MAT244--2019F > Term Test 2

Problem 1 (noon)

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Victor Ivrii:
(a) Find the general solution of
$$
y''-3y'+2y=\frac{e^{3t}}{e^{2t}+1}.
$$

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

Yiheng Bian:
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:
$$
y=-0.25\pi*e^{2t}+0.5ln2e^t+e^{2t}*arctan(e^t) - 0.5e^{t}*ln(e^{2t}+1)
$$

NANAC:
Please see the attachment for the answer

OK.

But No snapshots!

xilin zhang:
I got a different y' in part b.

baixiaox:
Answer for question1

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