Question

@hartnn

Answer 1

let z be a homogeneous function of x and y of order n so we can write--
\[\large z=x^n*f(\frac{ y }{ x })\]

Answer 2

right

Answer 3

now diffrentiating in partially respect to x
\[\frac{ \partial z }{ \partial x }=nx^{n-1}f(\frac{ y }{ x })+x^n*f'(\frac{ y }{ x })*\frac{ -y }{ x^2 }\\~~~~~~~~~=nx^{n-1}f(\frac{ y }{ x })-x^{n-2}y*f'(\frac{ y }{ x })\]

Answer 4

ok?

Answer 5

in same way i would find other terms ?

da z/da y , da^2 z/da x^2 , da^2z/da y^2 and others ?

Answer 6

da = \(\partial\)

Answer 7

now multiply this eq. by x
\[x \frac{ \partial z }{ \partial x }=nx^n*f(\frac{ y }{ x })-x^{n-1}y*f'(\frac{ y }{ x })\]

Answer 8

ok, pretty much same thing for all terms right ?

Answer 9

just mention steps for
x^2 (∂^2 u)/(∂x^2 )

i guess other steps will be similar

Answer 10

yeah
now in the same way i have to find
\[y*\frac{ \partial z }{ \partial y }\]
then just add them up
then u'll get
\[x \frac{ \partial z }{ \partial x }+y \frac{ \partial z }{ \partial y }=nz -----(1)\\so ~~for ~~x^2\frac{ \partial^2u }{ \partial x^2 }~~u ~~just~~have~~\to ~~diffrentiate (1)~w.r.t ~~x \\then ~~multi[ply~~x~~with ~~i~t.\]

Answer 11

so
find the value of
\(x^2 (∂^2 u)/(∂x^2 ) +2xy (∂^2 u)/∂x∂y+y^2 (∂^2 u)/(∂y^2 )+x ∂u/∂x+y ∂u/∂y\)
for
\(u=e^{(x+y)}+\log (x^3+y^3-x^2 y-xy^2)\)


from the proof, i got \(\Large n^2 u\)
here, n = 3, right ?

but what about e^(x+y) term ?
will the final answer be still "9u" ??

Answer 12

also for the log term it'll be 3 log x + f(y/x) ...