# Differential Equations and Variation of Parameters

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• Apr 5th 2007, 08:50 PM
tukeywilliams
Differential Equations and Variation of Parameters
• Apr 6th 2007, 12:10 AM
CaptainBlack
Quote:

Originally Posted by tukeywilliams

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 AM
tukeywilliams
how would I do the second one? just guess and check?
• Apr 6th 2007, 04:47 AM
CaptainBlack
Quote:

Originally Posted by tukeywilliams
how would I do the second one? just guess and check?

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 AM
tukeywilliams
Is it possible to solve the last one by hand?
• Apr 6th 2007, 08:52 AM
Soroban
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 AM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank

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

I think you made a mistake.

It should be,

y = Ate^{-t} + at^2+bt+c
• Apr 6th 2007, 09:45 AM
tukeywilliams
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 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I think you made a mistake.

It should be,

y = Ate^{-t} + at^2+bt+c

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 AM
ThePerfectHacker
What textbook you use?
• Apr 6th 2007, 10:08 AM
Jhevon
Quote:

Originally Posted by CaptainBlack
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

"infelicities"

what impressive vocabulary you have Captain!

(or is that a really common word that only I have never seen before?)
• Apr 6th 2007, 10:10 AM
ThePerfectHacker
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).
Yes, I did make a mistake. My first one in about 7 months.
• Apr 6th 2007, 10:14 AM
CaptainBlack
Quote:

Originally Posted by Jhevon
"infelicities"

what impressive vocabulary you have Captain!

(or is that a really common word that only I have never seen before?)

When you have been in this business as long as I have you build up an
impressive list of synonyms for mistake or error.

RonL
• Apr 6th 2007, 10:16 AM
Jhevon
Quote:

Originally Posted by CaptainBlack
When you have been in this business as long as I have you build up an
impressive list of synonyms for mistake or error.

RonL

indeed!
• Apr 6th 2007, 10:17 AM
tukeywilliams
Differential Equations & Linear Algebra by Farlow Hall Mcdill West
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