# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-1 => Topic started by: xuanzhong on September 27, 2019, 03:06:48 PM

Title: TUT0601 quiz 1
Post by: xuanzhong on September 27, 2019, 03:06:48 PM
Find the solution of the given initial value problem.
$$\frac{dy}{dx}-2y=e^{2t},y(0)=2$$
$$\mu(t)=e^{\int-2dt}=e^{-2t}$$

Multiplying both sides by $\mu(t)$:
$$e^{-2t}*\frac{dy}{dx}\ -\ 2e^{-2t}y=e^{-2t}{*e}^{2t}$$
$$\frac{d}{dx}(e^{-2t}y)=1$$

By integrating both sides:
$$e^{-2t}y=t+c$$
$$y=\frac{t+c}{e^{-2t}}$$
$$y=(t+c)e^{2t}$$

since $y\left(0\right)=2$, then $2=\left(0+c\right)\ast e^0=c\ast1=c$

Hence the solution to the initial value problem is $y=(t+2)e^{2t}$