Question

Find the general solution and the solution that satisfies the initial condition y(0) = 0.
dy/dt + y = t^3 + sin(3t)

Answer 1

What I have is \[dy/dt = t^3 + \sin(3t)\]
\[y _{h}(t) = e ^{-t}\]

My problem is getting \[y _{p}(t)\]

So my guess would be \[y _{p}(t) = ct^4 + \alpha \cos(3t) + \beta \sin(3t)\]

but I can't get the correct answer. I looked at chegg and they used\[y _{p}(t) = at^3 + bt^2 + ct + d + \alpha \cos(3t) + \beta \sin(3t)\]
I have no idea how they got that. Can anyone explain this to me?

Answer 2

Well if you have y' and y when added together leaving you with t^3 and sin(3t) then you have to consider these functions and their derivatives. So really all that they're using for their guess is just all the possible derivatives of these functions with arbitrary constants. Luckily sine has a cyclic derivative that comes back to sine and polynomials eventually go to zero after taking their derivatives enough times.

Answer 3

Would I normally use all possible derivatives of the function if it's a polynomial?

Answer 4

do you see that superposition is working hidden in the background ?

Answer 5

y' + y = t^3

y' + y = sin(3t)

Answer 6

you can solve above separately and combine the solution i guess

Answer 7

but aren't you allowed to solve it using standard method for solving linear de ?

Answer 8

@Hammafer Yeah you have to use all the derivatives because although you want to say that y only has t^3 terms, now if you add a t^2 term it should take care of it. However when you take the derivative you now have a t^1 term to get rid of! So every time you add a term to account for the last term, you get a new term that has to go away. Eventually though you get down to the end and since constants go away when you take the derivative the cycle ends.

Answer 9

Method of undetermined coefficients only works if the right-hand side Q(x) has a *finite* number of linearly independent derivatives, and the "guess" yp should be Q(x) and *all* its linearly independent derivatives.