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Question: The solution of $\frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y}$, $-1<x<1$ is given by:...

The solution of dydx+x1−x2y=xy\frac{dy}{dx} + \frac{x}{1-x^2}y = x\sqrt{y}, −1<x<1-1<x<1 is given by:

A

3y+(1−x2)=c(1−x2)1/43\sqrt{y} + (1-x^2) = c(1-x^2)^{1/4}

B

32y+(1−x2)=c(1−x2)3/2\frac{3}{2}\sqrt{y} + (1-x^2) = c(1-x^2)^{3/2}

C

3y−(1−x2)=c(1−x2)3/23\sqrt{y} - (1-x^2) = c(1-x^2)^{3/2}

D

3y−(1−x2)=c(1−x2)1/43\sqrt{y} - (1-x^2) = c(1-x^2)^{1/4}

Answer

3y+(1−x2)=c(1−x2)1/43\sqrt{y} + (1-x^2) = c(1-x^2)^{1/4}

Explanation

Solution

The given differential equation is a Bernoulli equation. We transform it into a linear first-order differential equation using the substitution v=yv = \sqrt{y}. After finding the integrating factor and solving the linear equation for vv, we substitute back v=yv=\sqrt{y} and rearrange the terms to match the given options.