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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.

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    THIS IS a very simple example for a DE which can be solved by integrating factor method :() – 2017-01-26
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    I am asked to find the general solution and then determine how solutions behave as t approaches infinity. – 2017-01-26
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    https://en.wikipedia.org/wiki/Integrating_factor very helpful to understand the integration factor method – 2017-01-26

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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?

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    Do I need to separate the variables so that I have $y dy$ on one side and $dt$ on the other side? – 2017-01-26
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    @user1917407 I've updated my answer. – 2017-01-26
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    This helped me, thank you very much. I think I need to revisit the product rule a bit. – 2017-01-26
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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 $$y = Ce^{-t/2} + 3t - 6. $$

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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 $y=3t-6$.

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    Can you include how $t\mapsto 3t-6 ?$ – 2017-01-26
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    So proud to tchat with Ramanujan :) If we look for a particular solution of the form $t\mapsto at+b$, it must verify, for all $t\in\mathbb{R}$ : $2a+(at+b)=3t$, which leads to $a=3$ and $b=-6$. So $f_0:t\mapsto3t-6$ is a solution of $y'+2y=3t$. Now, if $f$ is any solution of $y'+2y=3t$, then by subtracting the two equalities $2\,f'(t)+f(t)=3t$ and $2\,f_0'(t)+f_0(t)=3t$ we obtain that $f-f_0$ is a solution of $2y'+y=0$. – 2017-01-26
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$2y'+y=3t$

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 $v'=\frac{v}{2}$.