Question

Solve the differential equation

Answer 1

Step-by-step explanation:

I hope that its the true answerand you got the answer and method of doing it

Answer 2

$\bf\large\red{\underline{\underline{Question}}}:$

To solve the differential equation : $\dfrac{dy}{dx}+ \dfrac{2x}{1-x^2}y=xy^2$

$\bf\large\red{\underline{\underline{Solution}}}:$

$\longmapsto\dfrac{dy}{dx}+ \dfrac{2x}{1-x^2}y=xy^2$

As we can see given equation is not proper linear equation. So we have to reduce it to linear equation.

$\longmapsto\rm\dfrac{dy}{dx}+ \dfrac{2x}{1-x^2}y=xy^2$

Divide all terms by y², we got :

$\rm\implies\dfrac{1}{y^2}\dfrac{dy}{dx}+ \dfrac{2x}{1-x^2}\dfrac{1}{y}=x$

Now ,

put 1/y = t

So that, $\bf\dfrac{1}{y^2}\dfrac{dy}{dx} = -\dfrac{dt}{dx}$

So,

Our equation will be : $\dfrac{dt}{dx} -\dfrac{2x}{1-x^2}t = x$

Comparing with  $\dfrac{dt}{dx} + Pt = Q$, We get , P = -2x/(1-x²) and Q = x

I.F. = $e^{\int \dfrac{-2x}{1-x^2}} = e^{\log(1-x^2)} = 1-x^2$

∴ Sol. of the eq. is (1-x²).t = ∫(1-x²).x dx + c

(1-x²).t = ∫(x-x³)dx + c

(1-x²).t = ∫x dx -∫x³dx + c

$\rm (1-x^2).t = \dfrac{x^2}{2} - \dfrac{x^4}{4} + c$

$\rm t = \dfrac{x^2}{2(1-x^2)} - \dfrac{x^4}{4(1-x^2)} + (1-x^2)^{-1}c$

t = 1/y

$\rm \dfrac{1}{y} = \dfrac{x^2}{2(1-x^2)} - \dfrac{x^4}{4(1-x^2)} + (1-x^2)^{-1}c$