Question

prove that if y³-3ax²+x³=0,d²y/dx²+2a²x²/y²y³​

Answer 1

Question:

If $\displaystyle{\sf\:y^3\:-\:3ax^2\:+\:x^3\:=\:0}$, prove that $\displaystyle{\sf\:\dfrac{d^2y}{dx^2}\:+\:\dfrac{2a^2x^2}{y^2\:y^3}\:=\:0}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{d^2y}{dx^2}\:+\:\dfrac{2a^2x^2}{y^2\:y^3}\:=\:0\:}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:y^3\:-\:3ax^2\:+\:x^3\:=\:0}$

We have to prove that,

$\displaystyle{\sf\:\dfrac{d^2y}{dx^2}\:+\:\dfrac{2a^2x^2}{y^2\:y^3}\:=\:0}$

Now,

$\displaystyle{\sf\:y^3\:-\:3ax^2\:+\:x^3\:=\:0}$

Differentiating both sides w.r.t x, we get,

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:(\:y^3\:-\:3ax^2\:+\:x^3\:)\:=\:\dfrac{d}{dx}\:(\:0\:)}$

$\displaystyle{\implies\sf\:\dfrac{d}{dx}\:(\:y^3\:)\:-\:\dfrac{d}{dx}\:(\:3a\:x^2\:)\:+\:\dfrac{d}{dx}\:(\:x^3\:)\:=\:0}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{du}{dv}\:=\:\dfrac{du}{dx}\:\times\:\dfrac{dx}{dv}\:}}}$

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d}{dx}\:(\:ky\:)\:=\:k\:\dfrac{d}{dx}\:(\:y\:)\:}}}$

$\displaystyle{\boxed{\green{\sf\:\dfrac{d}{dx}\:(\:x^n\:)\:=\:n\:x^{n\:-\:1}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{d}{dy}\:(\:y^3\:)\:\times\:\dfrac{d}{dx}\:(\:y\:)\:-\:3a\:\dfrac{d}{dx}\:(\:x^2\:)\:+\:3\:x^{3\:-\:1}\:=\:0}$

$\displaystyle{\implies\sf\:3\:y^{3\:-\:1}\:\times\:\dfrac{dy}{dx}\:-\:3a\:\times\:2\:x^{2\:-\:1}\:+\:3x^2\:=\:0}$

$\displaystyle{\implies\sf\:3y^2\:\dfrac{dy}{dx}\:-\:6ax\:+\:3x^2\:=\:0}$

$\displaystyle{\implies\sf\:3y^2\:\dfrac{dy}{dx}\:=\:-\:3x^2\:+\:6ax}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{-\:3x^2\:+\:6ax}{3y^2}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{-\:\cancel{3}\:x^2}{\cancel{3}\:y^2}\:+\:\dfrac{\cancel{6}\:ax}{\cancel{3}\:y^2}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{-\:x^2}{y^2}\:+\:\dfrac{2ax}{y^2}}$

$\displaystyle{\implies\:\boxed{\blue{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{-\:x^2\:+\:2ax}{y^2}}}}$

Differentiating both sides w.r.t x, we get,

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{d}{dx}\:\left(\:\dfrac{-\:x^2\:+\:2ax}{y^2}\:\right)}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d}{dx}\:\left(\:\dfrac{u}{v}\:\right)\:=\:\dfrac{u'v\:-\:uv'}{v^2}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{\dfrac{d}{dx}\:(\:-\:x^2\:+\:2ax\:)\:y^2\:-\:(\:-\:x^2\:+\:2ax\:)\:\dfrac{d}{dx}\:(\:y^2\:)}{(\:y^2\:)^2}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{\left(\:\dfrac{d}{dx}\:(\:-\:x^2\:)\:+\:\dfrac{d}{dx}\:(\:2ax\:)\:\right)\:y^2\:-\:(\:-\:x^2\:+\:2ax\:)\:\left(\:\dfrac{d}{dy}\:(\:y^2\:)\:\times\:\dfrac{d}{dx}\:(\:y\:)\:\right)}{y^4}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{(\:-\:2x\:+\:2a\:)\:y^2\:-\:(\:x^2\:+\:2ax\:)\:\left(\:2y\:\dfrac{dy}{dx}\:\right)}{y^4}}$

By substituting the value of $\displaystyle{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{-\:x^2\:+\:2ax}{y^2}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{(\:-\:2x\:+\:2a\:)\:y^2\:-\:(\:-\:x^2\:+\:2ax\:)\:\left[\:2\:\cancel{y}\:\times\:\left(\:\dfrac{-\:x^2\:+\:2ax}{y^{\cancel{2}}}\:\right)\:\right]}{y^4}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{\dfrac{(\:-\:2x\:+\:2a\:)\:y^3\:-\:2\:(\:-\:x^2\:+\:2ax\:)\:(\:-\:x^2\:+\:2ax\:)}{y}}{y^4}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{(\:-\:2x\:+\:2a\:)\:y^3\:-\:2\:(\:-\:x^2\:+\:2ax\:)^2}{y^5}}$

Now,

$\displaystyle{\sf\:y^3\:-\:3ax^2\:+\:x^3\:=\:0}$

$\displaystyle{\implies\sf\:y^3\:=\:3ax^2\:-\:x^3}$

By using this value,

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{(\:-\:2x\:+\:2a\:)\:(\:3ax^2\:-\:x^3\:)\:-\:2\:(\:x^4\:+\:4a^2x^2\:-\:4ax^3\:)}{y^5}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{(\:-\:6ax^3\:+\:2x^4\:+\:6a^2x^2\:-\:2ax^3\:)\:-\:(\:2x^4\:+\:8a^2x^2\:-\:8ax^3\:)}{y^5}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{-\:6ax^3\:+\:\cancel{2x^4}\:+\:6a^2x^2\:-\:2ax^3\:-\:\cancel{2x^4}\:-\:8a^2x^2\:+\:8ax^3}{y^5}}$

$\displaystyle{\implies\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{-\:\cancel{6ax^3}\:+\:6a^2x^2\:-\:8a^2x^2\:+\:\cancel{6ax^3}}{y^5}}$

$\displaystyle{\implies\:\boxed{\pink{\sf\:\dfrac{d^2y}{dx^2}\:=\:\dfrac{-\:2a^2x^2}{y^5}}}}$

Now,

$\displaystyle{\sf\:\dfrac{d^2y}{dx^2}\:+\:\dfrac{2a^2x^2}{y^2\:y^3}\:=\:0}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{d^2y}{dx^2}\:+\:\dfrac{2a^2x^2}{y^2\:y^3}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{-\:2a^2x^2}{y^5}\:+\:\dfrac{2a^2x^2}{y^5}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{-\:2a^2x^2\:+\:2a^2x^2}{y^5}}$

$\displaystyle{\implies\sf\:LHS\:=\:0}$

$\displaystyle{\sf\:RHS\:=\:0}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:LHS\:=\:RHS\:}}}}$

Hence proved!