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- Apr 5th 2007, 08:50 PMtukeywilliamsDifferential Equations and Variation of Parameters
- Apr 6th 2007, 12:10 AMCaptainBlack
y'' - 4y' + 3y = e^-t + t.

The general solution to this is the sum of the general solution to

y'' - 4y' + 3y = 0,

and a particular integral of the original equation.

To find a particular integral look at a trial solution:

y(t) = A e^{-t} + ax^2 + bx +c.

Determine the coefficients by differentiating twice and substituting into

the original equation.

You will now have the general solution and you need only fit the initial conditions to find the solution you require.

RonL - Apr 6th 2007, 12:23 AMtukeywilliams
how would I do the second one? just guess and check?

- Apr 6th 2007, 04:47 AMCaptainBlack
No, use exactly the same method as before, find the general solution of

y^(4) = 0

(which is the general cubic), and a particular solution of

y^(4) = e^t,

which we might as well take to be y=e^t. So the general solution is:

y(t) = e^t + ax^3 + bx^2 + cx +d

Now fit the initial conditions.

RonL - Apr 6th 2007, 05:49 AMtukeywilliams
Is it possible to solve the last one by hand?

- Apr 6th 2007, 08:52 AMSoroban
Hello, tukeywilliams!

Here's the second one . . .

Quote:

y'''' = e^t, . y(0) = 1, .y'(0) = y''(0) = y'''(0) = 0.

We have: .y'''' .= .e^t

. . Integrate: .y''' .= .e^t + C1

. . Since y'''(0) = 0, we have: .e^0 + C1 .= .0 . → . C1 = -1

. . Hence: .y'''(t) .= .e^t - 1

Integrate: .y''(t) .= .e^t - t + C2

. . Since y''(0) = 0, we have: .e^0 - 0 + C2 .= .0 . → . C2 = -1

. . Hence: .y''(t) .= . e^t - t - 1

Integrate: .y'(t) .= . e^t - t²/2 - t + C3

. . Since y'(0) = 0, we have: .e^0 - 0²/2 - 0 + C3 .= .0 . → . C3 = -1

. . Hence: .y'(t) .= .e^t - t²/2 - t - 1

Integrate: .y(t) .= .e^t - t³/6 - t²/2 - t + C4

. . Since y(0) =1, we have: .e^0 - 0³/6 - 0²/2 - 0 + C4 .= .1 . → . C4 = 0

Therefore: .y(t) .= .e^t - t³/6 - t²/2 - t

- Apr 6th 2007, 09:36 AMThePerfectHacker
- Apr 6th 2007, 09:45 AMtukeywilliams
no I think he is correct. This is not the serpent in paradise.

you use y_p = Ate^2t only if we get L(y_p) = 0. - Apr 6th 2007, 10:02 AMCaptainBlack
Well I did actualy do a bit more than I presented in my post and am

fairly comfortable with the idea that my trial solution will give a particular

integal (that is I did the work and have it here - of course there could be

some algebraic infelicities which would lead to it being wrong, but I don't

see any).

[-1 is not a root of the indical equation]

RonL - Apr 6th 2007, 10:07 AMThePerfectHacker
What textbook you use?

- Apr 6th 2007, 10:08 AMJhevon
- Apr 6th 2007, 10:10 AMThePerfectHacker
For the last solution,

y'''+y'=sec t

You can let z=y'

To get,

z''+z=sec t

The general solution to homogenous equation is:

z=C_1sin(t)+C_2cos(t)

Then you can find particular via Lagrange's Variation of Parameters techiqnue.

Then you need to integrate the entire solution for z tto get y.

Quote:

Well I did actualy do a bit more than I presented in my post and am

fairly comfortable with the idea that my trial solution will give a particular

integal (that is I did the work and have it here - of course there could be

some algebraic infelicities which would lead to it being wrong, but I don't

see any).

- Apr 6th 2007, 10:14 AMCaptainBlack
- Apr 6th 2007, 10:16 AMJhevon
- Apr 6th 2007, 10:17 AMtukeywilliams
Differential Equations & Linear Algebra by Farlow Hall Mcdill West