Question

$x^p{y}^q=(x+y{)}^{p+q}$.

Find $y^1$.

Answer 1

Consider the given function:

$x^p{y}^q={(x+y)}^{p+q}$

$p{x}^{p-q}{y}^q+x^{p}q{y}^{q-p}{y}^1=(p+q){(x+y)}^{(p+q)}(1+y^1)$

$p{x}^{p-1}{y}^q-(p+q){(x+y)}^{p+q-1}=[(p+q){(x+y)}^{p+q+1}-x^{p}q{y}^{q-1}]y^1$

$y^1=\frac{p}{x}\frac{{(x+y)}^{p+q}(p+q){(x+y)}^{p+q-1}}{(p+q){(x+y)}^{p+q-1}}-\frac{q}{y}{(x+y)}^{p+q}$

$y^1=\frac{\frac{p}{x}(x+y)-(p+q)}{(p+q)-\frac{q}{y}(x+y)}$

$y^1=\frac{p+\frac{py}{x}-p-q}{p+q-\frac{qx}{y}-q}$

$y^1=\frac{(py-qx)y}{(py-qx)x}$

$y^1=\frac{y}{x}$

Hence this is the answer.