Answer to Question #110097 in Differential Equations for ABHISHEK TOMAR

Question

Solve the differential equation
dy/dx + (x/1-x^2)y = x√y, y(0) = 1.

Answer 1

Given

$y' +\left( \frac{x}{1-x^2}\right)y=xy^{1/2}$

This is a Bernoli equation , multiply both sides by $y^{-1/2}$ then

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

Let $z=y^{1/2},\ \ \ \ \ ,z'= \frac{1}{2}y^{-1/2}y'$ , then

$2z' +\left( \frac{x}{1-x^2}\right)z=x\\ z' +\frac{1}{2}\left( \frac{x}{1-x^2}\right)z=\frac{x}{2}$

This is a linear equation with integral coefficient

$\mu= e^{\int\left( \frac{ x}{2(1-x^2)}\right)dx }\\ \ \ \ \ = e^{-\frac{1}{4}\ln (1-x^2)}\\ \ \ \ \ =(1-x^2)^{-1/4}$

Then the general solution is

$\mu z \ \ \ \ \ \ \ \ = \frac{1}{2} \int \mu (x)dx\\ (1-x^2)^{-1/4}z= \frac{1}{2} \int x(1-x^2)^{-1/4}dx\\ (1-x^2)^{-1/4}z=-\frac{1}{3}\left(1-x^2\right)^{\frac{3}{4}}+C\\ (1-x^2)^{-1/4}\sqrt{y}=-\frac{1}{3}\left(1-x^2\right)^{\frac{3}{4}}+C$

Since $y(0)=1\ \ \ \to C= \frac{4}{3}$

$(1-x^2)^{-1/4}\sqrt{y}=-\frac{1}{3}\left(1-x^2\right)^{\frac{3}{4}}+\frac{4}{3}\\ \ \ \ \ \ \ \ \ \ \ \ \sqrt{y}= \frac{-1}{3}(1-x^2) +\frac{4}{3}(1-x^2)^{1/4}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y=\left[ \frac{-1}{3}(1-x^2) +\frac{4}{3}(1-x^2)^{1/4}\right]^2$