The solution of the differential equation ({x | vedclass

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Question

(A) The given differential equation is $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$.Rearranging the terms,we get $x^2(1 - y)\frac{dy}{dx}

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$\log \left( {\frac{x}{y}} \right) = \frac{1}{x} + \frac{1}{y} + c$

Vedclass · Answer

(A) The given differential equation is $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$.Rearranging the terms,we get $x^2(1 - y)\frac{dy}{dx} = -y^2(1 + x)$.Separating the variables,we have $\frac{1 - y}{y^2} dy = -\frac{1 + x}{x^2} dx$.This can be written as $(\frac{1}{y^2} - \frac{1}{y}) dy = -(\frac{1}{x^2} + \frac{1}{x}) dx$.Integrating both sides,$\int (y^{-2} - y^{-1}) dy = -\int (x^{-2} + x^{-1}) dx$.$-y^{-1} - \log|y| = -(-x^{-1} + \log|x|) + C$.$-\frac{1}{y} - \log|y| = \frac{1}{x} - \log|x| + C$.Rearranging the terms,$\log|x| - \log|y| = \frac{1}{x} + \frac{1}{y} + C$.Thus,$\log \left( \frac{x}{y} \right) = \frac{1}{x} + \frac{1}{y} + C$.

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

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