The solution of the differential equation x^2 | vedclass

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Question

(C) Given the differential equation: $x^2(y+1) \frac{dy}{dx} + y^2(x+1)^2 = 0$. Rearranging the terms to separate the variables: $\frac{y+1}{y^2} dy =

Vedclass · Accepted Answer

$\log |\frac{1}{2} x^2 y| = \frac{1}{x} + \frac{1}{y} - x - \frac{1}{2}$

Vedclass · Answer

(C) Given the differential equation: $x^2(y+1) \frac{dy}{dx} + y^2(x+1)^2 = 0$. Rearranging the terms to separate the variables: $\frac{y+1}{y^2} dy = -\frac{(x+1)^2}{x^2} dx$. Integrating both sides: $\int (\frac{1}{y} + \frac{1}{y^2}) dy = -\int (\frac{x^2 + 2x + 1}{x^2}) dx$. $\int (\frac{1}{y} + y^{-2}) dy = -\int (1 + \frac{2}{x} + \frac{1}{x^2}) dx$. $\log |y| - \frac{1}{y} = -(x + 2 \log |x| - \frac{1}{x}) + C$. $\log |y| - \frac{1}{y} = -x - 2 \log |x| + \frac{1}{x} + C$. $\log |y| + 2 \log |x| = \frac{1}{x} + \frac{1}{y} - x + C$. $\log |x^2 y| = \frac{1}{x} + \frac{1}{y} - x + C$. Given $y(1) = 2$,substitute $x=1$ and $y=2$: $\log |1^2 	imes 2| = \frac{1}{1} + \frac{1}{2} - 1 + C$. $\log 2 = 1 + 0.5 - 1 + C \implies \log 2 = 0.5 + C \implies C = \log 2 - 0.5$. Substituting $C$ back: $\log |x^2 y| = \frac{1}{x} + \frac{1}{y} - x + \log 2 - 0.5$. $\log |x^2 y| - \log 2 = \frac{1}{x} + \frac{1}{y} - x - 0.5$. $\log |\frac{x^2 y}{2}| = \frac{1}{x} + \frac{1}{y} - x - 0.5$. This matches option $C$.

The solution of the differential equation $x^2(y+1) \frac{dy}{dx} + y^2(x+1)^2 = 0$,given $y(1) = 2$,is

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