Partial diff. equations problem (non-homogeneous)

[netcrush]

Hi everyone,
I need help in solving a Non-homogeneous first-order partial differential equations problem:

$\displaystyle z\frac{\partial{z}}{\partial{x}} -y\frac{\partial{z}}{\partial{y}} = z - y sin(y)$

I have, just a few steps and the final result included, but I don't get the same result when I try to solve it on my own. I tried again and again and I can't see what I am doing wrong. Please help.

This is the final result I should get:
$\displaystyle F \left(\frac{z}{y}, x(y+z) + cos(y+z)\right) = 0$

That means I have to get two functions (two "first integrals" that would be the solution of the problem). First function (and thus the "first first integral") is $\displaystyle \frac{z}{y} = C_{1}$ and the second (and also "second first integral") is $\displaystyle x(y+z) + cos(y+z)) = C_{2}$

My solving steps:
1. I add an "associated system in symmetrical form" (not sure if this is how it's called in English):

$\displaystyle \frac{dx}{z} = \frac{dy}{-y} = \frac{dz}{z - y sin(y)}$

2. Solving this system like any nonlinear system of differential equations, to get the "first integrals":
- I have used second part of the equation from step 1:
$\displaystyle \frac{dy}{-y} = \frac{dz}{z - y sin(y)}$

$\displaystyle \frac{z - y sin(y)}{-y} = \frac{dz}{dy}$

$\displaystyle \frac{dz}{dy} = \frac{z - y sin(y)}{-y}$

$\displaystyle \frac{dz}{dy} = -\frac{1}{y}z + sin(y)$

$\displaystyle z' + \frac{1}{y}z = sin(y)$

3. So I got one ordinary linear differential equation and I am solving it using a formula:

$\displaystyle z' + p(y)z = q(y)$
$\displaystyle z = e^{-\int p(y) dy}\cdot [C + \int q(y) \cdot e^{\int p(y) dy}dy]$

In my case $\displaystyle p(y) = \frac{1}{y}$ (thus $\displaystyle \int p(y)dy = \int \frac{dy}{y} = ln|y|$) and $\displaystyle q(y) = sin(y)$ (thus $\displaystyle \int q(y) \cdot e^{\int p(y) dy} = \int y sin(y) dy = sin(y) - y cos(y)$) Adding this in formula, I get:
$\displaystyle z = e^{-ln|y|}\cdot [C + sin(y) - y cos(y)]$
$\displaystyle z = \frac{1}{y} \cdot [C + sin(y) - y cos(y)]$

$\displaystyle z y - sin(y) + y cos(y)= C$

and this should be the first integral, but (as you can see in final result above) their first integral is

$\displaystyle \frac{z}{y} = C$

What am I doing wrong? And how did they get this first integral?
Thanks all.

 

[HallsofIvy]

Hi everyone,
I need help in solving a Non-homogeneous first-order partial differential equations problem:

$\displaystyle z\frac{\partial{z}}{\partial{x}} -y\frac{\partial{z}}{\partial{y}} = z - y sin(y)$

I have, just a few steps and the final result included, but I don't get the same result when I try to solve it on my own. I tried again and again and I can't see what I am doing wrong. Please help.

This is the final result I should get:
$\displaystyle F \left(\frac{z}{y}, x(y+z) + cos(y+z)\right) = 0$

That means I have to get two functions (two "first integrals" that would be the solution of the problem). First function (and thus the "first first integral") is $\displaystyle \frac{z}{y} = C_{1}$ and the second (and also "second first integral") is $\displaystyle x(y+z) + cos(y+z)) = C_{2}$

My solving steps:
1. I add an "associated system in symmetrical form" (not sure if this is how it's called in English):

$\displaystyle \frac{dx}{z} = \frac{dy}{-y} = \frac{dz}{z - y sin(y)}$

2. Solving this system like any nonlinear system of differential equations, to get the "first integrals":
- I have used second part of the equation from step 1:
$\displaystyle \frac{dy}{-y} = \frac{dz}{z - y sin(y)}$

$\displaystyle \frac{z - y sin(y)}{-y} = \frac{dz}{dy}$

$\displaystyle \frac{dz}{dy} = \frac{z - y sin(y)}{-y}$

$\displaystyle \frac{dz}{dy} = -\frac{1}{y}z + sin(y)$

$\displaystyle z' + \frac{1}{y}z = sin(y)$

3. So I got one ordinary linear differential equation and I am solving it using a formula:


In my case $\displaystyle p(y) = \frac{1}{y}$ (thus $\displaystyle \int p(y)dy = \int \frac{dy}{y} = ln|y|$) and $\displaystyle q(y) = sin(y)$ (thus $\displaystyle \int q(y) \cdot e^{\int p(y) dy} = \int y sin(y) dy = sin(y) - y cos(y)$) Adding this in formula, I get:
$\displaystyle z = e^{-ln|y|}\cdot [C + sin(y) - y cos(y)]$
$\displaystyle z = \frac{1}{y} \cdot [C + sin(y) - y cos(y)]$

$\displaystyle z y - sin(y) + y cos(y)= C$

and this should be the first integral, but (as you can see in final result above) their first integral is

$\displaystyle \frac{z}{y} = C$

What am I doing wrong? And how did they get this first integral?
Thanks all.

Your "constants of integration" are really constants. But if you integrate a function of f(x,y, z), say, with respect to x, the result may involve a "constant of integration" that is really an unknown function of y and z.