# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Quiz-1 => Topic started by: Victor Ivrii on January 25, 2018, 08:15:02 AM

Title: Q1-T0201
Post by: Victor Ivrii 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*}
Title: Re: Q1-T0201
Post by: Junya Zhang on January 25, 2018, 09:13:56 AM
(http://i64.tinypic.com/10x6mhc.jpg)
Title: Re: Q1-T0201
Post by: Meng Wu on January 25, 2018, 10:03:38 AM

This answer can not be improved anymore I think :O
Title: Re: Q1-T0201
Post by: Meng Wu 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
$$y=2te^{2t}-2e^{2t}+3e^{t}$$