The general solution of the differential equa | vedclass

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Question

(D) Given the differential equation: $\frac{1}{x} \frac{dy}{dx} = 	an^{-1} x$ Separating the variables,we get: $dy = x 	an^{-1} x dx$ Integrating bo

Vedclass · Accepted Answer

$y = \frac{x^2 	an^{-1} x}{2} - \frac{1}{2}(x - 	an^{-1} x) + c$,where $c$ is a constant of integration.

Vedclass · Answer

(D) Given the differential equation: $\frac{1}{x} \frac{dy}{dx} = 	an^{-1} x$ Separating the variables,we get: $dy = x 	an^{-1} x dx$ Integrating both sides: $y = \int x 	an^{-1} x dx$ Using integration by parts,$\int u dv = uv - \int v du$,where $u = 	an^{-1} x$ and $dv = x dx$: $y = 	an^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{1+x^2} \cdot \frac{x^2}{2} dx$ $y = \frac{x^2 	an^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{1+x^2} dx$ Adding and subtracting $1$ in the numerator: $y = \frac{x^2 	an^{-1} x}{2} - \frac{1}{2} \int \frac{x^2+1-1}{1+x^2} dx$ $y = \frac{x^2 	an^{-1} x}{2} - \frac{1}{2} \left( \int 1 dx - \int \frac{1}{1+x^2} dx ight)$ $y = \frac{x^2 	an^{-1} x}{2} - \frac{1}{2} (x - 	an^{-1} x) + c$

The general solution of the differential equation $\frac{1}{x} \frac{dy}{dx} = \tan^{-1} x$ is

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