Problem 1 (morning)

[Victor Ivrii]

(a) Find the general solution of
$$
y''-y=\frac{12}{e^{t}+1}.
$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$

[Yiheng Bian]

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
$$
y(t)=(6-6ln2)e^t+6ln2e^{-t}6e^t[-e^{-t}-t+ln(e^t+1)] - 6e^{-t}ln(e^t+1)
$$

[Lan Cheng]

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


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

[Yiran Wang]

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))

[Mingdi Xie]

Attached is my solution, more detail on the integral parts.