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Question

(B) Given differential equation: $(1+y^2)(1+\log x) dx + x dy = 0$ Rearranging the terms: $(1+y^2)(1+\log x) dx = -x dy$ Separating the variables: $\f

Vedclass · Accepted Answer

$\log x + \frac{1}{2}(\log x)^2 + 	an^{-1} y = \frac{\pi}{4}$

Vedclass · Answer

(B) Given differential equation: $(1+y^2)(1+\log x) dx + x dy = 0$ Rearranging the terms: $(1+y^2)(1+\log x) dx = -x dy$ Separating the variables: $\frac{1+\log x}{x} dx = -\frac{1}{1+y^2} dy$ Integrating both sides: $\int \frac{1+\log x}{x} dx = -\int \frac{1}{1+y^2} dy$ Let $1+\log x = t$,then $\frac{1}{x} dx = dt$. Substituting this: $\int t dt = -	an^{-1} y + C$ $\frac{t^2}{2} = -	an^{-1} y + C$ $\frac{(1+\log x)^2}{2} = -	an^{-1} y + C$ At $x=1, y=1$: $\frac{(1+\log 1)^2}{2} = -	an^{-1}(1) + C$ $\frac{1}{2} = -\frac{\pi}{4} + C \implies C = \frac{1}{2} + \frac{\pi}{4}$ Substituting $C$ back: $\frac{(1+\log x)^2}{2} = -	an^{-1} y + \frac{1}{2} + \frac{\pi}{4}$ $\frac{1 + 2\log x + (\log x)^2}{2} = -	an^{-1} y + \frac{1}{2} + \frac{\pi}{4}$ $\frac{1}{2} + \log x + \frac{(\log x)^2}{2} = -	an^{-1} y + \frac{1}{2} + \frac{\pi}{4}$ $\log x + \frac{(\log x)^2}{2} + 	an^{-1} y = \frac{\pi}{4}$

The particular solution of the differential equation $(1+y^2)(1+\log x) dx + x dy = 0$ at $x=1, y=1$ is

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