# Thread: Stuck on a particular integral

1. ## Stuck on a particular integral

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

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

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

Normally I would make $\displaystyle y = px + q$, 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 $\displaystyle y$ function in the original equation, only derivatives of $\displaystyle y$.

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

Thanks

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

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

Thanks

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

However, $\displaystyle y=x^2+x-a\exp(-x)+b$, where $\displaystyle a,b\in\mathbb{R}$ 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..

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

However, $\displaystyle y=x^2+x-a\exp(-x)+b$, where $\displaystyle a,b\in\mathbb{R}$ 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:

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

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

6. 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 $\displaystyle y=px^2+qx+r$ with $\displaystyle p,q\in\mathbb{R}$ constants..
Could you explain how I've not got all the solutions if you don't mind?

7. 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 $\displaystyle y=px^2+qx+r$", but since you arrived at this:
Originally Posted by craig
Here's my final general solution:

$\displaystyle y = A + Be^{-x} + x^2 + x$
then my post was useless

8. Originally Posted by AMI
I simply didn't (and still don't) get what you meant when you said "set $\displaystyle y=px^2+qx+r$"
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

9. Originally Posted by craig
Hi, I've got the following question I need a bit of help on.

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

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

Normally I would make $\displaystyle y = px + q$, 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 $\displaystyle y$ function in the original equation, only derivatives of $\displaystyle y$.

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

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

But $\displaystyle y = b$ 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:

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

10. Of course, the auxiliary equation has a root at zero, there getting rid of the $\displaystyle e$ 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 $\displaystyle y = b$ 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:

$\displaystyle y_p = x(ax + b) = ax^2 + bx$.
So is it best not to use something like $\displaystyle ax^2 + bx + c$, just $\displaystyle ax^2 + bx$?