Question

Solve the differential equation:
dy/dx+(x÷1-x²)y=x√y,y(0)=1

Answer 1

Given differential equation is $\frac{dy}{dx}+\frac{x}{1-x²}y=x\sqrt{y}$

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

Now, let $\sqrt{y}=t \implies \frac{1}{2\sqrt{y}} {dy}{dx} = \frac{dt}{dx}$

So, differential equation becomes: $2\frac{dt}{dx} + \frac{x}{1-x^2} t = x \implies \frac{dt}{dx} +\frac{1}{2} \frac{x}{1-x^2} t = \frac{x}{2}$,

this is linear differential equation of form $\frac{dt}{dx} + P(x) t = Q(x)$.

So, Integrating factor(I.F.) = $e^{{\int} \frac{1}{2} \frac{x}{1-x^2} dx} = e^{\frac{-1}{4} log(1-x^2)} = (1-x^2)^{\frac{-1}{4}}$

Hence solution is $t (1-x^2)^{\frac{-1}{4}} = \int {\frac{x}{2} (1-x^2)^{\frac{-1}{4}}} dx + c$

$\implies t (1-x^2)^{\frac{-1}{4}} = (\frac{1}{-4}) (\frac{4}{3}) (1-x^2)^{\frac{3}{4}} + c$

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

Now given y(0)=1

$\implies 1 = \frac{-1}{3} + c \implies c = \frac{4}{3}$.

Hence the final solution is $\sqrt{y} = (\frac{-1}{3}) (1-x^2) + \frac{4}{3} (1-x^2)^{\frac{1}{4}}$.