Author Topic: TT2--P1  (Read 3756 times)

Victor Ivrii

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TT2--P1
« on: March 21, 2018, 02:54:56 PM »
a. Find general solution of
$$
y''-y=\frac{4}{e^{2t}+1}.$$

b. Find solution, such that $y(0)=0$, $y'(0)=\pi$.

Meng Wu

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Re: TT2--P1
« Reply #1 on: March 21, 2018, 03:06:17 PM »
$\underline{\text{Solution:}}$$\\$
Part(a) $\\$
First consider homogeneous equation: $$y''-y=0$$
characteristic equation: $$r^2-1=0 \implies \cases{r_1=1\\r_2=-1}$$
Thus, the complementary solution $$y_c(t)=c_1e^t+c_2e^{-t}$$
Now consider the nonhomogeneous equation $$y''-y={4\over e^{2t}+1}$$
Since $y_1(t)=e^t$ and $y_2(t)=e^{-t}$, Wronskain $$W=[y_1,y_2](t)=\begin{array}{|c c|}e^t &e^{-t}\\e^t&-e^{-t}\end{array}=e^t\cdot(-e^{-t})-e^{-t}\cdot e^t=-2 \neq0$$
Therefore, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.$\\$
Use the Method of Variation of Parameters: $\\$
The particular solution $$\begin{align}Y(t)&=-y_1(t)\int_{t_0}^{t}{y_2(s)g(s)\over W[y_1,y_2](s)}ds+y_2(t)\int_{t_0}^{t}{y_1(s)g(s)\over W[y_1,y_2](s)}ds\\&=-e^t\int_{t_0}^{t}{e^{-s}\cdot {4\over e^{2s}+1}\over -2}ds+e^{-t}\int_{t_0}^{t}{e^s \cdot {4\over e^{2s}+1}\over -2}ds\\&=2e^t\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}ds-2e^{-t}\int_{t_0}^{t}{e^s\over e^{2s}+1}ds\end{align}$$
For integral $\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}ds$:$\\$
Let $$u=e^t \implies \cases{du=e^tdt \implies dt=e^{-t}du\\e^{-t}={1\over u}}$$
Thus $$\begin{align}\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}ds&=\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}e^{-s}du\\&=\int_{t_0}^{t}{1\over u^2(u^2+1)}du\\&=\int_{t_0}^{t}({1\over u^2}-{1\over u^2+1})du\\&=\int_{t_0}^{t}{1\over u^2}du-\int_{t_0}^{t}{1\over u^2+1}du\\&=-{1\over u}-\arctan(u)\\&=-\arctan(e^t)-e^{-t}\end{align}$$
For integral $\int_{t_0}^{t}{e^s\over e^{2s}+1}ds$:$\\$
$$\int_{t_0}^{t}{e^s\over e^{2s}+1}ds=\arctan(e^t)$$
Thus $$Y(t)=2e^t[-\arctan(e^t)-e^{-t}]-2e^{-t}[\arctan(e^t)]$$
Therefore, the general solution $$\begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1e^t+c_2e^{-t}+2e^t[-\arctan(e^t)-e^{-t}]-2e^{-t}[\arctan(e^t)]\end{align}$$
Part(b)$\\$
$$\begin{align}y(0)&=c_1e^0+c_2e^0+2e^0[-\arctan(e^0)-e^0]-2e^0[\arctan(e^0)]=0\\&=c_1+c_2+2(-{\pi\over4}-1)-2{\pi \over4}=0\end{align}$$
$$\implies c_1+c_2=2+\pi$$
Note
$$y'(t)=c_1e^t-c_2e^{-t}-2e^t\arctan(e^t)-{2e^{2t}\over e^{2t}+1}+2e^{-t}\arctan(e^t)- {2\over e^{2t}+1}$$
$$\begin{align}y'(0)&=c_1e^0-c_2e^0-2e^0 \arctan(e^0)-{2e^0\over e^0+1}+2e^0\arctan(e^0)-{2\over e^0+1}=0\\&=c_1-c_2-2{\pi\over4}-1+2{\pi\over 4}-1\end{align}$$
$$\implies c_1-c_2=2$$
Thus, $$\cases{c_1=2+{\pi \over 2}\\c_2=-{\pi\over2}}$$
Therefore, general solution of the $IVP$ is $$y(t)=(2+{\pi \over 2})e^t-{\pi\over2}e^{-t}-2(e^t+e^{-t})\arctan(e^t)-2$$

Meng Wu

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Re: TT2--P1
« Reply #2 on: March 21, 2018, 03:08:05 PM »
Don't know if my answer is correct or not. My brain was totally frozen for this question's integral during the test  :( :-[ :-\

Meng Wu

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Re: TT2--P1
« Reply #3 on: March 21, 2018, 03:18:38 PM »
I think $(15)$ should be $$\implies c_1-c_2-2=\pi$$
Therefore, $$\cases{c_1=2+\pi\\c_2=0}$$
(Maybe I'm wrong again :o)

Zihan Wan

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Re: TT2--P1
« Reply #4 on: March 22, 2018, 04:51:13 PM »
y(0)=0,y'(0)=pi
but you are solving like y'(0)=0

Victor Ivrii

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Re: TT2--P1
« Reply #5 on: March 24, 2018, 10:18:44 AM »
Indeed, the second attempt to (b) was correct  $c_1=2+\pi,c_2=0$