I have the system:
$x'(t)=3x+2y+2e^{-t}$
$y'(t)=-2x+y+e^{-t}$
which I solve as:
$x'-3x+2y=2e^{-t}$
$-2x+y'+y=e^{-t}$
which gives:
$(D-3)x+2y=2e^{-t}$
$-2x+(D+1)y=e^{-t}$
We multiply the uper eqn. by 2, and the lower by (D-3) to eliminate the x terms:
$4y+(D^2-2D-3)y=3^{-t}$
Since D is the differential operator, this gives the equation:
$y''-2y'+y=3e^{-t}$
The general solution of this is simply
$q=\frac{2\pm\sqrt{4-4}}{2}=1$
which we plug in the form $y(t)=ce^{qt}\longrightarrow ce^t$.
- Now we have to find the specific solution, but the Wronskian, based on the two components of the general solution $c_1y_1(x)+c_2y_2(x)$:
W=\begin{vmatrix}
y_1(x) & y_2(x)\\
y_1'(x) & y_2'(x)
\end{vmatrix}
is based (un)fortunately on only one function, that is $ce^t$. So the Wronskian (determinant of this matrix) is 0.
So how do I proceed by variation of parameters when W=0?
Thanks!
Best Answer
$$y''-2y'+y=3e^{-t}$$ Rewrite it as: $$(y(t)e^{-t})''=3e^{-2t}$$ Integrate twice.
Note that the Wronskian is not zero: $$y_1(t)=e^t \implies y_2(t)=te^t$$ $$W(y_1,y_2)=W(e^t,te^t)$$ $$W(e^t,te^t)=e^t(e^t+te^t)-te^{2y}=e^{2t}$$