y'' - 4y' + 3y = e^-t + t.Originally Posted by tukeywilliams
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
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
Hello, tukeywilliams!
Here's the second one . . .
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
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
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.
Yes, I did make a mistake. My first one in about 7 months.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).
  |
LinkBack URL
About LinkBacks
