I've got a very simple question which I'm failing to solve . Here it is:
$x'=4x+2y, y'=2x+4y, x(0) =-1, y(0)=2$
I'm just starting to learn Laplace Transformation and still very weak at it. Can anyone help me by showing steps which will lead me to the final solution?
$(x+y)'=6(x+y)\implies x+y=ce^{6t}$. So $x'=4x+2(ce^{6t}-x)=2x+2ce^{6t}$. Write $x=c_0e^{2t}+c_1e^{6t}$, then $4c_1=2c\implies c_1=\dfrac c2$. Hence $x=c_0e^{2t}+\dfrac c2e^{6t}$, which determines the solution for y. The constants may be determined by plugging in the initial values.