Question

if y= x^x . prove that d^2y/dx^2 - 1/y (dy/dx)^2 - y/x = 0

Answer 1

and d^2y/dx^2-1/y(dy/dx)^2-y/x=0

Answer 2

Answer:

Step-by-step explanation:

$y=x^{x}$

Taking log on both sides,

$logy=xlogx$

Differentiating with respect to x, we have

$\frac{1}{y}(\frac{dy}{dx})=x{\times}\frac{1}{x}+logx$

$\frac{dy}{dx}=y+ylogx$

$\frac{dy}{dx}=y(1+logx)$

Again differentiating with respect to x,

$\frac{d^{2}y}{dx^{2}}=y{\times}\frac{1}{x}+(1+logx)\frac{dy}{dx}$

$\frac{d^{2}y}{dx^{2}}=\frac{y}{x}+y(1+logx)^{2}$

Now, the given equation is:

$\frac{d^{2}y}{dx^{2}}-\frac{1}{y}(\frac{dy}{dx})^{2}-\frac{y}{x}$

=$\frac{y}{x}+y(1+logx)^{2}- y(1+logx)^{2}-\frac{y}{x}$

=$0$