MAT244-2018S > Quiz-1



Victor Ivrii:
Find the solution of the given initial value problem.
y' - y = 2te^{2t},\qquad y(0) = 1.

Junya Zhang:

Meng Wu:

This answer can not be improved anymore I think :O

Meng Wu:
Since given differential equation has the form
Hence $p(t)=-1$ and $g(t)=2te^{2t}$$\\$
First, we find the integrating factor $\mu(t)$ $\\$
As we know. $\mu(t)=\exp^{\int{p(t)dt}}$ $\\$
Thus, $\mu(t)=\exp^{\int{-1dt}}=e^{-t}$$\\$
Then mutiply $\mu(t)$ to both sides of the equation, we get:
$$e^{-t}y'-e^{-t}y=e^{-t}\cdot 2te^{2t}=2te^{t}$$
and $(e^{-t}y)'=2te^{t}$ $\\$
Integrating both sides:
Thus, $$e^{-t}y=\int{2te^{t}}$$
For $\int{2te^{t}}$, we use Integration By Parts:$\\$
Let $u=2t, dv=e^{t}$.$\\$
Then $du=2dt, v=e^{t}$$\\$
Hence, $$\int{2te^{t}}=uv-\int{vdu}$$
Thus $$e^{-t}y=2te^{t}-2e^{t}+c$$
where $c$ is arbitrary constant.$\\$
Now we divide both sides by $e^{-t}$, we get the general solution:
To satisfy the initial condition, we set $t=0$ and $y=1$$\\$
Hence, $$1=2(0)e^{2\cdot 0}-2e^{2\cdot 0}+ce^{0}$$
so $$c=3$$
Therefore, the solution of the initial problem is


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