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

[KevinD6]

Homework Statement

y'' + 9y = 3sin(3x) + 3 + e^{3x}

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:
r^2 + 9 = 0
Therefore, I get the two roots {0 \pm 3i}

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

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 y_p
So:
y_p = A(cos(3x)) + B(sin(3x)) + C + D(e^{3x})
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:
y'' = -9A cos(3x) - 9B sin(3x) + 9D e^{3x}
That equation is added onto 9y, which is:
9y = 9Acos(3x) + 9B sin(3x) + 9C + 9D e^{3x}

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

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

y'' + 9y = 3sin(3x) + 3 + e^{3x}

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:
r^2 + 9 = 0
Therefore, I get the two roots {0 \pm 3i}

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

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 y_p
So:
y_p = A(cos(3x)) + B(sin(3x)) + C + D(e^{3x})
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:
y'' = -9A cos(3x) - 9B sin(3x) + 9D e^{3x}
That equation is added onto 9y, which is:
9y = 9Acos(3x) + 9B sin(3x) + 9C + 9D e^{3x}

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

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.