Author Topic: TUT0601 quiz 1  (Read 627 times)

xuanzhong

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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 »