Use greens theorem

Question

Use greens theorem

ksaimouli · Answer

$$y''+y=t+1$$

ksaimouli · Answer

so, the general solution is $$C1 \cos(t)+ C2 \sin (t)$$ and let $$g(0)=0 ,g'(0)=1$$

ksaimouli · Answer

so the green's function is $$g(t)= \sec(t)$$

ksaimouli · Answer

$$\int\limits_{0}^{t} \sin(t-s) (s+1) ds$$

ksaimouli · Answer

@Loser66

ganeshie8 · Answer

@nerdguy2535  @Concentrationalizing

loser66 · Answer

I am sorry to say that I forgot how to do it.

ksaimouli · Answer

its fine but how to evaluate this integral $$\int\limits_{0}^{t}(\sin(t) \cos(s)-\sin(t)\cos(s))\frac{ 1 }{ \cos(s)} ds$$

ksaimouli · Answer

That is the result from greens function

loser66 · Answer

I don't get your question, is it not that sin ( t) cos (s) - sin(t) cos (s) =0?

ksaimouli · Answer

That's what I thought, but the answer is different. $$tsin(t) +\cos(t) \ln(\cos(t))$$

anonymous · Answer

Can't say I've really worked with boundary-value problems yet. But if we're just looking at the integral you came up with:
$$\int\limits_{0}^{t}\sin(t-s)(s+1)ds$$we can just do this by parts since we're only integrating with respect to s. 
u = (s+1) du = 1
dv = sin(t-s)  v = cos(t-s)
$$\int\limits_{0}^{t}\sin(t-s)(s+1)ds = (s+1)\cos(t-s) - \int\limits_{0}^{t}\cos(t-s)ds$$=
$$(s+1)\cos(t-s) + \sin(t-s)$$ from 0 to t
=
$$t+1-cost-sint$$ Somehow I doubt this is what you were looking for, but at least what I can offer with my limited knowledge.