Toronto Math Forum
MAT2442018S => MAT244Tests => Quiz1 => Topic started by: Yingqi Wang on January 26, 2018, 12:43:48 AM

My solution is in the attachment.

$$\cases{
t^3y'+4t^2y=e^{t}, & t<0\cr
y(1)=0}$$
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}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{{4\over t}dt}}=e^{4lnt}=t^4$$\\$
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, $$0={(1)e^{(1)}e^{(1)}+c\over (1)^4}$$
$$0=ee+c$$
so $$c=0$$
Therefore, the solution of the initial problem is
$$y=(te^{t}e^{t})/t^4, \space t<0$$