Question

find the partial differential equation of z=1/x *(f(x-ay)+f(x+ay))
where a is an arbitrary constant..

Answer 1

here z=f(x,y)....i.e;
z is a function in x and y..

Answer 2

you mean\[z=\frac{1}{x}(f(x-ay)+f(x+ay))\] ?

Answer 3

yup..

Answer 4

answer......plz..:(

Answer 5

\[z_{x}=−\frac{1}{x^2}(f(x−ay)+f(x+ay))+\frac{1}{x}(f'(x−ay)+f'(x+ay))\]

Answer 6

\[z_{y}=\frac{1}{x}(-af'(x−ay)+af'(x+ay))\]

Answer 7

k...next?

Answer 8

\[z_{xx}=\frac{2}{x^3}(f'(x−ay)+f'(x+ay))-\frac{1}{x^2}(f''(x−ay)+f''(x+ay))+z_{x}\]

Answer 9

\[z_{yy}=\frac{a^2}{x}(f''(x−ay)+f''(x+ay))\]

Answer 10

k..

Answer 11

\[f''(x−ay)+f''(x+ay)=\frac{1}{a^2} x z_{yy}\]

Answer 12

hmn...i got it....thanq very much...@mukushla ...i can do it now...

Answer 13

\[f'(x−ay)+f'(x+ay)=xz_{x}+z\]

Answer 14

ok