# Math Help - Finding yp(x) Method of Undertermined Co-efficients (Clarification please!)

1. ## Finding yp(x) Method of Undertermined Co-efficients (Clarification please!)

$
y'' - 6 y' + 9 y = -2.5 e^{3 x}$

Is my question and its only asking for yp(x).

So I know that to solve for that I use the method of undetermined coefficients solving for A, B and C.

I got up to the point of 9Ax^(2)+(-12A-9B)x+(2A+6B+9C) = 2.5e^(3x)...
Then I realized, what is 2.5e^(3x) considered?

I need to make it equal to x^(2) or x.

ie. 9A = 2.5e^(3x) then I'll solve using that.

I'm thinking its a x variable? But then what do I do with that?

2. Originally Posted by Lolcats
$
y'' - 6 y' + 9 y = -2.5 e^{3 x}$

Is my question and its only asking for yp(x).

So I know that to solve for that I use the method of undetermined coefficients solving for A, B and C.

I got up to the point of 9Ax^(2)+(-12A-9B)x+(2A+6B+9C) = 2.5e^(3x)...
Then I realized, what is 2.5e^(3x) considered?

I need to make it equal to x^(2) or x.

ie. 9A = 2.5e^(3x) then I'll solve using that.

I'm thinking its a x variable? But then what do I do with that?
This is more differential equations then it is calculus.

With this method what we are trying to do is guess an appropriate Yp by looking at the right hand side of the equation.

In this case, we have

$
y'' - 6 y' + 9 y = -2.5 e^{3 x}$

So an appropriate Yp would be

$Yp = Ce^{3x}$

But let us note that the characteristic equation

$\lambda^2 - 6 \lambda + 9 =0$

Has the roots

$3$ and $3$

Therefore we must select a multiple of Yp because with our initial guess we will get an equation that does not work. So let us guess

$Yp=Cxe^{3x}$

Solve to find C. There is no need to have A, B and C!

Edit- Like HallsofIvey pointed out, we have 2 roots that are identical (I should have noticed!) so we need a Yp of the form

$Yp=Cx^2e^{3x}$

3. THanks!

So then in that case I would have to solve the rest of the equation to find out what exact C is for yp(x)? And I thought we were given an easy one for once...

4. Originally Posted by Lolcats
THanks!

So then in that case I would have to solve the rest of the equation to find out what exact C is for yp(x)? And I thought we were given an easy one for once...
This actually is fairly simple. You should look up the chapter on undetermined co-efficients in your text, it will show you how to appropriately guess Yp (this is the hardest part!).

What you want to get is

$Yp^\prime$
$Yp^{\prime \prime}$

Then sub this into the equation and this will yield the value of C. Also remember to find the roots of the characteristic equation and add those values into Yh if you are asked for the general solution (and not just the particular solution).

5. OOOOOOOH wait this is like Euler's equations right?

6. No, it's not anything like Euler's equation. It is a linear equation with constant coefficients. AllenCuz almost had it. Normally if the right side of a d.e. has $e^{ax}$, you would try a solution of the form $Ae^{ax}$ but if a is already a solution of the characteristic equation, $e^{ax}$already satisfies the homogeneous equation so you try $Axe^{ax}$.

However, if a is a double root of the characteristic equation, both $e{ax}$ and $xe^{ax}$ are solutions to the homogeneous equation so try $y= Ax^2e^{ax}$

Here, with a= 3, you should try a solution of the form $y= Ax^2e^{ax}$.

With $y= Ax^2e^{3x}$, $y'= 3Ax^2e^{3x}+ 2Axe^{ax}= (3Ax^2+ 2Ax)e^{3x}$

$y"= 3(3Ax^2+ 2Ax)e^{3x}+ (6Ax+ 2A)e^{3x}= (9Ax^2+ 12Ax+ 2A)e^{3x}$.

Put those into the d.e.

7. Originally Posted by HallsofIvy
No, it's not anything like Euler's equation. It is a linear equation with constant coefficients. AllenCuz almost had it. Normally if the right side of a d.e. has $e^{ax}$, you would try a solution of the form $Ae^{ax}$ but if a is already a solution of the characteristic equation, $e^{ax}$already satisfies the homogeneous equation so you try $Axe^{ax}$.

However, if a is a double root of the characteristic equation, both $e{ax}$ and $xe^{ax}$ are solutions to the homogeneous equation so try $y= Ax^2e^{ax}$

Here, with a= 3, you should try a solution of the form $y= Ax^2e^{ax}$.

With $y= Ax^2e^{3x}$, $y'= 3Ax^2e^{3x}+ 2Axe^{ax}= (3Ax^2+ 2Ax)e^{3x}$

$y"= 3(3Ax^2+ 2Ax)e^{3x}+ (6Ax+ 2A)e^{3x}= (9Ax^2+ 12Ax+ 2A)e^{3x}$.

Put those into the d.e.
Thanks! I forgot that we had double roots there. Of course if we were to actually go ahead with what I had, we would find that 0=something and notice that we would need another x.