General Solution of the first order differential equation

[Yr11Kid]

dy/dt + y = $$\infty$$ $$\sumSin(nt)/n^2$$ n=1


Ok still a bit new with all these symbols and stuff but that is the basic jist of it.

y(t) = yh(t) + yp(t) it what i thought about using to start off with, yh(t) = Acos2t + Bsin2t.
Then subbing yp(t) into the differential equation. Not really sure about how to do this, help much appreciated :)

 

[Yr11Kid]

Hey everyone,
This question kind of confused me, i think the general idea is to use y(t) = yh(t) + yp(t) and basically saying yh(t) = Acos2t + Bsin2t

to find yp(t) subbing into the differential equation and equating.

help much appreciated :)

 

[rmehta]

Here's an idea. It shouldn't be too hard to find particular solutions to dy/dt + y = sin(nt)/n^2, where n is an unspecified constant. Do that and then sum the solutions. You'll end up with a weird infinite series, but it should be convergent because of the n^2 in the denominator. And this is probably the best you can do, since the problem itself has a series as the nonhomogeneous term.

 

[hunt_mat]

Do you mean:
$$\frac{dy}{dt}+y=\sum_{n=1}^{\infty}\frac{\sin (nt)}{n^{2}}$$
This is an integrating factor question.

 

[HallsofIvy]

The general solution to the equation y'+ y= 0 is $y(t)= Ce^{-t}$

To find the general solution to the entire equation, look for a solution of the form
$$\sum_{n=1}^\infty A_n sin(nt)+ B_n cos(nt)$$

That will give you as sequence of equations to solve for $A_n$ and $B_n$. Once you have that, add to $Ce^{-t}$.

 

[HallsofIvy]

I am combining the two threads on the same thing.