first-order ODE
• form $\frac{dy}{dx} = F(x,y)$

if $F(x,y) = X(x)Y(y)$, then the equation can be resolved by
$\int{\frac{dy}{Y(y)}}=\int{X(x)dx}+c$

• form $M(x,y)dx+N(x,y)dy=0$

$du=\frac{{\partial}u}{{\partial}x}dx+\frac{{\partial}u}{{\partial}y}dy=0$
$u(x,y)=c$

if $\frac{{\partial}M}{{\partial}y}=\frac{{\partial}N}{{\partial}x}$ ($\frac{{\partial}M}{{\partial}y}=\frac{{{\partial}^2}u}{{\partial}y{\partial}x}$, $\frac{{\partial}N}{{\partial}x}=\frac{{{\partial}^2}u}{{\partial}x{\partial}y}$) then

(1)
\begin{align} \frac{{\partial}u}{{\partial}x}=M, \frac{{\partial}u}{{\partial}y}=N \end{align}
(2)
\begin{align} u=\int{Mdx}+k(y) \end{align}

differences it and compares the result with $N=\frac{{\partial}u}{{\partial}y}$, we can find $\frac{dk}{dy}$.

## form Homogeneous $y'+p(x)y=0$

(7)
\begin{align} y(x)=ce^{-\int p(x)dx} \end{align}

## form ### Integrating Factor $F(x)=e^{\int{p(x)dx}}$ ($\frac{1}{F}\frac{dF}{dx}=p(x)$)

if $\frac{dy}{dx}+p(x)y(x)=q(x)$ is multiplied by Integrating Factor
then

(8)
\begin{align} \frac{d}{dx}[e^{\int{p(x)dx}}y(x)]=e^{\int{p(x)dx}}[\frac{dy(x)}{dx}+p(x)y(x)]=e^{\int{p(x)dx}}q(x) \end{align}

integrates each side, we can get

(9)
\begin{align} \mu{y(x)}=\int{\mu{q(x)}dx+c} \end{align}

so, page revision: 8, last edited: 08 Jun 2010 03:57
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License