# Particular solution to a differential equations

• Apr 6th 2009, 11:49 AM
Particular solution to a differential equations
Find the particular solution to $\displaystyle y^{(4)}+5y''+4y = \sin x + \cos 2x$

First, I found that the general solution is $\displaystyle y_h=$$\displaystyle C_1 \cos (2x) + C_2 \sin (2x) + C_3 \cos (-2x) +$$\displaystyle C_4 \sin (-2x) + C_5 \cos (x) + C_6 \sin (x) + C_7 \cos (-x) + C_8 \sin (-x)$

So the list of terms that I obtain here is: $\displaystyle \{ \cos (2x), \sin (2x), \cos (-2x) , \sin (-2x) , \cos (x) , \sin (x) , \cos (-x) , \sin (-x) \}$

Now, for my guess of the particular solution, I have list of terms $\displaystyle \{ \sin x , \cos (2x) \}$

Since I have repeats, I would take the derivative, so I have new list $\displaystyle \{ \sin x , \cos x , \cos (2x) , \sin (2x) \}$

So my particular solution is: $\displaystyle y_p = A \sin x + B \cos x + C \cos (2x) + D \sin (2x)$

I know I'm missing something, I think I'm suppose to multiply x by one of the list, but I just can't remember what the method was, please help. Thanks.
• Apr 6th 2009, 02:51 PM
Jester
Quote:

Find the particular solution to $\displaystyle y^{(4)}+5y''+4y = \sin x + \cos 2x$

First, I found that the general solution is $\displaystyle y_h=$$\displaystyle C_1 \cos (2x) + C_2 \sin (2x) + C_3 \cos (-2x) +$$\displaystyle C_4 \sin (-2x) + C_5 \cos (x) + C_6 \sin (x) + C_7 \cos (-x) + C_8 \sin (-x)$

So the list of terms that I obtain here is: $\displaystyle \{ \cos (2x), \sin (2x), \cos (-2x) , \sin (-2x) , \cos (x) , \sin (x) , \cos (-x) , \sin (-x) \}$

Now, for my guess of the particular solution, I have list of terms $\displaystyle \{ \sin x , \cos (2x) \}$

Since I have repeats, I would take the derivative, so I have new list $\displaystyle \{ \sin x , \cos x , \cos (2x) , \sin (2x) \}$

So my particular solution is: $\displaystyle y_p = A \sin x + B \cos x + C \cos (2x) + D \sin (2x)$

I know I'm missing something, I think I'm suppose to multiply x by one of the list, but I just can't remember what the method was, please help. Thanks.

First off, you have some redundancy. The complimentary solution is only

$\displaystyle y = c_1 \sin x + c_2 \cos x + c_3 \sin 2x + c_4 \cos 2x$

Second, since the homogeneous terms appears as part of your complimentary solution you would need to seek a particluar solution of the form

$\displaystyle y = c_1 x \sin x + c_2 x \cos x + c_3 x \sin 2x + c_4 x \cos 2x$
• Apr 6th 2009, 07:00 PM
So if the same terms appear, I just need to multiply everything by x?
• Apr 7th 2009, 05:38 AM
Jester
Quote:

So if the same terms appear, I just need to multiply everything by x?

Yes or higher. For example

$\displaystyle y'' - 2y' + y = e^x$

For the particular solution you'll need $\displaystyle y_p = Ax^2 e^x$ as $\displaystyle e^x\;\text{and}\; x e^x$ are a part of the complimentary solution.
• Apr 7th 2009, 06:14 AM
So for $\displaystyle y''+2y'-3y=1-xe^x$

I have the complimentary solution $\displaystyle y = C_1e^{-3x}+C_2e^x$

So my terms are $\displaystyle \{ 1 , xe^x \}$

But this time there are no repeats, and I tried with $\displaystyle y=A+Bxe^x$, but things went wrong, everything cancel, what should I do next?
• Apr 7th 2009, 06:37 AM
Jester
Quote:

So for $\displaystyle y''+2y'-3y=1-xe^x$

I have the complimentary solution $\displaystyle y = C_1e^{-3x}+C_2e^x$

So my terms are $\displaystyle \{ 1 , xe^x \}$

But this time there are no repeats, and I tried with $\displaystyle y=A+Bxe^x$, but things went wrong, everything cancel, what should I do next?

Because $\displaystyle e^x$ is a part of your complimentary solution and $\displaystyle x e^x$ is a part of your homogeneous term, then you would seek a particular solution of the form

$\displaystyle y_p = A x e^x + B x^2 e^x$.
• Apr 7th 2009, 05:41 PM
Okay, I tried with this particular solution, but weird things happened:

After plugging in everything to the original equations, I have the following:

$\displaystyle y=Axe^x+Bx^2e^x$
$\displaystyle y'=Ae^x+(A+2B)xe^x+Bx^2e^x$
$\displaystyle y''=(A+2B)e^x+(A+2B)xe^x+Bx^2e^x$

So I have

$\displaystyle (B+2b-3B)x^2e^x+(A+2B+2A+4B-3A)xe^x+(A+2B+2A)e^x=1+xe^x$
$\displaystyle (3A+2B)e^x+(6B)xe^x=1+(1)xe^x$

But I don't have anything on the left to equal to the 1 on the right, what am I doing wrong? I really feel stupid now because I'm actually tutoring this class, I took DEQ like 3 years ago, I JUST CAN'T UNDERSTAND WHY I AM NOT GETTING THIS RIGHT AS A GRADUATE STUDENT!!!!

Perhaps I need $\displaystyle y_p=A+Bxe^x+Cx^2e^x$ because of the constant term in the RHS?
• Apr 7th 2009, 05:57 PM
mr fantastic
Quote:

So for $\displaystyle y''+2y'-3y=1-xe^x$
I have the complimentary solution $\displaystyle y = C_1e^{-3x}+C_2e^x$

So my terms are $\displaystyle \{ 1 , xe^x \}$

But this time there are no repeats, and I tried with $\displaystyle y=A+Bxe^x$, but things went wrong, everything cancel, what should I do next?

Quote:

Okay, I tried with this particular solution, but weird things happened:

After plugging in everything to the original equations, I have the following:

$\displaystyle y=Axe^x+Bx^2e^x$
$\displaystyle y'=Ae^x+(A+2B)xe^x+Bx^2e^x$
$\displaystyle y''=(A+2B)e^x+(A+2B)xe^x+Bx^2e^x$

So I have

$\displaystyle (B+2b-3B)x^2e^x+(A+2B+2A+4B-3A)xe^x+(A+2B+2A)e^x=1+xe^x$
$\displaystyle (3A+2B)e^x+(6B)xe^x=1+(1)xe^x$

But I don't have anything on the left to equal to the 1 on the right, what am I doing wrong? I really feel stupid now because I'm actually tutoring this class, I took DEQ like 3 years ago, I JUST CAN'T UNDERSTAND WHY I AM NOT GETTING THIS RIGHT AS A GRADUATE STUDENT!!!!

Perhaps I need $\displaystyle y_p=A+Bxe^x+Cx^2e^x$ because of the constant term in the RHS?

Quote:

Originally Posted by danny arrigo
Because $\displaystyle e^x$ is a part of your complimentary solution and $\displaystyle x e^x$ is a part of your homogeneous term, then you would seek a particular solution of the form

$\displaystyle y_p = A x e^x + B x^2 e^x$.

This is how to find the particular solution corresponding to the $\displaystyle -x e^x$ term on the RHS of the DE.

To find the particular solution corresponding to the constant term on the RHS of the DE you try one of the form y = a and get a = -1/3.

Then you add these two particular solutions together to get the overall particular solution.