# Thread: Very Urgent!! Systems of Differential Equations: Method of Elimination

1. ## Very 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

2. Originally Posted by googoogaga
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
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