})^n\)

Solve as a linear equation.

Substitute back in for $u$ and $u’$.
Reduction to Separable
$y’ = f(Ax + By + C)$
To solve:

Find a substitution of the form $u = Ax + By + C$

Solve $u$ in terms of $y$ and $y’$: \(u = Ax + By + C \Rightarrow By = u  Ax  C\) \(y = \frac{1}{B}u  \frac{A}{B}x  \frac{C}{B}\) \(y' = \frac{1}{B}u'  \frac{A}{B}\)

Substitute in $y$ and $y’$

Solve as a separable equation.

Substitute back in for $u$
Homogeneous Equations
Not to be confused with homogeneous linear equations when $f(x) = 0$.
Any equation (not just a DE) is homogeneous to the degree $\alpha$ if: $f(tx, ty) = t^{\alpha}f(x, y)$
$M(x, y)dx + N(x, y)dy = 0$
If the equation has the form of an exact equation, but $M_y = N_x$, this method can be used if $M(x, y)$ and $N(x, y)$ are homogeneous to the same degree.
To solve:

Substitute $y = ux$ and $y’ = udx + xdu$ or $x = vy$ and $x’ = vdy + ydv$, whichever is easier.

Solve as a separable equation.

Substitute back in for $u$ or $v$.