Question

solve the differential equation dy/dx + (x/1-x^2)y=xy^1/2 , y(0)=1

correct question

Answer 1

the given equation can be written as

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

this is of the form f'(y) y' + P f(y) = Q

where P = $\frac{x}{2(1-x^2)}$

Q= x

f(y) = 2 $\sqrt{y}$

let v= 2$\sqrt{y}$

differentiating,

$\frac{1}{\sqrt{y}} \frac{dy}{dx} = \frac{dv}{dx}$

(1) becomes

$\frac{dv}{dx} + \frac{xv}{2(1-x^2)} = x$

this is of the form $\frac{dv}{dx} + Av = B$

$A = \frac{x}{2(1-x^2)} and B=x$

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

the general solution is given by:

v (1-x²)^-1/4 = $\int$ x. (1-x²)^-1/4 dx + c

2$\sqrt{y}$ = -2/3 (1-x²) + c(1-x²)^1/4

now, given y(0)=1

2 = -2/3 + c

c= 8/3

so, the solution is

2$\sqrt{y}$ = -2/3 (1-x²) + 8/3(1-x²)^1/4