Question

Please solve this differentiation.​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{2ay=x\bigg(b+a\dfrac{dy}{dx}\bigg)}$

$\sf{2ay=bx+ax\dfrac{dy}{dx}}$

$\sf{\implies\dfrac{dy}{dx}=\dfrac{2ay-bx}{ax}\,\,\,\,\,\,\,\,...(1)}$

Now, we have,

$\sf{bx+ax\dfrac{dy}{dx}=2ay}$

Differentiating both sides w.r.t. x,

$\sf{b+a\dfrac{dy}{dx}+ax\dfrac{d^2y}{dx^2}=2a\dfrac{dy}{dx}}$

$\sf{\implies\,b+ax\dfrac{d^2y}{dx^2}=a\dfrac{dy}{dx}}$

Again,

$\sf{b+ax\dfrac{d^2y}{dx^2}=a\dfrac{dy}{dx}}$

Differentiate both sides w.r.t x,

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

$\sf{\implies\,ax\dfrac{d^3y}{dx^3}=0}$

$\sf{\implies\,\dfrac{d^3y}{dx^3}=0}$