Solving 1st Order ODE

\begin{aligned}
\frac{dy}{dx} &= f(y)\\\\
\int \frac{1}{f(y)} dy &= \int 1\;dx
\end{aligned}

\begin{aligned}
by' + ay &= 0\\\\
\frac{dy}{dx} &= -\frac{a}{b} y\\
\int \frac{1}{y} dy &= \int -\frac{a}{b} dx\\
\ln{y} &= -\frac{a}{b}x + c\\\\
y &= Ae^{\lambda x}, \text{ where } A = e^c, \lambda = -\frac{a}{b}
\end{aligned}

\begin{aligned}
\frac{dy}{dx} &= f(x)\\\\
\int 1\;dy &= \int f(x)\;dx
\end{aligned}

\begin{aligned}
\frac{dy}{dx} &= g(x)h(y)\\\\
\int \frac{1}{h(y)} dy &= \int g(x)\;dx
\end{aligned}

\begin{aligned}
bxy' + ay &= 0\\\\
\frac{dy}{dx} &= -\frac{a}{b} \left(\frac{y}{x}\right)\\
\int \frac{1}{y} dy &= -\frac{a}{b}\int \frac{1}{x} dx\\
\ln{y} &= \lambda \ln x + c\\
\ln{y} &= \ln (x^\lambda e^c)\\\\

y &= Ax^{\lambda}, \text{ where } A = e^c, \lambda = -\frac{a}{b}
\end{aligned}

\begin{aligned}
\frac{dy}{dx} + P(x)y &= Q(x)\;\;\text{(P(x) can be a constant)}\\\\
\text{Let }R(x) &= e^{\int P(x) dx} \\\\
y &= \frac{\int {RQ}\; dx}{R}
\end{aligned}

\begin{aligned}
R(x) &= e^{\int P(x)\;dx}\\
\ln R(x) &=\int P(x)\;dx\\
P(x) &= \frac{R'}{R}\\\\
y &= \frac{\int {RQ}\;dx}{R}\\\\
\frac{dy}{dx} &= \frac{RQR - R'\int {RQ}\;dx}{R^2}\\
&= Q - \frac{R'(Ry)}{R^2}\\
&= Q - \frac{R'y}{R}\\
&= Q - Py
\end{aligned}