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Exercise: Homogeneous Differential Eq... · Q20

Q.Solve the differential equation dydx=y2xy−x2\dfrac{dy}{dx} = \dfrac{y^2}{xy - x^2}.

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dydx=y2xy−x2\dfrac{dy}{dx}=\dfrac{y^2}{xy-x^2}; both numerator and denominator are homogeneous of degree 22. Substitute y=vxy=vx: numerator =v2x2=v^2x^2, denominator =x(vx)−x2=x2(v−1)=x(vx)-x^2=x^2(v-1), so RHS =v2v−1=\dfrac{v^2}{v-1}.

v+xdvdx=v2v−1  ⟹  xdvdx=v2−v(v−1)v−1=vv−1.v+x\frac{dv}{dx}=\frac{v^2}{v-1} \implies x\frac{dv}{dx}=\frac{v^2-v(v-1)}{v-1}=\frac{v}{v-1}.

Separate: v−1v dv=dxx  ⟹  (1−1v)dv=dxx\dfrac{v-1}{v}\,dv=\dfrac{dx}{x} \implies \left(1-\dfrac1v\right)dv=\dfrac{dx}{x}. Integrating: v−ln⁡∣v∣=ln⁡∣x∣+C  ⟹  v=ln⁡∣vx∣+Cv-\ln|v| = \ln|x|+C \implies v = \ln|vx|+C (combining logs). Since vx=yvx=y: v=ln⁡∣y∣+Cv=\ln|y|+C, and v=y/xv=y/x, so

yx=ln⁡∣y∣+C.\frac{y}{x} = \ln|y| + C. …

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