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Exercise: Homogeneous Differential Eq... · Q22

Q.Solve the differential equation (x−y) dy=(x+y) dx(x-y)\,dy = (x+y)\,dx.

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(x−y) dy=(x+y) dx  ⟹  dydx=x+yx−y=1+v1−v(x-y)\,dy=(x+y)\,dx \implies \dfrac{dy}{dx}=\dfrac{x+y}{x-y}=\dfrac{1+v}{1-v} (with y=vxy=vx), homogeneous.

v+xdvdx=1+v1−v  ⟹  xdvdx=(1+v)−v(1−v)1−v=1+v21−v.v+x\frac{dv}{dx}=\frac{1+v}{1-v} \implies x\frac{dv}{dx}=\frac{(1+v)-v(1-v)}{1-v}=\frac{1+v^2}{1-v}.

Separate: 1−v1+v2 dv=dxx\dfrac{1-v}{1+v^2}\,dv=\dfrac{dx}{x}, i.e. (11+v2−v1+v2)dv=dxx\left(\dfrac1{1+v^2}-\dfrac{v}{1+v^2}\right)dv=\dfrac{dx}{x}. Integrating: arctan⁡v−12ln⁡(1+v2)=ln⁡∣x∣+C\arctan v - \dfrac12\ln(1+v^2) = \ln|x|+C. Substituting v=y/xv=y/x and combining the logarithmic terms −12ln⁡(1+y2/x2)+ln⁡∣x∣-\tfrac12\ln(1+y^2/x^2)+\ln|x| into a single −12ln⁡(x2+y2)-\tfrac12\ln(x^2+y^2):

arctan⁡ ⁣(yx)−12ln⁡(x2+y2)=C.\arctan\!\left(\frac{y}{x}\right) - \frac12\ln(x^2+y^2) = C. …

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