Solving a system of non-homogeneous differential equations....

[Vraellie]

So, it's my first time posting here, and I would be incredibly grateful for an extra pair of eyes to tell me if I'm understanding this correctly. The problem is as follows:



After putting my equations into standard form and using the modified kramer's method to find both solution's for x and y, I'm pretty unnerved by the answer, because it is the same answer, and although these equations are non-homogeneous, there is no particular solution. Usually, we have to express one of the constants, either c or d, in

 

[HallsofIvy]

The set of equations is
$\displaystyle \frac{dx}{dt}- \frac{dy}{dt}- x= 2e^t$ and
$\displaystyle \frac{d^2x}{dt^2}- \frac{dy}{dt}+ 3x- y= 4e^t$.

The first thing I would do is solve the first equation for $\displaystyle \frac{dy}{dt}$:
$\displaystyle \frac{dy}{dt}= \frac{dx}{dt}- x- 2e^t$. Replace $\displaystyle \frac{dy}{dt}$ by that in the second equation: $\displaystyle \frac{d^2x}{dt^2}- \left(\frac{dx}{dt}- x- 2e^t\right)+ 3x- y= 4e^t$ or $\displaystyle \frac{d^2x}{dt^2}-\frac{dx}{dt}+ 4x- y= 2e^t$.

Now, differentiate that equation again: $\displaystyle \frac{d^3x}{dt^3}- \frac{d^2x}{dt^2}+ 4\frac{dx}{dt}- \frac{dy}{dt}= 2e^t$. And, finally, replace that "dy/dt" with $\displaystyle \frac{dx}{dt}- x- 2e^t$ again: $\displaystyle \frac{d^3x}{dt^3}- \frac{d^2x}{dt^2}+ 4\frac{dx}{dx}- \frac{dx}{dt}+ x+ 2e^t= 2e^t$.

$\displaystyle \frac{d^3x}{dt^3}- \frac{d^2x}{dt^2}+ 3\frac{dx}{dt}+ x= 0$.

That is a linear homogeneous equation with constant coefficients. It has characteristic equation $\displaystyle r^3- r^2+ 3r+ 1= 0$. I will leave you to finish it. (Check my arithmetic!)