Question

If $x^m{y}^n=(x+y{)}^{m+n}$. Prove that $\frac{d^{2}y}{d{x}^2}=0$.

Answer 1

Given: $x^m{y}^n=(x+y{)}^{m+n}$

Take $\log$ on both side

$\log x^m+\log y^n+(m+n)\log =y(x+y)$

$m\log x+n\log y=(m+n)\log (x+y)$

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

$m.\frac{1}{x}+n\frac{1}{y}\frac{dy}{dx}=(m+n).\frac{1}{(x+y)}(1+\frac{dy}{dx})$

$\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=\frac{m+n}{x+y}(1+\frac{dy}{dx})$

$\frac{dy}{dx}(\frac{n}{y}-\frac{m+n}{x+y})=\frac{m+n}{x+y}-\frac{m}{x}$

$\frac{dy}{dx}\left(\frac{n(x+y)-y(m+n)}{y(x+y)}\right)=\frac{x(m+n)-m(x+y)}{x(x+y)}$

On simplifying we get,

$\frac{dy}{dx}\left(\frac{nx-my}{y}\right)=\frac{nx-my}{x}$

$\therefore \frac{dy}{dx}=\frac{y}{x}$

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

Apply quotient rule

$\frac{d^{2}y}{d{x}^2}=\frac{x.\frac{dy}{dx}-y.1}{x^2}$

$\Rightarrow \frac{x.\frac{dy}{dx}-y}{x^2}$

This can be written as

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

But $\frac{dy}{dx}=\frac{y}{x}$

So substituting this we get,

$\frac{y}{x^2}-\frac{y}{x^2}=0$

Hence proved