Author Topic: TUT0601 quiz 1  (Read 482 times)

xuanzhong

  • Jr. Member
  • **
  • Posts: 12
  • Karma: 1
    • View Profile
TUT0601 quiz 1
« 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 $y=(t+2)e^{2t}$


« Last Edit: September 27, 2019, 04:43:55 PM by xuanzhong »