System of First Order ODE

[arepeace]

Find the solutions of
x' = x + 4y - e^t
y' = x + y + 2e^t
Given x(0)=4 and y(0)=5/4


Solution:
x' = x + 4y - e^t
(D-1)x - 4y = -e^t -------- (1)

y' = x + y + 2e^t
(D-1)y - x = 2e^t -----------(2)

Solve (1) and (2) I got
x" - 2x' - 3x = 8e^t ---------(3)
y" - 2y' - 3y = -e^t --------- (4)


Solve (3) and (4) using homogeneous equation and particular solution x = x_h + x_p​ and y = y_h +y_p I got
x = ae^3t + be^-t - 2e^t --------(5)
y = ae^3t + be^-t + 1/4*e^t ---(6)


Plug in x(0)=4 and y(0)=5/4 I got
(5): a + b = 6
(6): a + b = 1
Then of course I can't proceed. Can somebody tell me what went wrong? Thank you.

NOTE: the solutions are:
x = 4e^3t + 2e^-t - 2e^t
y = 2e^3t - e^-t + 1/4*e^t

 

[Ishuda]

Find the solutions of
x' = x + 4y - e^t
y' = x + y + 2e^t
Given x(0)=4 and y(0)=5/4


Solution:
x' = x + 4y - e^t
(D-1)x - 4y = -e^t -------- (1)

y' = x + y + 2e^t
(D-1)y - x = 2e^t -----------(2)

Solve (1) and (2) I got
x" - 2x' - 3x = 8e^t ---------(3)
y" - 2y' - 3y = -e^t --------- (4)


Solve (3) and (4) using homogeneous equation and particular solution x = x_h + x_p​ and y = y_h +y_p I got
x = ae^3t + be^-t - 2e^t --------(5)
y = ae^3t + be^-t + 1/4*e^t ---(6)


Plug in x(0)=4 and y(0)=5/4 I got
(5): a + b = 6
(6): a + b = 1
Then of course I can't proceed. Can somebody tell me what went wrong? Thank you.

NOTE: the solutions are:
x = 4e^3t + 2e^-t - 2e^t
y = 2e^3t - e^-t + 1/4*e^t

You found the correct eigenvalues (-1 and 3) of the appropriate matrix and the correct particular solution. What you did wrong was to continue to 'steps 2 and 3' properly for the homogeneous equation, see
http://en.wikipedia.org/wiki/Matrix_differential_equation#Solved_example_of_a_matrix_ODE

That is, your homogeneous solution as written is only partially correct. You must find the eigenvectors v₁ and v₂ to write the homogeneous solution as
\(\displaystyle \begin{pmatrix} x\\ y \end{pmatrix}\, =\, a\, e^{3t}\, \boldsymbol{v_1}\, +\, b\, e^{-t}\, \boldsymbol{v_2}\),
and use the initial conditions to solve for a and b

 

[arepeace]

Thanks. I was confused by expression e^t. Jeez, this is actually very easy though. :lol: