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Miscellaneous · Q30

Q.Solve the differential equation (x2−y2) dx+2xy dy=0(x^2 - y^2)\,dx + 2xy\,dy = 0.

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(x2−y2) dx+2xy dy=0  ⟹  dydx=y2−x22xy=v2−12v(x^2-y^2)\,dx+2xy\,dy=0 \implies \dfrac{dy}{dx}=\dfrac{y^2-x^2}{2xy}=\dfrac{v^2-1}{2v} (with y=vxy=vx), homogeneous.

v+xdvdx=v2−12v  ⟹  xdvdx=v2−1−2v22v=−(v2+1)2v.v+x\frac{dv}{dx}=\frac{v^2-1}{2v} \implies x\frac{dv}{dx}=\frac{v^2-1-2v^2}{2v}=\frac{-(v^2+1)}{2v}.

Separate: 2vv2+1 dv=−dxx\dfrac{2v}{v^2+1}\,dv=-\dfrac{dx}{x}. Integrating: ln⁡(v2+1)=−ln⁡∣x∣+C1  ⟹  ln⁡[(v2+1)x]=C1  ⟹  (v2+1)x=K\ln(v^2+1)=-\ln|x|+C_1 \implies \ln[(v^2+1)x]=C_1 \implies (v^2+1)x=K.

Substituting v=y/xv=y/x: (y2x2+1)x=K  ⟹  y2+x2x=K  ⟹  x2+y2=Kx\left(\dfrac{y^2}{x^2}+1\right)x=K \implies \dfrac{y^2+x^2}{x}=K \implies x^2+y^2=Kx, i.e. x2+y2=Cxx^2+y^2=Cx (renaming KK as CC). …

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