Author Topic: Q1-T0401  (Read 1437 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

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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
$$y=(-te^{-t}-e^{-t})/t^4, \space t<0$$