The solution of the differential equation y^{ | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}$.This is a homogeneous differential equation. Le

Vedclass · Accepted Answer

$y^2 = x^2 \log x^2 + 4x^2$

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(D) Given the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{1}{v} + v$.Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{1}{v}$.Separating variables: $v \, dv = \frac{1}{x} \, dx$.Integrating both sides: $\int v \, dv = \int \frac{1}{x} \, dx \Rightarrow \frac{v^2}{2} = \log |x| + C$.Substituting $v = \frac{y}{x}$: $\frac{y^2}{2x^2} = \log |x| + C \Rightarrow y^2 = 2x^2 \log |x| + 2x^2 C$.Since $\log |x| = \frac{1}{2} \log x^2$,we have $y^2 = x^2 \log x^2 + 2x^2 C$.Using the condition $y(1) = -2$: $(-2)^2 = (1)^2 \log(1)^2 + 2(1)^2 C \Rightarrow 4 = 0 + 2C \Rightarrow C = 2$.Substituting $C = 2$ into the general solution: $y^2 = x^2 \log x^2 + 4x^2$.

The solution of the differential equation $y^{\prime} = \frac{x^2 + y^2}{xy}$,with the initial condition $y(1) = -2$,is given by:

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