If the integrating factor of x(1 - x^2)dy + ( | vedclass

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(D) Given the differential equation: $x(1 - x^2)dy + (2x^2y - y - ax^3)dx = 0$. Divide the entire equation by $x(1 - x^2)dx$: $\frac{dy}{dx} + \frac{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the differential equation: $x(1 - x^2)dy + (2x^2y - y - ax^3)dx = 0$. Divide the entire equation by $x(1 - x^2)dx$: $\frac{dy}{dx} + \frac{2x^2y - y - ax^3}{x(1 - x^2)} = 0$. Rearranging the terms to fit the standard linear form $\frac{dy}{dx} + Py = Q$: $\frac{dy}{dx} + \frac{(2x^2 - 1)y - ax^3}{x(1 - x^2)} = 0$. $\frac{dy}{dx} + \frac{2x^2 - 1}{x(1 - x^2)}y = \frac{ax^3}{x(1 - x^2)}$. Comparing this with the standard form $\frac{dy}{dx} + Py = Q$,we identify $P = \frac{2x^2 - 1}{x(1 - x^2)}$.

If the integrating factor of $x(1 - x^2)dy + (2x^2y - y - ax^3)dx = 0$ is $e^{\int Pdx}$,then $P$ is equal to

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