Method of undertermined coefficients (IVP)

[jwxie]

Homework Statement

Find the solution of the given initial value problem.

$\[y''-2y'-3y=3te^{2t}\]$

Homework Equations

The Attempt at a Solution

(1) the homogenous solution is given by
$\[y_{h}=c_{1}e^{3t}+c_{2}e^{-t}\]$

(2) the particular solution is in the form
$\[y_{p}=(At+B)e^{2t}\]$

(3) The first and second derivatives are then, respectively
$\[y'_{p}=Ae^{2t}+2Ate^{2t}+2Be^{2t}\]$
$\[y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}\]$

(3) substitutions and line up
$\[y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}\]$
$\[-2y'_{p}=-2Ae^{2t}-4Ate^{2t}-4Be^{2t}\]$
$\[-3y_{p}=-3Ate^{2t}-3Be^{2t}\]$

(4) combine like terms
$\[te^{t}\left [ 4A-4A-3A \right ]=1<br /> \]$
$\[e^{t}\left [ 4A+4B-2A-4B-3B \right ]=0\]$

(5) I got A = -1/3
but this is wrong.
According to the book, Ate^{2t) has a leading coefficient of -1.
However, coincidentally, my B is 2/3, while the book gives -2/3.

The complete solution to this problem with the given IV is
$y=e^{3t}+(2/3)e^{-t}-(2/3)e^{2t}-te^{2t}$

I have been working on this and other problems for hours and apparently I have been getting many answers... but I have been checking and redoing. Has anyone catch my mistake yet?

Thank you very much!

 

[micromass]

Fixed latex. You have to include everything in [t e x]...[/t e x]-boxes (without spaces) to make it work.

Homework Statement

Find the solution of the given initial value problem.

$$y''-2y'-3y=3te^{2t}$$

Homework Equations

The Attempt at a Solution

(1) the homogenous solution is given by
$$y_{h}=c_{1}e^{3t}+c_{2}e^{-t}$$

(2) the particular solution is in the form
$$y_{p}=(At+B)e^{2t}$$

(3) The first and second derivatives are then, respectively
$$y'_{p}=Ae^{2t}+2Ate^{2t}+2Be^{2t}$$
$$y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}$$

(3) substitutions and line up
$$y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}$$
$$-2y'_{p}=-2Ae^{2t}-4Ate^{2t}-4Be^{2t}$$
$$-3y_{p}=-3Ate^{2t}-3Be^{2t}$$

(4) combine like terms
$$te^{t}(4A-4A-3A)=1$$
$$e^{t}( 4A+4B-2A-4B-3B)=0$$

(5) I got A = -1/3
but this is wrong.
According to the book, Ate^{2t) has a leading coefficient of -1.
However, coincidentally, my B is 2/3, while the book gives -2/3.

The complete solution to this problem with the given IV is
$$y=e^{3t}+(2/3)e^{-t}-(2/3)e^{2t}-te^{2t}$$

I have been working on this and other problems for hours and apparently I have been getting many answers... but I have been checking and redoing. Has anyone catch my mistake yet?

Thank you very much!

 

[LCKurtz]

Homework Statement

Find the solution of the given initial value problem.

$$y''-2y'-3y=3te^{2t}$$
(1) the homogenous solution is given by
$$y_{h}=c_{1}e^{3t}+c_{2}e^{-t}$$

(2) the particular solution is in the form
$$y_{p}=(At+B)e^{2t}$$

(3) The first and second derivatives are then, respectively
$$y'_{p}=Ae^{2t}+2Ate^{2t}+2Be^{2t}$$
$$y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}$$

(3) substitutions and line up
$$y''_{p}=2Ae^{2t}+2Ae^{2t}+4Ate^{2t}+4Be^{2t}$$
$$-2y'_{p}=-2Ae^{2t}-4Ate^{2t}-4Be^{2t}$$
$$-3y_{p}=-3Ate^{2t}-3Be^{2t}$$

(4) combine like terms
$$te^{t}\left [ 4A-4A-3A \right ]=1$$
$$e^{t}\left [ 4A+4B-2A-4B-3B \right ]=0$$

(5) I got A = -1/3
but this is wrong.
According to the book, Ate^{2t) has a leading coefficient of -1.
However, coincidentally, my B is 2/3, while the book gives -2/3.

The complete solution to this problem with the given IV is
$$y=e^{3t}+(2/3)e^{-t}-(2/3)e^{2t}-te^{2t}$$

I have been working on this and other problems for hours and apparently I have been getting many answers... but I have been checking and redoing. Has anyone catch my mistake yet?

Thank you very much!

I have fixed your tex tags for you. Use tex instead of latex in your tags and don't start with \[ and end with \]. I will check your arithmetic in a while unless someone beats me to it.

[Edit] I see micromass already did this.
[Edit II] And somehow, editing removed the corrections
[Edit III] And now they are back. Go figure.

 

[micromass]

OK, your mistake lies in step (4). You've written

$$4A-4A-3A=1$$

But this should be

$$4A-4A-3A=3$$

I suspect the other equation in (4) is also wrong...

Edit: the other equation isn't wrong. You should get the correct answer now.

Also, be careful with your notation, writing $te^t(4A-4A-3A)=1$ is very wrong. You should not write $te^t$ in front.

 

[jwxie]

wooo. i must be stupid at the time
thanks to both of you!
sorry! i was on my school computer and i thought it was because of the javascript.
you guys are awesome!

Also, be careful with your notation, writing tet(4A−4A−3A)=1 is very wrong. You should not write tet in front.

I actually mean the like terms of t*e^t. Would that affect my calculation?

:] thanks i will be careful next time