Author Topic: Q1-T0201  (Read 2121 times)

Victor Ivrii

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Q1-T0201
« on: January 25, 2018, 08:15:02 AM »
Find the solution of the given initial value problem.
\begin{equation*}
y' - y = 2te^{2t},\qquad y(0) = 1.
\end{equation*}

Junya Zhang

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Re: Q1-T0201
« Reply #1 on: January 25, 2018, 09:13:56 AM »
« Last Edit: January 25, 2018, 09:19:12 AM by Junya Zhang »

Meng Wu

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Re: Q1-T0201
« Reply #2 on: January 25, 2018, 10:03:38 AM »

This answer can not be improved anymore I think :O

Meng Wu

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Re: Q1-T0201
« Reply #3 on: January 25, 2018, 11:19:50 AM »
Since given differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)=-1$ and $g(t)=2te^{2t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{-1dt}}=e^{-t}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$e^{-t}y'-e^{-t}y=e^{-t}\cdot 2te^{2t}=2te^{t}$$
and $(e^{-t}y)'=2te^{t}$ $\\$
Integrating both sides:
$$\int{(e^{-t}y)'}=\int{2te^{t}}$$
Thus, $$e^{-t}y=\int{2te^{t}}$$
For $\int{2te^{t}}$, we use Integration By Parts:$\\$
Let $u=2t, dv=e^{t}$.$\\$
Then $du=2dt, v=e^{t}$$\\$
Hence, $$\int{2te^{t}}=uv-\int{vdu}$$
$$\int{2te^{t}}=2te^{t}-\int{2e^{t}dt}$$
$$\int{2te^{t}}=2te^{t}-2e^{t}+c$$
Thus $$e^{-t}y=2te^{t}-2e^{t}+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $e^{-t}$, we get the general solution:
$$y=2te^{2t}-2e^{2t}+ce^{t}$$
To satisfy the initial condition, we set $t=0$ and $y=1$$\\$
Hence, $$1=2(0)e^{2\cdot 0}-2e^{2\cdot 0}+ce^{0}$$
$$1=0-2+c$$
so $$c=3$$
Therefore, the solution of the initial problem is
$$y=2te^{2t}-2e^{2t}+3e^{t}$$