# Thread: PDE - Yet another

1. ## PDE - Yet another

This is my third question regarding PDE solvable as ODE and I am still confused because the book I am following (Advanced Engineering Mathematics by Erwin Kreyszig) has only two very elementary examples where as I find the exercise problems to be very difficult, especially in their approach.

Currently I am having problem in Q5 of Problem Set 12.1.

$\displaystyle u_y + u = e^{xy}$

How do I approach this one. I write $\displaystyle u_y$ in differential form but then I do not know how to handle the $\displaystyle e^{xy}$ part. Can someone please explain the solutiion to this one.

And please can I be provided with some links to online examples for PDEs solvable as ODEs ?

2. Originally Posted by Altair
This is my third question regarding PDE solvable as ODE and I am still confused because the book I am following (Advanced Engineering Mathematics by Erwin Kreyszig) has only two very elementary examples where as I find the exercise problems to be very difficult, especially in their approach.

Currently I am having problem in Q5 of Problem Set 12.1.

$\displaystyle u_y + u = e^{xy}$

How do I approach this one. I write $\displaystyle u_y$ in differential form but then I do not know how to handle the $\displaystyle e^{xy}$ part. Can someone please explain the solutiion to this one.

And please can I be provided with some links to online examples for PDEs solvable as ODEs ?
Not sure on links to PDEs that can be solved as ODEs but I'll comment on the one you have here. You'll notice that your PDE has no x derivative so x is treated as a constant in your PDE. This problem is now like the following ODE

$\displaystyle u' + u = e^{cy}$

It's linear and can be solved using an integrating factor. Remember, put back the x( $\displaystyle c \to x$) and with the constant of integration, it's a fucntion of integration.

3. A perhaps daft question to tag on the end here, but i thought the integrating factor method would only work for "proper" derivatives, not partial derivatives? I don't know where ive got this idea from, perhaps someone could clarify?

4. Originally Posted by OnkelTom
A perhaps daft question to tag on the end here, but i thought the integrating factor method would only work for "proper" derivatives, not partial derivatives? I don't know where ive got this idea from, perhaps someone could clarify?
The key thing here is the absence of the $\displaystyle x$ derivative. So the integrating factor for

$\displaystyle u_y + u$ is $\displaystyle \mu = e^{\int 1dy} = e^y$

and sure enough

$\displaystyle e^y\left(u_y+u\right) = \left(e^y u\right)_y$

5. Since the partial derivative respect to $\displaystyle x$ is absent , we can consider the equation written as...

$\displaystyle u^{'} + u = e^{x\cdot y}$ (1)

... where the independent variable is $\displaystyle y$ and $\displaystyle x$ is a constant. The (1) is a standard linear DE of first order and its solution is...

$\displaystyle u(y) = c\cdot e^{-y} + \varphi(y)$ (2)

... where $\displaystyle \varphi(*)$ is a particular solution of (1) that is found solving an indefinite integral...

$\displaystyle \varphi(y)= e^{-y}\cdot \int e^{(1+x)x\cdot y}\cdot dy = \frac{e^{x\cdot y}}{1+x}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Originally Posted by chisigma
Since the partial derivative respect to $\displaystyle x$ is absent , we can consider the equation written as...

$\displaystyle u^{'} + u = e^{x\cdot y}$ (1)

... where the independent variable is $\displaystyle y$ and $\displaystyle x$ is a constant. The (1) is a standard linear DE of first order and its solution is...

$\displaystyle u(y) = c\cdot e^{-y} + \varphi(y)$ (2)

... where $\displaystyle \varphi(*)$ is a particular solution of (1) that is found solving an indefinite integral...

$\displaystyle \varphi(y)= e^{-y}\cdot \int e^{(1+x)x\cdot y}\cdot dy = \frac{e^{x\cdot y}}{1+x}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
I think you mean $\displaystyle \varphi(x,y)= \frac{e^{x\cdot y}}{1+x}$. What you have doesn't make sense. Further the general solution of the PDE is

$\displaystyle u = \frac{e^{x\cdot y}}{1+x} + f(x) e^{-y}$ where $\displaystyle f(x)$ is arbitrary (not a constant as you have it).

