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Qihui Huang

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TUT 0702 question
« on: September 27, 2019, 02:26:19 PM »
Quiz tut0702

QUESTION: $$y'+y=te^{-t}+1$$

follow the form $$y'+p(t)y=g(t)$$
$$p(t)=1$$

find the integrating factor: $$u(t)=e^{\int1dt}=e^t$$

mutiply u(t) on both sides $$e^ty'+e^ty=te^te^{-t}+e^t$$

integrating both sides $$\int{(e^ty')}=\int{te^0+e^t dt}$$
$$e^ty=\int{t dt}+\int{e^t dt}$$
$$e^ty=\frac{1}{2}t^2+e^t+c$$
$$y=\frac{1}{2}t^2e^{-t}+1+ce^{-t}$$

So, as $t\rightarrow \infty$, $y \rightarrow 1$, since the term with $e^{-t}$ goes to zero.

« Last Edit: September 27, 2019, 02:30:21 PM by Qihui Huang »