# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:15:32 AM

Title: Problem 1 (noon)
Post by: Victor Ivrii on November 19, 2019, 04:15:32 AM
(a) Find the general solution of
$$y''-3y'+2y=\frac{e^{3t}}{e^{2t}+1}.$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$
Title: Re: Problem 1 (noon)
Post by: Yiheng Bian on November 19, 2019, 04:33:17 AM
No double-dipping

(a):
We can solve homo firstly:
$$r^2-3r+2=0\\ (r-2)(r-1)=0\\ r_1=2,r_2=1$$
Therefore:
$$y=c_1e^{2t}+c_2e^t$$
So we can get:
$$W=\begin{vmatrix} e^{2t} & e^t \\ 2e^{2t} & e^t \end{vmatrix}=-e^{3t}\\ W_1=\begin{vmatrix} 0 & e^{t} \\ 1 & e^{t} \end{vmatrix}=-e^{t}\\ W_2=\begin{vmatrix} e^{2t} & 0 \\ 2e^{2t} & 1 \end{vmatrix}=e^{2t}$$
So we can get:
$$Y(t)=e^{2t}\int{\frac{-e^{s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds + e^{t}\int{\frac{e^{2s}*\frac{e^{3s}}{e^{s2}+1}}{-e^{3s}}}ds\\ Y(t)=e^{2t}\int{\frac{e^{s}}{e^{2s}+1}}ds - e^{t}\int{\frac{e^{2s}}{e^{2s}+1}}ds\\ Y(t)=e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)$$
Finally:
$$y(t)=c_1e^{2t}+c_2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)$$

(b):
So we can get y'(t):
$$y'=2c_1e^{2t}+c_2e^t+2e^{2t}arctan(e^t)+e^{2t}*\frac{e^t}{e^{2t}+1}-0.5e^t*ln(e^{2t}+1)-0.5e^t*\frac{e^{2t}}{e^{2t}+1}$$
We take y(0)=y'(0)=0,so we can get:
$$2c_1+2c_2+0.5\pi-ln2=0\\ 2c_1+c_2+0.5\pi-0.5ln2=0$$
So
$$c_1=-0.25\pi,c_2=0.5ln2$$
Therefore:
$$y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)$$
Title: Re: Problem 1 (noon)
Post by: NANAC on November 19, 2019, 09:04:42 AM

OK.

But No snapshots!
Title: Re: Problem 1 (noon)
Post by: xilin zhang on November 19, 2019, 09:11:19 AM
I got a different y' in part b.
Title: Re: Problem 1 (noon)
Post by: baixiaox on November 19, 2019, 05:34:40 PM
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+c_1 \Bigr)e^{t} + \Bigl( \arctan (e^t)+c_2\Bigr)e^{2t}. }$$
$$\boxed{ y= \Bigl(-\frac{1}{2}\ln (e^{2t}+1)+\frac{1}{2}\ln (2) \Bigr)e^{t} + \Bigl( \arctan (e^t)-\frac{\pi}{4}\Bigr)e^{2t}. }$$