Solving Differential Equations with Undetermined Coefficients

[EugP]

Homework Statement

1) y''' - y' = 2\sin{t}, find the general solution.

2) x^2y'' + 3xy' + 5y = 0, find the general solution.

I'm pretty sure I did #2 correct, but I'm stuck on #1. I can't find the particular solution. I would, however, like to know if I did #2 right.

Homework Equations

F(r) = r^2 + (\alpha - 1)r + \beta = 0

For r_1 \ and \ r_2 Complex conjugates:

y = C_1x^{\lambda}\cos(\mu\ln{x}) + C_2x^{\lambda}\cos(\mu\ln{x})

The Attempt at a Solution

1) First I found the homogenous solution:

y''' - y' = 0

r^3 - r = 0

r(r + 1)(r - 1) = 0

r_1 = 0, \ r_2 = -1, \ r_3 = 1

y = C_1 + C_2e^{-t} + C_3e^t

Now I try finding the particular solution by using the method of undetermined coefficients. Since y''' - y' = 2\sin{t}, I will assume y_1 = A_1tcost + A_2tsint, from this:

y_1' = (A_1 + A_2t)\cos{t} + (A_2 - A_1t)\sin{t}

y_1'' = (2A_2 - A_1t)\cose{t} - (2A_1 + A_2t)\sin{t}

y_1''' = (A_1t - 3A_2)\sin{t} - (3A_1 + A_2t)\cos{t}

Now I plug that into the original equation, simplify and get:

(2A_1t - 4A_2)\sint{t} - 2(2A_1 + A_2t)\csot{t} = 2\sin{t}

This is where I'm stuck. How do I find A_1 and A_2?

2) x^2y'' + 3xy' + 5y = 0

\alpha = 3, \ \beta = 5

So:

F(r) = r^2 + (\alpha - 1)r + \beta = 0

r^2 + 2r + 5 = 0

r_1 = -1 + 2i, \\ r_2 = 1 - 2i

So the general solution is:

y = C_1x^{-1}\cos(2\ln{x}) + C_2x^{-1}\sin(2\ln{x})

Any help would be greatly appreciated.

EDIT: I changed my assumption in #1. Since cos t and sin t are not solutions of the homogenous equation, I chose:

y_1 = A_1\sin {t} + A_2\cos{t}

y_1' = A_1\cos{t} - A_2\sin {t}

y_1'' = -A_1\sin {t} - A_2\cos{t}

y_1''' = -A_1\cos{t} + A_2\sin {t}

Now I plug back in and get:

-A_1\cos{t} + A_2\sin{t} - (A_1\cos{t} - A_2\sin {t}) = 2\sin{t}

2A_2\sin {t} - 2A_1\cos{t} = 2\sin {t}

A_2 - A_1\cot{t} = 1

So now I get A_1 = 0 and A_2 = 1

Therefore:

y = C_1 + C_2e^{-t} + C_3e^t + \cos{t}[/tex]

Is this correct?

 

[daveb]

Have you tried using just A1*sin(t) +A2*cos(t)?

 

[EugP]

Have you tried using just A1*sin(t) +A2*cos(t)?

Yes I just edited my first post to show that I did. Thanks.

 

[HallsofIvy]

Try differentiating y(t)= C₁+ C²e^{-t+ C₃et+ cos(t) three times and plug the result into the differential equation!}