I got stuck, then got cofused while doing this homework problem last night. My answers compared to the book was totally wrong. Could someone please help?:(

Find the General Solution to the system

{x' = 4x + y + 2t

{y' = -2x+ y

- Aug 3rd 2007, 03:57 AMgoogoogagaVery Urgent!! Systems of Differential Equations: Method of Elimination
I got stuck, then got cofused while doing this homework problem last night. My answers compared to the book was totally wrong. Could someone please help?:(

Find the General Solution to the system

{x' = 4x + y + 2t

{y' = -2x+ y - Aug 3rd 2007, 06:03 AMtopsquark
I assume $\displaystyle y^{\prime} = \frac{dy}{dt}$ and the same for $\displaystyle x^{\prime}$?

First the bottom equation becomes:

$\displaystyle 2x = -y^{\prime} + y$

Then, from the bottom equation again:

$\displaystyle y^{\prime \prime} = -2x^{\prime} + y^{\prime}$

Inserting the value of $\displaystyle x^{\prime}$ from the top equation gives:

$\displaystyle y^{\prime \prime} = -2(4x + y + 2t) + y^{\prime} = -8x - 2y - 4t + y^{\prime}$

Now, $\displaystyle 8x = 4(2x) = 4(-y^{\prime} + y) = -4y^{\prime} + 4y$, so

$\displaystyle y^{\prime \prime} = -(-4y^{\prime} + 4y) - 2y - 4t + y^{\prime}$

$\displaystyle y^{\prime \prime} = 4y^{\prime} - 4y - 2y - 4t + y^{\prime}$

$\displaystyle y^{\prime \prime} - 5y^{\prime} + 6y = - 4t$

The solution for y is:

$\displaystyle y(t) = Ae^{3t} + Be^{2t} - \frac{2}{3}t - \frac{5}{9}$

Thus

$\displaystyle x(t) = \frac{1}{2} \cdot ( -y^{\prime} + y) = \frac{1}{2} \cdot \left ( -3Ae^{3t} - 2Be^{2t} + \frac{2}{3} + Ae^{3t} + Be^{2t} - \frac{2}{3}t - \frac{5}{9} \right )$

$\displaystyle x(t) = -Ae^{3t} - \frac{B}{2}e^{2t} - \frac{2t}{3} + \frac{1}{9}$

-Dan