Separable Differential Equations
\[\frac{{\rm d}y}{{\rm d}x}=\frac{f(x)}{g(y)},g(y)\ne0
\]
Solution
\[\begin{align}
\frac{{\rm d}y}{{\rm d}x}=\frac{f(x)}{g(y)}&\Longleftrightarrow g(y)\frac{{\rm d}y}{{\rm d}x}=f(x)\\
&\Longleftrightarrow\int g(y)\frac{{\rm d}y}{{\rm d}x}{\rm d}x=\int f(x){\rm d}x\\
&\Longleftrightarrow\int g(y){\rm d} y=\int f(x){\rm d}x\\
\end{align}
\]
First-Order Linear Differential Equations
\[\frac{{\rm d}y}{{\rm d}x}+P(x)y=Q(x)
\]
Solution
Let
\[\mu(x)=e^{\int P(x){\rm d}x}
\]
Then
\[\begin{align}
\frac{{\rm d}y}{{\rm d}x}+P(x)y=Q(x)&\Longleftrightarrow\frac{{\rm d}y}{{\rm d}x}\mu(x)+\mu(x)P(x)y=\mu(x)Q(x)\\
&\Longleftrightarrow\frac{{\rm d}y}{{\rm d}x}\mu(x)+\frac{{\rm d}\mu}{{\rm d}x}y=\mu(x)Q(x)\\
&\Longleftrightarrow\frac{\rm d}{{\rm d}x}(\mu(x)y)=\mu(x)Q(x)\\
&\Longleftrightarrow\int\frac{\rm d}{{\rm d}x}(\mu(x)y){\rm d}x=\int\mu(x)Q(x){\rm d}x\\
&\Longleftrightarrow\mu(x)y=\int\mu(x)Q(x){\rm d}x\\
&\Longleftrightarrow y=\frac{1}{\mu(x)}\int\mu(x)Q(x){\rm d}x\\
\end{align}
\]
Exact Differential Equations
\[M(x,y){\rm d}x+N(x,y){\rm d}y=0,\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}
\]
Solution
\[\begin{align}
\frac{\partial F}{\partial x}=M(x,y)&\Longleftrightarrow\exist h(y),F(x,y)=\int M(x,y){\rm d}x+h(y)\\
&\Longleftrightarrow\exist h(y),\frac{\partial F}{\partial y}=\frac{\partial}{\partial y}\int M(x,y){\rm d}x+\frac{{\rm d}h}{{\rm d}y}
\end{align}
\]
Therefore, if \(\frac{\partial F}{\partial x}=M\),
\[\begin{align}
\frac{\partial F}{\partial y}=N(x,y)&\Longleftrightarrow\exist h(y),\frac{\partial}{\partial y}\int M(x,y){\rm d}x+\frac{{\rm d}h}{{\rm d}y}=N(x,y)\\
&\Longleftrightarrow\exist h(y),\frac{{\rm d}h}{{\rm d}y}=N(x,y)-\frac{\partial}{\partial y}\int M(x,y){\rm d}x
\end{align}
\]
Since
\[\begin{align}
\frac{\partial}{\partial x}\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y){\rm d}x\right)&=\frac{\partial N}{\partial x}-\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}\int M(x,y){\rm d}x\right)\\
&=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\\
&=0
\end{align}
\]
It follows that
\[\exist h(y),\frac{{\rm d}h}{{\rm d}y}=N(x,y)-\frac{\partial}{\partial y}\int M(x,y){\rm d}x
\]
Therefore,
\[\exist F(x,y),\frac{\partial F}{\partial x}=M(x,y)\land\frac{\partial F}{\partial y}=N(x,y)
\]
Therefore,
\[\begin{align}
M(x,y){\rm d}x+N(x,y){\rm d}y=0&\Longleftrightarrow{\rm d}F=0\\
&\Longleftrightarrow F(x,y)=C
\end{align}
\]
Homogeneous Differential Equations
\[\frac{{\rm d}y}{{\rm d}x}=f\left(\frac{y}{x}\right)
\]
Solution
Let
\[u(x)=\frac{y}{x}
\]
Then
\[\begin{align}
u(x)=\frac{y}{x}&\Longleftrightarrow y=u(x)x\\
&\Longleftrightarrow\frac{{\rm d}y}{{\rm d}x}=\frac{{\rm d}u}{{\rm d}x}x+u(x)\\
\end{align}
\]
Therefore,
\[\begin{align}
\frac{{\rm d}y}{{\rm d}x}=f\left(\frac{y}{x}\right)&\Longleftrightarrow\frac{{\rm d}u}{{\rm d}x}x+u(x)=f(u(x))\\
&\Longleftrightarrow\frac{{\rm d}u}{{\rm d}x}=\frac{f(u(x))-u(x)}{x}
\end{align}
\]
which is a separable differential equation.
Bernoulli Equations
\[\frac{{\rm d}y}{{\rm d}x}+P(x)y=Q(x)y^n,n\ne0\land n\ne1
\]
Solution
Let
\[u(x)=y^{1-n}
\]
Then
\[\frac{{\rm d}u}{{\rm d}x}=(1-n)y^{-n}\frac{{\rm d}y}{{\rm d}x}\Longleftrightarrow\frac{{\rm d}y}{{\rm d}x}=\frac{y^n}{1-n}\frac{{\rm d}u}{{\rm d}x}
\]
Therefore,
\[\begin{align}
\frac{{\rm d}y}{{\rm d}x}+P(x)y=Q(x)y^n&\Longleftrightarrow\frac{y^n}{1-n}\frac{{\rm d}u}{{\rm d}x}+P(x)y=Q(x)y^n\\
&\Longleftrightarrow\frac{{\rm d}u}{{\rm d}x}+(1-n)P(x)u(x)=(1-n)Q(x)
\end{align}
\]
which is a first-order linear differential equation.
\(f(Ax+By+C)\) Differential Equations
\[\frac{{\rm d}y}{{\rm d}x}=f(Ax+By+C),B\ne0
\]
Solution
Let
\[u(x)=Ax+By+C
\]
Then
\[\frac{{\rm d}u}{{\rm d}x}=A+B\frac{{\rm d}y}{{\rm d}x}\Longleftrightarrow\frac{{\rm d}y}{{\rm d}x}=\frac{1}{B}\left(\frac{{\rm d}u}{{\rm d}x}-A\right)
\]
Therefore,
\[\begin{align}
\frac{{\rm d}y}{{\rm d}x}=f(Ax+By+C)&\Longleftrightarrow\frac{1}{B}\left(\frac{{\rm d}u}{{\rm d}x}-A\right)=f(u(x))\\
&\Longleftrightarrow\frac{{\rm d}u}{{\rm d}x}=A+Bf(u(x))
\end{align}
\]
which is a separable differential equation.