Method of Undetermined Coefficients

**petition Edgecombe** · November 13th 2008, 06:36 PM

For the equation $y''+4y=t^2+3e^t$ : I have split up the particular solution into two parts, so that I have

$Y_p=At^2+Bt+C=t^2$ and $Y_{p2}=Ae^t=3e^t$

which gives me

$Y_p=\frac{1}{4}t^2$ and $Y_{p2}=\frac{1}{5}e^t$

The solution of the corresponding homogeneous equation is of course $y=c_1cos(2t)+c_2sin(2t)$ . So for my answer I get $\frac{1}{4}t^2+\frac{1}{5}e^t+\frac{1}{5}cos(2t)+\ frac{9}{10}sin(2t)$

Where did I go wrong?

**mr fantastic** · November 13th 2008, 06:53 PM

Originally Posted by petition Edgecombe

For the equation $y''+4y=t^2+3e^t$ : I have split up the particular solution into two parts, so that I have

$Y_p=At^2+Bt+C=t^2$ and $Y_{p2}=Ae^t=3e^t$ Mr F says: It's ${\color{red}Y_{p2} = \frac{3}{5} e^t}$ . I realise you've got it almost correct down below, but it's still careless here.

which gives me

$Y_p=\frac{1}{4}t^2$ Mr F says: This is wrong. Show all your working on how you got it (when you do this you will probably realise for yourself why it's wrong).

and $Y_{p2}=\frac{1}{5}e^t$ Mr F says: Wrong. Show your working.

The solution of the corresponding homogeneous equation is of course $y=c_1cos(2t)+c_2sin(2t)$ . So for my answer I get $\frac{1}{4}t^2+\frac{1}{5}e^t+\frac{1}{5}cos(2t)+\ frac{9}{10}sin(2t)$

Where did I go wrong?

..

**petition Edgecombe** · November 13th 2008, 11:27 PM

Originally Posted by mr fantastic

and

Mr F says: It's

. I realise you've got it almost correct down below, but it's still careless here.

I did have $\frac{3}{5} e^t$ , but I forgot how I got it, so I changed it (that's why I edited my post).

I see what I did there. I included the 3 in the initial guess so that it looked like
$y_{p2}=3Ae^t$ instead of $y_{p2}=Ae^t$

Mr F says: This is wrong. Show all your working on how you got it (when you do this you will probably realise for yourself why it's wrong).

I never really learned about equating coefficients. I briefly used it in calculus for partial fraction decomposition, that's it.

$y_{p}=At^2+Bt+C$

$y_{p}'=2At+B$

$y_{p}''=2A$

$2A+4At^2+4Bt+4C=t^2+0t+0c$

$4At^2=t^2$

$A=1/4$

$4Bt=0t$

$B=0$

$C=0c$

$C=0$

**mr fantastic** · November 13th 2008, 11:36 PM

Originally Posted by petition Edgecombe

I did have $\frac{3}{5} e^t$ , but I forgot how I got it, so I changed it (that's why I edited my post).

I see what I did there. I included the 3 in the initial guess so that it looked like
$y_{p2}=3Ae^t$ instead of $y_{p2}=Ae^t$

I never really learned about equating coefficients. I briefly used it in calculus for partial fraction decomposition, that's it.

$y_{p}=At^2+Bt+C$

$y_{p}'=2At+B$

$y_{p}''=2A$

$2A+4At^2+4Bt+4C=t^2+0t+0c$ Mr F says: This can be re-arranged as 4At^2 + 4Bt + (2A + 4C) = t^2 + 0t + 0. The constant term on the left hand side is what you've failed to see properly ..... See below.

$4At^2=t^2$

$A=1/4$

$4Bt=0t$

$B=0$

$C=0c$ Mr F says: Wrong.

This is where I knew the mistake would be. I was hoping you would have realised it while typing your work out. The correct equation is 2A + 4C = 0. Substitute A = 1/4 and solve for C.

$C = 0$ Mr F says: No. C = ....

..

**petition Edgecombe** · November 14th 2008, 12:11 AM

$c=-\frac{1}{8}$

And $c_1=-\frac{19}{40}$

So my problem was with the method of equating coefficients, as I had thought.

**Jason Bourne** · November 14th 2008, 02:15 AM

$y=c_1cos(2t)+c_2sin(2t) + \frac{1}{4}t^2 - \frac{1}{8} + \frac{3}{5}e^t$

Thread: Method of Undetermined Coefficients

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