General solution for the differential equation $2y' + y = 3t$? I don't understand how to use the integrating factor to solve this equation. I just need a walkthrough with this simple example so I can do the rest of my problems.
 A: remember the product rule fro derivatives? it says that $(ay)' = ay' + a'y.$ the idea of the integrating factor is to figure out what a multiplication factor $a$ is so that $a(2y'+y)$ is $2(ay)'.$ so we are solving for $$ 2ay'+ay = (ay)' = 2ay'+2a'y \text{ for } a. $$ canceling $2a'y$ and dividing through by $y$ leaves you with $$a = 2a' \to a = e^{t/2} \text{ is one solution for }a.  $$
now, go back to the equation $$2y'+y = 3t\to a(2y'+y) = 3at \to (ay)'=3at $$ and has the general solution $$e^{t/2}y = 3\int t e^{t/2} dt + C$$ 
i will let you do the integration by parts to find the rhs.
note that it is much is easier to do this using the facts:
(a) solution to the hgs problem $2y' + y = 0 \text{ has the solution } y = Ce^{-t/2} $  and
(b) by looking for a particular solution of $2y' + y = 3t\text{ in the form of  } y = at + b $ and finding out that $a = 3, b = -6$ 
(c) general solution is $$y = Ce^{-t/2} + 3t - 6. $$
A: Clearly, $t\mapsto3t-6$ is a solution. Therefore, the solutions are given by $t\mapsto3t-6+K\,e^{-t/2}$ where $K\in\mathbb{R}$ is an arbitrary constant.
This prove that, for any solution, we have : $\lim_{t\to\infty}f(t)=+\infty$
It should be added that the graphs of all solutions share a common asymptote, namely the line $y=3t-6$.
A: $2y'+y=3t$
Divide both sides by 2
$y'+\frac{1}{2}y=\frac{3}{2}t$
Now you need to multiply both sides by a function $v$ not equaled to 0.
$vy'+\frac{v}{2} y=v \frac{3}{2}t$
We need to choose $v$ such that we can write $vy'+\frac{v}{2}y$ as $(vy)'$ 
Recall $(vy)'=vy'+v'y$
So comparing $vy'+\frac{v}{2}y$ to $vy'+v'y$ we need to select $v$
such that it satisfies $v'=\frac{v}{2}$.
A: The integrating factor can be applied to a differential equation in the form:
$$\frac{dy}{dt}+P(t)y=Q(t)$$
Therefore, we will rewrite our differential equation as follows:
$$\frac{dy}{dt}+\frac{1}{2}y=\frac{3}{2}t$$
The integrating factor is given by:
$$\mu(t)=e^{\int P(t) dt}=e^{\int \frac{1}{2} dt}=e^{\frac{t}{2}}$$
Multiply the differential equation by $\mu(t)$:
$$e^{\frac{t}{2}} \frac{dy}{dt}+\frac{e^{\frac{t}{2}}}{2}y=\frac{3}{2}te^{\frac{t}{2}}$$
Now, you can substitute $\frac{d}{dt} \left(e^{\frac{t}{2}}\right)=\frac{e^{\frac{t}{2}}}{2}$
$$e^{\frac{t}{2}} \frac{dy}{dt}+\frac{d}{dt}\left(e^{\frac{t}{2}}\right)y=\frac{3}{2}te^{\frac{t}{2}}$$
Note that we can now use the product rule $f\frac{dg}{dt}+\frac{df}{dt}g=\frac{d}{dt} (f\cdot g)$ on the left hand side to obtain (let $f=e^{\frac{t}{2}}$ and $g=y$):
$$\frac{d}{dt}(e^{\frac{t}{2}}y)=\frac{3}{2}te^{\frac{t}{2}}$$
We can now integrate both sides with respect to $t$:
$$\int \frac{d}{dt}(e^{\frac{t}{2}}y)~dt=\int \frac{3}{2}te^{\frac{t}{2}}~dt$$
Can you continue?
