# Partial Derivative Equation as solutions

#### zain0097

I have to solve PDE as ODE, but my lecturer didn't give me enough help and tutorial, to solve these problems.

The problems are:

i have to solve PDE as ODE where $$\displaystyle u = u(x,y)$$

$$\displaystyle u_y + u = e^{xy}$$
$$\displaystyle u_{xx} = 4y^2u$$
$$\displaystyle u_y = 2*x*y*u$$

#### romsek

MHF Helper
There must be some typo here. By subtracting the first equation from the third you are able to solve for $u(x,y)= \dfrac{e^{x y}}{2 x y+1}$
by simple algebra. But that solution does not satisfy the second equation.

topsquark

#### topsquark

Forum Staff
I had figured this to be three problems, not one?

They don't separate but I can't believe that your instructor would just give you a set of problems that haven't (and won't) be discussed either in class or the reading.

Skim the chapter and see if you find this: Method of Characteristics.

-Dan

Actually, the second two equations do separate. Just remember than when integrating over, say x, we take y to be a constant.
For example, in the second one we have
Let u(x, y) = X(x)Y(y)
$$\displaystyle u_{xx} = 4y^2u \implies X''Y = 4y^2 XY \implies \dfrac{X''}{X} = 4y^2 \text{ and } Y \neq 0$$
You would treat the y^2 as a constant.

Since these last two are separable and you haven't gone further than that(?) is it possible that the first has a typo? Possibly $$\displaystyle u_y + u = e^{x + y}$$ ? This does separate.

-Dan

Last edited:
romsek

#### romsek

MHF Helper
I had figured this to be three problems, not one?
oh duh, I guess the word problem(s) is a giveaway.

topsquark

#### zain0097

I had figured this to be three problems, not one?

They don't separate but I can't believe that your instructor would just give you a set of problems that haven't (and won't) be discussed either in class or the reading.

Skim the chapter and see if you find this: Method of Characteristics.

-Dan

Actually, the second two equations do separate. Just remember than when integrating over, say x, we take y to be a constant.
For example, in the second one we have
Let u(x, y) = X(x)Y(y)
$$\displaystyle u_{xx} = 4y^2u \implies X''Y = 4y^2 XY \implies \dfrac{X''}{X} = 4y^2 \text{ and } Y \neq 0$$
You would treat the y^2 as a constant.

Since these last two are separable and you haven't gone further than that(?) is it possible that the first has a typo? Possibly $$\displaystyle u_y + u = e^{x + y}$$ ? This does separate.

-Dan
I know how to seperate, but the first and third don't, how do i solve it?

#### topsquark

Forum Staff
I know how to seperate, but the first and third don't, how do i solve it?
The third one separates, too.

Again, let u(x,y) = X(x)Y(y)
$$\displaystyle u_y = 2xyu \implies XY'= 2xyXY$$

So
$$\displaystyle \dfrac{1}{y} \dfrac{Y'}{Y} = 2x$$ so long as $$\displaystyle X \neq 0$$.

As to the first check out the link. It's pretty much step by step.

-Dan