The Solution Of Differential Equation (x^2 + Y^2)dy = Xy Dx Is Y ... - Toppr

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SolveGuidesJoin / LoginUse appLogin0You visited us 0 times! Enjoying our articles? Unlock Full Access!Standard XIIMathematicsSolving Homogeneous Differential EquationsQuestionThe solution of differential equation (x2+y2)dy=xydx is y=y(x). Ify(1)=1 and y(x0)=e, then x0 is:
  1. √2(e2−1)
  2. √2(e2+1)
  3. √3e
  4. √e2+12
A√3eB√e2+12C√2(e2+1)D√2(e2−1)Open in AppSolutionVerified by Toppr

We have, (x2+y2)dy=xydx⇒dydx=xyx2+y2Substitute y=vx⇒dydx=v+xdvdx⇒v+xdvdx=v1+v2⇒xdvdx=v1+v2−v=−v31+v2⇒1+v2v3dv=−dxxIntegrating both sides we get, −12v2+logv=−logx+logc, where c is constant ⇒−12v2+log(vxc)=0⇒y=cex2/2y2Now using y(1)=1⇒c=e−1/2Also y(x0)=e⇒e=e−1/2+x20/2e2⇒1=−1/2+x20/2e2⇒x20=3e2⇒x0=√3e

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