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Solve $\frac{dy}{dx}=1-x(y-x)-x^3(y-x)^2$ .
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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$

$\frac{1}{v}e^{-\frac{x^2}{2}}=2e^{-\frac{x^2}{2}}\left(-\frac{x^2}{2}-1 \right)+c$

$\frac{1}{y-x}e^{-\frac{x^2}{2}}=2e^{-\frac{x^2}{2}}\left(-\frac{x^2}{2}-1 \right)+c$

$\frac{1}{y-x}=-x^2-2+ce^{-\frac{x^2}{2}}$

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