7. Excuse me sir!...

... explain me please what is the difference in the two cases...

a) $\displaystyle u = \frac {e^{x\cdot y}}{1+x} + c \cdot e^{-y}$ where $\displaystyle x$ and $\displaystyle c$ are two constants independent from $\displaystyle y$

b) $\displaystyle u = \frac {e^{x\cdot y}}{1+x} + f(x) \cdot e^{-y}$ where $\displaystyle x$ is an arbitrary constant independent from $\displaystyle y$ and $\displaystyle f(x)$ is an arbitrary function of the constant $\displaystyle x$

... a little mysterious question ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. Originally Posted by chisigma
Excuse me sir!...

... explain me please what is the difference in the two cases...

a) $\displaystyle u = \frac {e^{x\cdot y}}{1+x} + c \cdot e^{-y}$ where $\displaystyle x$ and $\displaystyle c$ are two constants independent from $\displaystyle y$

b) $\displaystyle u = \frac {e^{x\cdot y}}{1+x} + f(x) \cdot e^{-y}$ where $\displaystyle x$ is an arbitrary constant independent from $\displaystyle y$ and $\displaystyle f(x)$ is an arbitrary function of the constant $\displaystyle x$

... a little mysterious question ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Sure. $\displaystyle u = \frac {e^{x\cdot y}}{1+x} + f(x) \cdot e^{-y}$ is much more general than $\displaystyle u = \frac {e^{x\cdot y}}{1+x} + c \cdot e^{-y}$ and contains it as a special case (i.e. where $\displaystyle f(x) = c$)

Side note, x is not a constant but a variable as y is.

9. ... I understand ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

10. Originally Posted by danny arrigo
The key thing here is the absence of the $\displaystyle x$ derivative. So the integrating factor for

$\displaystyle u_y + u$ is $\displaystyle \mu = e^{\int 1dy} = e^y$

and sure enough

$\displaystyle e^y\left(u_y+u\right) = \left(e^y u\right)_y$
Can you please explain this method...As to how do I identify that what is going to be the I.F and what is the geenral form of this method and how do I know when to apply this ?

11. Originally Posted by chisigma
Since the partial derivative respect to $\displaystyle x$ is absent , we can consider the equation written as...

$\displaystyle u^{'} + u = e^{x\cdot y}$ (1)

... where the independent variable is $\displaystyle y$ and $\displaystyle x$ is a constant. The (1) is a standard linear DE of first order and its solution is...

$\displaystyle u(y) = c\cdot e^{-y} + \varphi(y)$ (2)

... where $\displaystyle \varphi(*)$ is a particular solution of (1) that is found solving an indefinite integral...

$\displaystyle \varphi(y)= e^{-y}\cdot \int e^{(1+x)x\cdot y}\cdot dy = \frac{e^{x\cdot y}}{1+x}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Same question goes for you chisigma

12. The kind I have solved with I.F are given in the following tutorial under point 3.

$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)$

Where as in this question (which I posted) there isn't an $\displaystyle Q(x)$ only on the right side of equality sign. Its $\displaystyle e^{xy}$. So even if $\displaystyle x$ is constant its still a $\displaystyle Q(y)$ and not $\displaystyle Q(x)$

http://www.mathhelpforum.com/math-he...-tutorial.html

13. Originally Posted by Altair
The kind I have solved with I.F are given in the following tutorial under point 3.

$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)$

Where as in this question (which I posted) there isn't an $\displaystyle Q(x)$ only on the right side of equality sign. Its $\displaystyle e^{xy}$. So even if $\displaystyle x$ is constant its still a $\displaystyle Q(y)$ and not $\displaystyle Q(x)$

http://www.mathhelpforum.com/math-he...-tutorial.html
Your PDE (an ODE now that we're setting x constant)

$\displaystyle \frac{du}{dy} + u = e^{cy}\;\;(1)$

$\displaystyle \frac{dy}{dx} + P(x)y = Q(x)\;\;(2)$