Skip to content
Exercise 10.4 · Q7

Q.Show that the differential equation representing the family of curves y2=2a(x+a2/3)y^2=2a\left(x+a^{2/3}\right), where aa is a positive parameter, is (y2−2xydydx)3=8(ydydx)5\left(y^2-2xy\dfrac{dy}{dx}\right)^3=8\left(y\dfrac{dy}{dx}\right)^5.

Puducherry TnboardTextbookSubjectiveImportance★★★★★
14% · 18/126 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 →

One differentiation gives aa directly in terms of y,y′y,y' (since only one constant aa needs eliminating); substituting back and cubing reproduces exactly the target differential equation.

Step 1. Differentiate y2=2a(x+a2/3)=2ax+2a5/3y^2=2a\left(x+a^{2/3}\right)=2ax+2a^{5/3} with respect to xx. 2ydydx=2a ⟹ a=ydydx2y\dfrac{dy}{dx}=2a\ \Longrightarrow\ a=y\dfrac{dy}{dx}.

Step 2. Substitute a=ydydxa=y\dfrac{dy}{dx} back into the original equation. y2=2(ydydx)x+2(ydydx)5/3 ⟹ y2−2xydydx=2(ydydx)5/3y^2=2\left(y\dfrac{dy}{dx}\right)x+2\left(y\dfrac{dy}{dx}\right)^{5/3}\ \Longrightarrow\ y^2-2xy\dfrac{dy}{dx}=2\left(y\dfrac{dy}{dx}\right)^{5/3}. …

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.