Question

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

Answer 1

the given equation can be written as:

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

let $v=2\sqrt{y}$

then, $v'=\frac{y'}{\sqrt{y}}$

substituting above, we get,

$v'+ \frac{xv}{2(1-x^2)} = x$ ........(1)

comparing with v'+Pv=Q where P and Q are functions of x

P= $\frac{x}{1-x^2}$ and Q= x

integrating factor= $e^{\int{Pdx}}= e^{\int{\frac{x}{(1-x^2)}dx}} = (1-x^2)^{-1/4}$

multiplying (1) by integrating factor,

$v. (1-x^2)^{-1/4} = \int{(1-x^2)^{-1/4}.x}dx$

$v. (1-x^2)^{-1/4} = -\frac{2}{3}(1-x^2)^{3/4} + c$

substituting value of v,

$2\sqrt{y}. (1-x^2)^{-1/4} = -\frac{2}{3}(1-x^2)^{3/4} + c$

$2\sqrt{y}= -\frac{2}{3}(1-x^2) + c(1-x^2)^{1/4}$

now, y(0)=1

$2=\frac{-2}{3}+c \\ c=\frac{8}{3}$

$2\sqrt{y}= -\frac{2}{3}(1-x^2) + \frac{8}{3}(1-x^2)^{1/4}$