MHB Method of undetermined coefficients

[ineedhelpnow]

Use the method of undetermined coefficients to find a general solution of the ODE:
$y''+3y'+2y=2x^{2}+4x+5$$r^{2}+3r+2=0$
$r=-2$ and $r=-1$
$y_{h}=c_{1}e^{-2x}+c_{2}e^{-x}$
I'm not sure how to get $y_{p}$ here

So here's what I've done so far. I have my final exam tomorrow and I have a few questions to go over as a review. I need help with the rest of this!

Here's another one:
$y''+3y''+3y'+y=e^{x}-x-1$

$r^{3}+3r^{2}+3r+1=0$
$r_{1,2,3}=-1$
$y_{h}=c_{1}e^{-x}+xe^{-x}(c_{2}+c_{3})$
Not really sure how to get $y_{p}$ here either

 

[MarkFL]

1.) You have correctly identified the homogeneous solution, and so if we consult a table, we find that the particular solution must take the form:

$$y_p(x)=Ax^2+Bx+C$$

Then compute the needed derivatives, and substitute into the original ODE, and the equate the corresponding coefficients to determine $A,\,B,\,C$.

Can you proceed?

2.) You have correctly identified that the characteristic equation has the root $r=-1$ of multiplicity 3, and so the homogeneous solution will take the form:

$$y_h(x)=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}$$

Consulting a table, and applying the principle of superposition, we find the particular solution must take the form:

$$y_p(x)=Ae^x+Bx+C$$

Can you proceed? :)

 

[ineedhelpnow]

I understand. But can you please post the table or a picture of it?! Because I can't find it anywhere in my book but I know there is one because I have referenced it before.

 

[MarkFL]

Here is the table (along with its justification):

http://mathhelpboards.com/math-notes-49/justifying-method-undetermined-coefficients-4839.html

 

[ineedhelpnow]

I don't really understand what Pn is. :(

 

[MarkFL]

I don't really understand what Pn is. :(

If $g(x)$ contains the $n$th degree polynomial $p_n(x)$, then the particular solution (in types I and IV) will contain the $n$th degree polynomial $P_n(x)$.

 

[ineedhelpnow]

the particular solution is chosen based on the RHS, correct?

 

[MarkFL]

the particular solution is chosen based on the RHS, correct?

Yes, if the ODE is of the form:

$$L[y](x)=g(x)$$

then we base the form of our particular solution on $g(x)$. If you prefer not to use a table, you can use the annihilator method to find the form of the particular solution:

http://mathhelpboards.com/questions-other-sites-52/jeromes-questions-yahoo-answers-regarding-inhomogeneous-linear-odes-4718.html

 

[ineedhelpnow]

I figured the first one out but I still can't get around the second.

 

[MarkFL]

Okay, the second one is:

$$y'''+3y''+3y'+y=e^{x}-x-1$$

We have already established the homogeneous solution is:

$$y_h(x)=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}=e^{-x}\left(c_3x^2+c_2x+c_1\right)$$

Next, we posit that the particular solution must take the form:

$$y_p(x)=Ae^x+Bx+C$$

So, we need to compute:

$$y_p'(x)=Ae^x+B$$

$$y_p''(x)=Ae^x$$

$$y_p'''(x)=Ae^x$$

Now, substituting these into the given ODE, there results:

$$Ae^x+3Ae^x+3(Ae^x+B)+Ae^x+Bx+C=e^{x}-x-1$$

$$8Ae^x+Bx+3B+C=e^{x}-x-1$$

By equating corresponding coefficients, we obtain:

$$8A=1\implies A=\frac{1}{8}$$

$$B=-1$$

$$3B+C=-1\implies C=2$$

Thus, the particular solution is:

$$y_p(x)=\frac{1}{8}e^x-x+2$$

And so the general solution is:

$$y(x)=y_h(x)+y_p(x)=e^{-x}\left(c_3x^2+c_2x+c_1\right)+\frac{1}{8}e^x-x+2$$