$x^p y^q=(x+y)^{p+q}$ what is the value of $y_2$

Question

If $x^p y^q=(x+y)^{p+q}$ then how to compute ${d^2 y}\over{dx^2}$

  The result is 0 

Answer 1

The change of variable $y=xz$ yields $z^q=(1+z)^{p+q}$ hence $z$ is constant, $y'=z$ and $y''=0$.

Answer 2

Applying log wrt $e,p\ln x+q\ln y=(p+q)\ln(x+y)$

Differentiating wrt $x$ we get $$\frac px+\frac qy\cdot y_1=\frac{(p+q)}{x+y}\left(1+ y_1\right)$$

$$ y_1\left(\frac qy-\frac{p+q}{x+y}\right)=\frac{p+q}{x+y}-\frac px$$

$$\implies y_1(qx-py)x=(qx-py)y \text{ (assuming }xy(x+y)\ne0)$$

$$\implies x\cdot y_1=y\text{ (assuming }qx-py\ne0)$$

Differentiating wrt $x,$ $$xy_2+y_1=y_1\text{ and }\cdots $$

If $qx-py=0, q-py_1=0\implies y_1=\frac qp \implies y_2=?$