### Author Topic: Q1 problem 2 (Night sections)  (Read 2309 times)

#### Victor Ivrii

• Elder Member
• Posts: 2563
• Karma: 0
##### Q1 problem 2 (Night sections)
« on: September 24, 2014, 10:19:51 PM »

2.6 p. 102, # 25 Solve the initial value problem
\begin{equation*}
\end{equation*}

#### Chang Peng (Eddie) Liu

• Full Member
• Posts: 19
• Karma: 7
##### Re: Q1 problem 2 (Night sections)
« Reply #1 on: September 24, 2014, 11:04:15 PM »
Firstly, I'd like to apologize for the really ugly notations and equations.. there is the solution! I rewrote it--V.I.

Find the solution of the given initial value problem.
\begin{gather}
y' - 2y = e^{2t},\label{A}\\
y(0) = 2.\label{B}
\end{gather}

First, we need to find the integrating factor, which is $I = e^{\int -2\,dt}= e^{-2t}$

so we multiple the entire equation by I, thus, giving us
$e^{-2t} y' - 2e^{-2t} y = 1$.
We see that the left side of the equation can be rewritten as $[e^{-2t}y]'$
and we see that the right side of the equation is in fact 1

therefore: $[e^{-2t} y]' = 1$

we now take the integral of both sides, giving us: $e^{-2t} y = t + C$   where $C$ is a constant. To find  $C$, we substitute $y = 2$ when $t$ = 0 (this information is given in the question). We find out that $C = 2$.

rearranging the formula, we come to the solution as
\begin{equation*}
y = (t+2)  e^{2t}.
\end{equation*}
« Last Edit: September 26, 2014, 09:36:55 AM by Victor Ivrii »

#### Weiyang Guo

• Newbie
• Posts: 2
• Karma: 0
##### Re: Q1 problem 2 (Night sections)
« Reply #2 on: September 25, 2014, 10:49:35 PM »
haha, I saw the answer was wrong this afternoon (without the minus sign on 2t) and I thought I should go home and reply about this. But it's fixed so quickly.

#### Bowen Cheng

• Newbie
• Posts: 1
• Karma: 0
##### Re: Q1 problem 2 (Night sections)
« Reply #3 on: September 26, 2014, 08:01:30 PM »
I dont use intergrating factor for this problem, instead I use an alternative  approach
y' - 2y = e ^(2t) ---------(1)

--> z' -2z = 0 (assume the R.H.E. =0), then dz\z=2dt --> âˆ«dz/z =2dt --> ln |z| = 2t+C1

-->|z| = C1e^(2t) --> z = Ce^(2t), C = Â±C1

-->now let C = C(t), such that y= Ce^(2t)--------------(2)

plug (2) into (1), we get C'e^(2t) + 2Ce^(2t)- 2Ce^(2t) = e^(2t) --->C'e^(2t)  = e^(2t)

--->C' = 1 ---> C = t + C2  ---> y = Ce^(2t) = ( t + C2)e^(2t)   ----(3)

now  plug in (0,2) into  (3) -->2 =C2 -->therefore y = ( t + 2)e^(2t)