PDE - Yet another

**Altair** · May 13th 2009, 05:49 AM

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.

$u_y + u = e^{xy}$

How do I approach this one. I write $u_y$ in differential form but then I do not know how to handle the $e^{xy}$ 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 ?

**Danny** · May 13th 2009, 08:30 AM

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.

$u_y + u = e^{xy}$

How do I approach this one. I write $u_y$ in differential form but then I do not know how to handle the $e^{xy}$ 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

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

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

**OnkelTom** · May 13th 2009, 01:18 PM

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?

**Danny** · May 13th 2009, 01:41 PM

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 $x$ derivative. So the integrating factor for

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

and sure enough

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

**chisigma** · May 14th 2009, 12:43 AM

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

**Danny** · May 14th 2009, 04:40 AM

Originally Posted by chisigma

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

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

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

**chisigma** · May 14th 2009, 06:04 AM

Excuse me sir!...

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

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

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

... a little mysterious question

...

Kind regards

$\chi$ $\sigma$

**Danny** · May 14th 2009, 06:19 AM

Originally Posted by chisigma

Excuse me sir!...

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

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

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

... a little mysterious question

...

Kind regards

$\chi$ $\sigma$

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

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

**chisigma** · May 14th 2009, 06:25 AM

... I understand

...

Kind regards

$\chi$ $\sigma$

**Altair** · May 14th 2009, 07:55 PM

Originally Posted by danny arrigo

The key thing here is the absence of the $x$ derivative. So the integrating factor for

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

and sure enough

$ 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 ?

**Altair** · May 14th 2009, 07:55 PM

Originally Posted by chisigma

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

Same question goes for you chisigma

**Altair** · May 14th 2009, 09:12 PM

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

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

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

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

**Danny** · May 15th 2009, 05:50 AM

Originally Posted by Altair

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

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

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

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

Your PDE (an ODE now that we're setting x constant)

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

Comparing with your form

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

all we need to do is identify the variables going from (2) to (1) (y is u and x is y).

Thread: PDE - Yet another

LinkBack

Thread Tools

Display

PDE - Yet another

Search Tags