Using Undetermined Coefficients to solve an equation for a particular solution?

[KevinD6]

Homework Statement

[itex] y'' + 9y = 3sin(3x) + 3 + e^{3x}[/itex]

Homework Equations

The Attempt at a Solution

This is my first post here so let me know if I've done anything wrong, I've been looking at questions here for a long time though ^^.

So the problem asks me to solve for one particular solution by using undetermined coefficients, I begin by solving for the homogeneous solution:
y'' + 9y = 0
The characteristic polynomial becomes:
[itex] r^2 + 9 = 0 [/itex]
Therefore, I get the two roots [itex] {0 \pm 3i} [/itex]

Solving for the homogeneous solution, which I'll call [itex] y_h [/itex] I get:
[itex] y_h = cos(3x) + sin(3x) [/itex]

Now, my professor hasn't gone through undetermined coefficients in his class yet but put it on the assignment (weird right?), so this is the part I might have screwed up on due to lack of knowledge.

Based on what I've found on the internet, first I predict a form of the particular solution, which I will denote [itex] y_p [/itex]
So:
[itex] y_p = A(cos(3x)) + B(sin(3x)) + C + D(e^{3x}) [/itex]
Afterwards, I plug this particular solution back into my original equation in order to solve for the coefficients A, B, C, and D.

However, taking the second derivative of my particular equation, I get:
[itex] y'' = -9A cos(3x) - 9B sin(3x) + 9D e^{3x} [/itex]
That equation is added onto 9y, which is:
[itex] 9y = 9Acos(3x) + 9B sin(3x) + 9C + 9D e^{3x} [/itex]

I'm left with the final equation of:
[itex] 9C + 18D = 3sin(3x) + 3 + e^{3x} [/itex]

Now what I'm confused about is the fact that A and B cancel out in this case, and I've been looking at internet explanations of undetermined coefficients but I can't find an answer to this question. Can anyone help me?

 

[LCKurtz]

Homework Statement

[itex] y'' + 9y = 3sin(3x) + 3 + e^{3x}[/itex]

Homework Equations

The Attempt at a Solution

This is my first post here so let me know if I've done anything wrong, I've been looking at questions here for a long time though ^^.

So the problem asks me to solve for one particular solution by using undetermined coefficients, I begin by solving for the homogeneous solution:
y'' + 9y = 0
The characteristic polynomial becomes:
[itex] r^2 + 9 = 0 [/itex]
Therefore, I get the two roots [itex] {0 \pm 3i} [/itex]

Solving for the homogeneous solution, which I'll call [itex] y_h [/itex] I get:
[itex] y_h = cos(3x) + sin(3x) [/itex]

Now, my professor hasn't gone through undetermined coefficients in his class yet but put it on the assignment (weird right?), so this is the part I might have screwed up on due to lack of knowledge.

Based on what I've found on the internet, first I predict a form of the particular solution, which I will denote [itex] y_p [/itex]
So:
[itex] y_p = A(cos(3x)) + B(sin(3x)) + C + D(e^{3x}) [/itex]
Afterwards, I plug this particular solution back into my original equation in order to solve for the coefficients A, B, C, and D.

However, taking the second derivative of my particular equation, I get:
[itex] y'' = -9A cos(3x) - 9B sin(3x) + 9D e^{3x} [/itex]
That equation is added onto 9y, which is:
[itex] 9y = 9Acos(3x) + 9B sin(3x) + 9C + 9D e^{3x} [/itex]

I'm left with the final equation of:
[itex] 9C + 18D = 3sin(3x) + 3 + e^{3x} [/itex]

Now what I'm confused about is the fact that A and B cancel out in this case, and I've been looking at internet explanations of undetermined coefficients but I can't find an answer to this question. Can anyone help me?

##\cos(3x)## and ##\sin(3x)## satisfy the homogeneous equation so they can't contribute to the non-homogeneous. That's why they are dropping out.

Use ##Ax\cos(3x) + Bx\sin(3x)## in your trial solution for ##y_p##.

 

[HaLAA]

because your yh is Acos3x+Bsin3x, your yp should be x(Acos3x+Bsin3x).

 

[HaLAA]

because your yh is Acos3x+Bsin3x, your yp should be x(Acos3x+Bsin3x).

x(Acos3x+Bsin3x)+C+De^3x

 

[KevinD6]

Ahh, I see, it was just my ignorance about this method. Thanks guys! I hope to be posting here more in the future.

 

[GFauxPas]

Paul's math notes has a good explanation of this method of multiplying throughout by a variable fixes problems sometimes.