written 6.1 years ago by | • modified 6.0 years ago |
put,
$y-x=v$
$\frac{dy}{dx}-1=\frac{dv}{dx} $
Hence the given equation reduces to
$\frac{dv}{dx}+xv=-x^3v^2 $
Dividing throughout by $-v^2$, we get
$-\frac{1}{v^2}\frac{dv}{dx}-\frac{x}{v}=x^3$ $\dots\dots(1)$
put
$ \frac{1}{v}=t \\ -\frac{1}{v^2}\frac{dv}{dx}=\frac{dt}{dx}$
Eqn(1) becomes
$\frac{dt}{dx}-xt=x^3$
This is a linear differential equation in t with $P=-x$ and $Q=x^3$
$I.F=e^{\int(-x)dx}=e^{-\frac{x^2}{2}} $
The solution is
$te^{-\frac{x^2}{2}}=\int e^{-\frac{x^2}{2}}x^3dx+c$
put
$-\frac{x^2}{2}=u$
$ x^2=-2u$
$2xdx=-2du$
$xdx=-du$
$ te^{-\frac{x^2}{2}}=\int e^u(-2u)(-du)+c$
$te^{\frac{-x^2}{2}}=2\int e^u udu+c $
Using integration by parts on R.H.S, we get
$ te^{-\frac{x^2}{2}}=2(ue^u-e^u)+c$
$te^{-\frac{x^2}{2}}=2\left( -\frac{x^2}{2}e^{-\frac{x^2}{2}}-e^{-\frac{x^2}{2}}\right)+c$
$ te^{-\frac{x^2}{2}}=2e^{-\frac{x^2}{2}}\left(-\frac{x^2}{2}-1 \right)+c$
$\frac{1}{v}e^{-\frac{x^2}{2}}=2e^{-\frac{x^2}{2}}\left(-\frac{x^2}{2}-1 \right)+c$
$\frac{1}{y-x}e^{-\frac{x^2}{2}}=2e^{-\frac{x^2}{2}}\left(-\frac{x^2}{2}-1 \right)+c$
$\frac{1}{y-x}=-x^2-2+ce^{-\frac{x^2}{2}}$