Question

Pls help me guys, I an stuck here.

Answer 1

can anyone find the n?

Answer 2

for df/dx ,treat y as constant

Answer 3

imranmeah91, i did that , I wanna solve the 2nd part to find n?

Answer 4

alright, did you find Df/dy yet?

Answer 5

yes, i find both

Answer 6

Oh, Finally.. I got n=-3/2

Answer 7

\[\frac{\text{Df}}{\text{Dy}}=\frac{1}{4} e^{-\frac{x^2}{4 y}} x^2 y^{-2+n}+e^{-\frac{x^2}{4 y}} n y^{-1+n}\]

\[\frac{\text{Df}}{\text{Dx}}=-\frac{1}{2} e^{-\frac{x^2}{4 y}} x y^{-1+n}\]

\[x^2\frac{\text{Df}}{\text{Dx}}=-\frac{1}{2} e^{-\frac{x^2}{4 y}} x^3 y^{-1+n}\]

\[\frac{1}{x^2}\frac{D}{\text{Dx}}\left[-\frac{1}{2} e^{-\frac{x^2}{4 y}} x^3 y^{-1+n}\right]=\frac{1}{4} e^{-\frac{x^2}{4 y}} x^2 y^{-2+n}-\frac{3}{2} e^{-\frac{x^2}{4 y}}\text{ }y^{-1+n}\]

\[\frac{1}{4} e^{-\frac{x^2}{4 y}} x^2 y^{-2+n}-\frac{3}{2} e^{-\frac{x^2}{4 y}}\text{ }y^{-1+n}=\frac{1}{4} e^{-\frac{x^2}{4 y}} x^2 y^{-2+n}+e^{-\frac{x^2}{4 y}} n y^{-1+n}\]