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Question 173 of 177

Q.Solve the differential equation: xdydx=x⋅tan⁡(yx)+yx\dfrac{dy}{dx}=x\cdot\tan\left(\dfrac{y}{x}\right)+y.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2025Subjective· 3mImportance★★★★★
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This is a homogeneous differential equation; substitute y=vxy=vx.

xdydx=xtan⁡(yx)+y  ⟹  dydx=tan⁡(yx)+yxx\frac{dy}{dx}=x\tan\left(\frac yx\right)+y \implies \frac{dy}{dx}=\tan\left(\frac yx\right)+\frac yx

This is homogeneous (RHS is a function of y/xy/x alone). Let y=vxy=vx, so dydx=v+xdvdx\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.

v+xdvdx=tan⁡v+v  ⟹  xdvdx=tan⁡vv+x\frac{dv}{dx}=\tan v+v \implies x\frac{dv}{dx}=\tan v

Separate variables:

dvtan⁡v=dxx  ⟹  ∫cot⁡v dv=∫dxx\frac{dv}{\tan v}=\frac{dx}{x} \implies \int\cot v\,dv=\int\frac{dx}{x} …

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