Question

xdy + (y - x² y² ) dx = 0

Answer 1

Solution

$xdy + (y - x^2y^2 ) dx = 0$

Divide through by $dx$ as follows

$x{dy\over dx} + (y - x^2y^2 ){ dx\over dx} = {0\over dx}$

We get

$x{dy\over dx} + (y - x^2y^2 ) =0$

$x{dy\over dx} + y - x^2y^2 =0$

This is a first-order nonlinear differential equation

Let $u=xy$

Use product rule of differentiation to differentiate $u=xy$ with respect to $x$

$u=xy$

${du\over dx}=({d\over dx}(x)).y+x.({d\over dx}(y))$

$\implies {du\over dx}=y+x{dy\over dx}$

From ${du\over dx}=y+x{dy\over dx}$ ,We can get expression for ${dy\over dx}$

${du\over dx}=y+x{dy\over dx}$

$x{dy\over dx}={du\over dx}-y$

Now, divide through by $x$ to get

${dy\over dx}={{du\over dx}-y \over x}={{du\over dx}\over x}-{y\over x}$

From $u=xy$ , $y={u\over x}$

$\therefore {dy\over dx}={{du\over dx}\over x}-{{u\over x}\over x} ={{du\over dx}\over x}-{u\over x^2}$

Now, let's substitute ${dy\over dx}$ and $y$ to the differential equation

$x{dy\over dx} + y - x^2y^2 =0$

$x({{du\over dx}\over x}-{u\over x^2})+ {u\over x} - x^2({u\over x})^2 =0$

${{du\over dx}}-{u\over x}+ {u\over x} - x^2.{u^2\over x^2} =0$

${{du\over dx}}- u^2=0$

$\implies {du\over dx}=u^2 \implies {du\over u^2}=dx$

Take integral on both sides of ${du\over u^2}=dx$

we have

$\int{du\over u^2}=\int dx \implies -{1\over u}=x+C \implies u=-{1\over x+C}$

Recall $y={u\over x} \implies y={-{1\over x+C}\over x}=-{1\over x+C}\times{1\over x}=-{1\over x(x+C)}$

$\therefore$ the general solution is thus $y=-{1\over x(x+C)}$