Toronto Math Forum
MAT2442014F => MAT244 MathTests => Quiz 1 => Topic started by: Victor Ivrii on September 24, 2014, 10:19:51 PM

Please post solution
2.6 p. 102, # 25 Solve the initial value problem
\begin{equation*}
y'2y=e^{2t},\qquad y(0)=2.
\end{equation*}

Firstly, I'd like to apologize for the really ugly notations and equations.. there is the solution! I rewrote itV.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*}

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. :)

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)

I dont use intergrating factor for this problem, instead I use an alternative approach
It is difficult to decrypt. Please rewrite (quote one of the previous posts to see how to display math properly), check spelling and fix your name