Question

If y= x^3log1/x prove that x×dy/dx + x^3 =3y​

Answer 1

Answer:

$y = x^3.\:log(\dfrac{1}{x})\\\\\\\text{ Differentiating w.r.t. x we get:}\\\\\\\implies \dfrac{dy}{dx} = 3x^2.\:log(\dfrac{1}{x}) + x^3.\:(x).\:(\dfrac{-1}{x^2})\\\\\\\implies \dfrac{dy}{dx} = 3x^2.\:log(\dfrac{1}{x}) + x^2.\:(-1)\\\\\\\implies \dfrac{dy}{dx} = 3x^2.\:log(\dfrac{1}{x}) - x^2$

To prove:

$x \times \dfrac{dy}{dx} + x^3 = 3y$

Substituting the value of (dy/dx) we get:

$\implies (x) \times [ 3x^2.\:log(\dfrac{1}{x}) - x^{2} ] + x^3\\\\\\\implies 3x^3.\:log(\dfrac{1}{x}) - x^3 + x^3\\\\\\\implies \boxed{\bf{3x^3.\:log(\dfrac{1}{x})}}$

This is the LHS.

Calculating RHS we get:

$\implies 3y = 3 \times [x^3.\:log(\dfrac{1}{x})]\\\\\\\implies \boxed{\bf{3x^3.\:log(\dfrac{1}{x})}} = LHS$

Hence LHS = RHS

Hence Proved.