### Author Topic: Q1-T0401  (Read 2779 times)

#### Yingqi Wang

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##### Q1-T0401
« on: January 26, 2018, 12:43:48 AM »
My solution is in the attachment.

#### Meng Wu

• Elder Member
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• MAT3342018F
##### Re: Q1-T0401
« Reply #1 on: January 26, 2018, 09:19:01 AM »
$$\cases{ t^3y'+4t^2y=e^{-t}, & t<0\cr y(-1)=0}$$
First, we divide both sides of given equation by $t^3$, we get:
$$y'+{4\over t}y={e^{-t}\over t^3}$$
Since given differential equation has the form
$$y'+p(t)y=g(t)$$
Hence $p(t)={4\over t}$ and $g(t)={e^{-t}\over t^3}$$\\ First, we find the integrating factor \mu(t) \\ As we know. \mu(t)=\exp^{\int{p(t)dt}} \\ Thus, \mu(t)=\exp^{\int{{4\over t}dt}}=e^{4ln|t|}=t^4$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$t^4y'+4t^3y=te^{-t}$$
and $$(t^4y)'=te^{-t}$$
Integrating both sides:
$$\int{(t^4y)'}=\int{te^{-t}}$$
Thus, $$t^4y=\int{te^{-t}}$$
For $\int{te^{-t}}$, we use Integration By Parts:$\\$
Let $u=t, dv=e^{-t}$.$\\$
Then $du=dt, v=-e^{-t}$$\\ Hence,$$\int{te^{-t}}=uv-\int{vdu}\int{te^{-t}}=-te^{-t}-\int{-e^{-t}dt}\int{te^{-t}}=-te^{-t}-e^{-t}+c$$Thus$$t^4y=-te^{-t}-e^{-t}+c$$where c is arbitrary constant.\\ Now we divide both sides by t^4, we get the general solution:$$y=(-te^{-t}-e^{-t}+c)/t^4$$To satisfy the initial condition, we set t=-1 and y=0$$\\$
Hence, $$0={-(-1)e^{-(-1)}-e^{-(-1)}+c\over (-1)^4}$$
$$0=e-e+c$$
so $$c=0$$
Therefore, the solution of the initial problem is
$$y=(-te^{-t}-e^{-t})/t^4, \space t<0$$