Question

Solve the initial value problem:
ty' + 2y = sin(t) -> y(π/2)=1, t>0

Answer 1

I've just started studying ODEs, so it'd great if anyone solved it by integrating factors method

Answer 2

\[\Large y'+\frac{2}{t} y=\frac{\sin(t)}{t}\]
Can you tell from this how an integration factor must look like?

Answer 3

If I had to guess (I'm completely new to this), I'd guess 2/t.

Answer 4

Well, basically there is (as for many things) a formula to find this integrating factor:

\[\Large \mu(x)=\exp\left(\int p(x)dx \right)\]
Where \(p(x)\) is the function in front the linear term when the DE is in standard form:

\[\Large y'+p(x)y=q(x) \]

Answer 5

most people compute it via the formula, because it's easy to remember, if you want to derive it yourself though by doing it manually, we could take a look at this too.

Answer 6

The basic idea behind is that the LHS of the DE looks almost like a product rule of differentiation, integrating factors take advantage of that.

Answer 7

I noticed it. Does this mean that I'll have to do this? \[μ(x)*\frac{ dy }{ dx } + p(x)*μ(x)*y(x) = q(x)*μ(x)\]

Answer 8

exactly, do you want to do the general case or applied to your problem already? But you're on the right track anyway, the next thing you would want to do is compute

\[\Large\left(\mu(x)y\right)'=\mu(x)'y+y'\mu(x) \]
And then match coefficients

Answer 9

With the general case explanation I think I got it.

Answer 10

Just one moment 'till I solve it.

Answer 11

very good then :-) ! no hurries.

Answer 12

All right, Now I have that (yt²)' = t*sin(t). I suppose now I'll just have to integrate both sides?

Answer 13

exactly, you got it!

Answer 14

The solution for this DE is \[\frac{ \sin(t) - t*\cos(t) + C }{ t² }\]
putting in the value for y gives C = 1/4(pi²-4)

Answer 15

looks right to me

Answer 16

Reached the answer, many thanks!

Answer 17

you are welcome