Stuck on a particular integral

**craig** · June 15th 2009, 05:22 AM

Hi, I've got the following question I need a bit of help on.

$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 2x + 3$

It's the particular integral that I'm not too sure on.

Normally I would make $y = px + q$ , differentiate that, put it into the equation to get my particular integral.

Have tried this and it didn't work, I'm guessing because there is no $y$ function in the original equation, only derivatives of $y$ .

Due to this, would I set $y = px^2 + qx + r$ for example?

Thanks

**craig** · June 15th 2009, 06:23 AM

Just found an answer and they have used $y = px^2 + qx + r$ as the particular integral.

I'm curious however, is this case because there is no $y$ , or is that just a coincidence?

Thanks

**AMI** · June 15th 2009, 06:44 AM

By taking $y=px^2+qx+r$ , you get the solution $y=x^2+x+r$ , where $r\in\mathbb{R}$ is a constant. (Is that right? I hope I didn't mess up the calculations..

)

However, $y=x^2+x-a\exp(-x)+b$ , where $a,b\in\mathbb{R}$ are constants is also a valid solution, but it doesn't fit in the form that you obtained. So, using that method, you don't arrive to all the solutions. Just thought I'd point that out, I know that's not what you asked..

**craig** · June 15th 2009, 06:49 AM

Originally Posted by AMI

By taking $y=px^2+qx+r$ , you get the solution $y=x^2+x+r$ , where $r\in\mathbb{R}$ is a constant. (Is that right? I hope I didn't mess up the calculations..

)

However, $y=x^2+x-a\exp(-x)+b$ , where $a,b\in\mathbb{R}$ are constants is also a valid solution, but it doesn't fit in the form that you obtained. So, using that method, you don't arrive to all the solutions. Just thought I'd point that out, I know that's not what you asked..

Here's my final general solution:

$y = A + Be^{-x} + x^2 + x$ , how is it I'm missing a solution, the $r$ does not get included due to there being no $y$ in the original question :S

Sorry if I've misread your post.

**AMI** · June 15th 2009, 06:54 AM

Well, I guess then I am the one that misunderstood something.. I thought you said you are looking for solutions in the form $y=px^2+qx+r$ with $p,q\in\mathbb{R}$ constants..

**craig** · June 15th 2009, 07:04 AM

Originally Posted by AMI

Well, I guess then I am the one that misunderstood something

You might not have, it could quite possibly be me being rather slow, todays not been a good day

Originally Posted by AMI

I thought you said you are looking for solutions in the form $y=px^2+qx+r$ with $p,q\in\mathbb{R}$ constants..

Could you explain how I've not got all the solutions if you don't mind?

Thanks for the reply btw

**AMI** · June 15th 2009, 07:13 AM

Well, actually you have. You wouldn't have had, if what I understood was right. Which, as it seems, is not. I simply didn't (and still don't) get what you meant when you said "set $y=px^2+qx+r$ ", but since you arrived at this:

Originally Posted by craig

Here's my final general solution:

$y = A + Be^{-x} + x^2 + x$

then my post was useless

**craig** · June 15th 2009, 07:19 AM

Originally Posted by AMI

I simply didn't (and still don't) get what you meant when you said "set $y=px^2+qx+r$ "

This is a particular integral, part of the general solution to a second order differential equation.

Think we got our wires crossed somewhere, I wouldn't worry bout it

**mr fantastic** · June 15th 2009, 07:52 PM

Originally Posted by craig

Hi, I've got the following question I need a bit of help on.

$\frac{d^2y}{dx^2} + \frac{dy}{dx} = 2x + 3$

It's the particular integral that I'm not too sure on.

Normally I would make $y = px + q$ , differentiate that, put it into the equation to get my particular integral.

Have tried this and it didn't work, I'm guessing because there is no $y$ function in the original equation, only derivatives of $y$ .

Due to this, would I set $y = px^2 + qx + r$ for example?

Thanks

Normally you'd try $y_p = ax + b$ .

But $y = b$ is part of the homogenous solution. When that happens the usual thing to try is multiplying your initial model for the particular solution by x:

$y_p = x(ax + b) = ax^2 + bx$ .

**craig** · June 15th 2009, 11:37 PM

Of course, the auxiliary equation has a root at zero, there getting rid of the $e$ in the equation.

Believe it or not I do know this, I just seem to have a habit of missing the obvious at times

Originally Posted by mr fantastic

But $y = b$ is part of the homogenous solution. When that happens the usual thing to try is multiplying your initial model for the particular solution by x:

$y_p = x(ax + b) = ax^2 + bx$ .

So is it best not to use something like $ax^2 + bx + c$ , just $ax^2 + bx$ ?

Thanks for the reply

**Kaitosan** · June 16th 2009, 07:26 AM

I find this differential equation problem interesting because I've never seen it before. I don't understand the above explanations of the problem. Can anyone please show me the steps of solving it?

By the way, this problem isn't in the domain of multivariable calculus, right?

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