# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 2 => Topic started by: Victor Ivrii on November 19, 2019, 04:14:47 AM

Title: Problem 1 (morning)
Post by: Victor Ivrii on November 19, 2019, 04:14:47 AM
(a) Find the general solution of
$$y''-y=\frac{12}{e^{t}+1}.$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$
Title: Re: Problem 1 (morning)
Post by: Yiheng Bian on November 19, 2019, 04:27:31 AM
You should not double-dip. V.I.
The first question
we solve homo firstly, so we get:
$$r^2 - 1=0\\ r^2 =1\\ r_1=1\\ r_2=-1\\ \text{Therefore: }y=c_1e^t+c_2e^{-t}$$
So we can get
$$W = \begin{vmatrix} e^t & e^{-t} \\ e^t & -e^{-t} \end{vmatrix}=-2\\ W_1=\begin{vmatrix} 0 & e^{-t} \\ 1 & -e^{-t} \end{vmatrix}=-e^{-t}\\ W_2=\begin{vmatrix} e^t & 0 \\ e^t& 1 \end{vmatrix}=e^t$$
So we can get
$$Y(t)=e^t\int{\frac{-e^{-s}*\frac{12}{e^s+1}}{-2}}ds + e^{-t}\int{\frac{e^{s}*\frac{12}{e^s+1}}{-2}}ds\\ Y(t)=6e^t\int{\frac{e^{-s}}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{1}{e^s*(e^s+1)}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{e^s+1-e^s}{e^s*(e^s+1)}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{1}{e^s}-\frac{1}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\ Y(t)=6e^t\int{\frac{1}{e^s}-\frac{e^s+1-e^s}{e^s+1}}ds - 6e^{-t}\int{\frac{e^{s}}{e^s+1}}ds\\$$
Finally:
$$Y(t)=6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)\\ y(t)=c_1e^t+c_2e^{-t}+6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)$$

The second question:
since we have y(t), so we can get y'(t)
$$y'(t)=c_1e^t-c_2e^{-t}+6e^t[-e^{-t}-t+ln(e^t+1)]+6e^t[e^{-t}-1+\frac{e^t}{e^t+1}]+6e^{-t}ln(e^t+1)-6e^{-t}*\frac{e*t}{e^t+1}$$
Then we take y(0)=y'(0)=0 into y and y'
Therefore:
$$c_1+c_2-6=0\\ c_1-c_2-6+12ln2=0\\ c_1=6-6ln2, c_2=6ln2$$
So
$$y(t)=(6-6ln2)e^t+6ln2e^{-t}6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)$$
Title: Re: Problem 1 (morning)
Post by: Lan Cheng on November 19, 2019, 05:57:54 AM
a) First, we should solve for $y"-y=0.$

Let $r^{2}-1=0,r_{1}=1,r_{2}=-1.$

thus, $y_{c}(t)=C_{1}e^{t}+C_{2}e^{-t}.$

Second, we should calculate W.

$W=\begin{bmatrix}e^{t} & e^{-t}\\ e^{t} & -e^{-t} \end{bmatrix}=-2,W_{1}=\begin{bmatrix}0 & e^{-t}\\ 1 & -e^{-t} \end{bmatrix}=-e^{-t},W_{2}=\begin{bmatrix}e^{t} & 0\\ e^{t} & 1 \end{bmatrix}=e^{t}.$

therefore, $y_{p}(t)=e^{t}\int\frac{W_{1}(t)g(t)}{W(t)}dt-e^{-t}\int\frac{W_{2}(t)g(t)}{W(t)}dt.$

$y_{p}(t)=6e^{t}\int\frac{e^{-t}}{e^{t}+1}dt+6e^{-t}\int\frac{e^{t}}{e^{t}+1}dt.$

$y_{p}(t)=6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).$

Thus, $y(t)=C_{1}e^{t}+C_{2}e^{-t}+6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).$

b) $y'(t)=C_{1}e^{t}-C_{2}e^{-t}+6e^{t}ln\left|e^{t}+1\right|+6e^{t}\frac{e^{t}}{e^{t}+1}-6te^{t}-6e^{t}-6e^{-t}ln\left|e^{t}+1\right|+6e^{-t}\frac{e^{t}}{e^{t}+1}.$

plug in, we can get $C_{1}=6-6ln(2),C_{2}=6ln(2).$

Therefore, $y(t)=(6-6ln(2))e^{t}+6ln(2)e^{-t}+6e^{t}(-e^{-t}-t+ln(e^{t}+1))-6e^{-t}ln(e^{t}+1).$

OK, except LaTeX sucks:

2)  "operators" should be escaped: \cos, \sin, \tan, \ln

$$\boxed{ y= 6\Bigl(-e^{-t}+\ln (1+e^{-t})+1-\ln(2) \Bigr)e^{t} + 6\Bigl(\ln (e^t+1)-\ln(2)\Bigr)e^{-t}. }$$

Title: Re: Problem 1 (morning)
Post by: Yiran Wang on November 19, 2019, 12:01:37 PM
r2-1=0  r2=1  r=1, -1 we have homogenous sol: y=C1e^-t+C2e^t
w=e-tet+e-tet=2  w1=-e^t  w2=e^-t
yp=e^-t *∫[-e^t * 12/(e^t+1)]/2 dt +e^-t *∫[e^-t * 12/(e^t+1)]/2 dt
=-6e^-t∫e^t /(e^t+1) dt + 6e^-t∫e^-t /(e^t+1) dt
=-6e^-t *ln(e^t+1)+6e^-t∫e^-t /(e^t+1) dt
let u=e^^t, e-t=1/u, du=e^tdt, dt=1/u du
∫e^-t /(e^t+1) dt=∫[1/u/(u+1)] *1/u du
=∫1/u2(u+1) du
=∫ (1+u-u)/ u2(u+1) du
=∫1/u2 – 1/u(u+1) du
= -1/u – lnu +ln(u+1)
=-1/e^t – lne^t+ln(^et+1)
Y= C1e^-t+C2e^t-6e^-t *ln(e^t+1) +6e^-t *(-1/e^t – t+ln(e^t+1))
Y(0)=C1+C2-6ln2-6+6ln2
Y’(0)=C1-C2-6ln2-6+6ln2
C1=0, C2=6
Y=6 e^t-6e^-t * ln(et+1) +6e^-t *(-1/e^t – t+ln(e^t+1))

Title: Re: Problem 1 (morning)
Post by: Mingdi Xie on November 19, 2019, 01:19:13 PM
Attached is my solution, more detail on the integral parts.