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Question 177 of 177

Q.Solve the differential equation x2⋅dydx=x2+xy+y2x^2\cdot\dfrac{dy}{dx}=x^2+xy+y^2.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2026Subjective· 4mImportance★★★★★
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This is a homogeneous D.E.; substitute y=vxy=vx to separate variables.

x2dydx=x2+xy+y2  ⟹  dydx=1+yx+(yx)2x^2\frac{dy}{dx}=x^2+xy+y^2 \implies \frac{dy}{dx}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2

This is homogeneous of degree 00 on the right. Let y=vxy=vx, so dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.

v+xdvdx=1+v+v2v+x\frac{dv}{dx}=1+v+v^2

xdvdx=1+v2x\frac{dv}{dx}=1+v^2

Separating variables:

dv1+v2=dxx\frac{dv}{1+v^2}=\frac{dx}{x}

Integrating both sides: …

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