Skip to content
Worked Examples · Example 6

Q.Solve the homogeneous differential equation dydx=x2+y22xy\displaystyle\frac{dy}{dx} = \frac{x^{2} + y^{2}}{2xy}.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
33% · 13/40 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The equation is homogeneous (right side a function of y/xy/x). Put y=vxy = vx, so dydx=v+xdvdx\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}:

v+xdvdx=x2+v2x22x⋅vx=1+v22v.v + x\frac{dv}{dx} = \frac{x^{2} + v^{2}x^{2}}{2x\cdot vx} = \frac{1 + v^{2}}{2v}.

Isolate the derivative term:

xdvdx=1+v22v−v=1+v2−2v22v=1−v22v.x\frac{dv}{dx} = \frac{1 + v^{2}}{2v} - v = \frac{1 + v^{2} - 2v^{2}}{2v} = \frac{1 - v^{2}}{2v}.

Separate and integrate:

2v1−v2 dv=dxx⟹−ln⁡∣1−v2∣=ln⁡∣x∣+c1.\frac{2v}{1 - v^{2}}\,dv = \frac{dx}{x} \quad\Longrightarrow\quad -\ln|1 - v^{2}| = \ln|x| + c_1.

So ln⁡∣1−v2∣+ln⁡∣x∣=−c1\ln|1 - v^{2}| + \ln|x| = -c_1, giving x(1−v2)=kx(1 - v^{2}) = k (with k=±e−c1k = \pm e^{-c_1}).

Back-substitute v=y/xv = y/x:

x(1−y2x2)=k  ⟹  x2−y2x=k  ⟹  x2−y2=kx.x\left(1 - \frac{y^{2}}{x^{2}}\right) = k \;\Longrightarrow\; \frac{x^{2} - y^{2}}{x} = k \;\Longrightarrow\; x^{2} - y^{2} = kx. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.