Question

pz - qz=z^2 + ( x + y)^2
solve the partial differential equation
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Answer 1

f(x+y) = [z^2 + (x+y)^2] e^2y

Answer 2

Answer:

$x+y=C$

$a=(z^2+(x+y)^2)^{\frac{1}{2}}e^y$

Step-by-step explanation:

We are given that

$pz-qz=z^2+(x+y)^2$

We have to solve partial differential equation.

Formula of solving Pp+Qq=R

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

By using this

Then , we get

$\frac{dx}{z}=\frac{dy}{-z}=\frac{dz}{z^2+(x+y)^2}$

Taking first two equations

$\frac{dx}{z}=\frac{dy}{-z}$

$dx=-dy$

Integration on both sides then we get

$x+y=C$

Now taking last two equations

$\frac{dy}{-z}=\frac{dz}{z^2+C}$

$dy=-\frac{zdz}{z^2+C}$

Integrate on both sides

$y=-\frac{1}{2}ln(z^2+C)+ln a$

$y=ln\frac{a}{\sqrt{z^2+C}}$

$e^y=\frac{a}{\sqrt{z^2+C}}$

$a=(z^2+C)^{\frac{1}{2}}e^y$

$a=(z^2+(x+y)^2)^{\frac{1}{2}}e^y$