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(C) Given differential equation is $x y \frac{dy}{dx} = x^2 + 2y^2$.Dividing by $xy$,we get $\frac{dy}{dx} = \frac{x^2+2y^2}{xy} = \frac{x}{y} + \frac

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

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(C) Given differential equation is $x y \frac{dy}{dx} = x^2 + 2y^2$.Dividing by $xy$,we get $\frac{dy}{dx} = \frac{x^2+2y^2}{xy} = \frac{x}{y} + \frac{2y}{x} \dots(i)$This is a homogeneous differential equation. Put $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx} \dots(ii)$Substituting $(ii)$ in $(i)$,we get $v + x \frac{dv}{dx} = \frac{1}{v} + 2v$.Subtracting $v$ from both sides,$x \frac{dv}{dx} = \frac{1}{v} + v = \frac{1+v^2}{v}$.Separating variables,$\frac{v}{1+v^2} dv = \frac{1}{x} dx$.Integrating both sides,$\frac{1}{2} \ln(1+v^2) = \ln|x| + C$.Multiplying by $2$,$\ln(1+v^2) = 2\ln|x| + 2C = \ln(x^2) + K$.Thus,$1+v^2 = c x^2$.Substituting $v = \frac{y}{x}$,we get $1 + \frac{y^2}{x^2} = c x^2$,which simplifies to $x^2 + y^2 = c x^4$.Given $y(1) = 0$,we have $1^2 + 0^2 = c(1)^4$,so $c = 1$.Therefore,the particular solution is $x^2 + y^2 = x^4$.

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

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