Finding the solution of the nonhomogenous linar system of differential equations

Question

I have the system $$\dot{x_1}=x_1+2x_2-6, \enspace \enspace x_1(0)=1 \\ \dot{x_2}=6x_1-3x_2+24 \enspace\enspace x_2(0)=-1$$ and want to find the solution of this IVP. I wrote the system as $$\dot{x}=Ax+b(t)$$ where $$ A = \left( \begin{matrix} 1 & 2 \\ 6 & -3 \end{matrix} \right) \text{ and } b(t) = \left( \begin{matrix} -6 \\ 24 \end{matrix} \right). $$

I am studying from Perko's book and it says the following. If $e^{At}$ is any fundamental matrix solution of the system $\dot{x}=Ax$ then the solution of the nonhomogeneous linear system has the form $$ x(t)=e^{At}x_0 + e^{At} \int_{0}^{t} e^{-A\tau} b(\tau) d\tau. $$

To find $e^{At}$, I diagonalize $A$ by finding its eigenvalues $\lambda_1,\lambda_2$ and then corresponding eigenvectors $v_1,v_2$. So that I construct the matrix $P$ which has columns $v_1$ and $v_2$. Thus, I could be able to write $e^{At}=Pe^{Bt}P^{-1}$ where $B=diag(\lambda_1,\lambda_2)$. However, I think that there must be some easier way to find the solution as $b(t)$ does not depend on $t$ actually.

To sum up,

1. How can I find the solution of these type of systems where the nonhomogeneous part does not depend on the parameter?
2. How can I convert the system to a (probably) second order differential equation?

Answer 1

$$\dot{x_1}=x_1+2x_2-6, \enspace \enspace x_1(0)=1 \\ \dot{x_2}=6x_1-3x_2+24 \enspace\enspace x_2(0)=-1$$ Substract both differential equations: $${x'_1}-x'_2=-5x_1+5x_2-30$$ $$({x_1}-x_2)'=-5(x_1-x_2)-30$$ $$(({x_1}-x_2)e^{5t})'=-30e^{5t}$$ It's easy to integrate: $$({x_1}-x_2)e^{5t}=-6e^{5t}+c_1$$ $${x_1}=x_2-6+c_1e^{-5t}$$ Note that since we have : $$x_1(0)=1, x_2(0)=-1 \implies c_1=8$$ Substitute this in the second differential equation and solve to get $x_2$. Here substitute $x_1$: $$\dot{x_2}=6\color {red}{x_1}-3x_2+24 $$ $$\dot{x_2}=6\color {red}{(x_2-6+8e^{-5t} )}-3x_2+24 $$ Then solve the equation, it's a first order DE: $$\dot{x_2}-3x_2=48e^{-5t}-12 $$ $$({x_2}e^{-3t})'=48e^{-8t}-12e^{-3t} $$

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2. How can I convert the system to a (probably) second order differential equation?

Just differentiate one of the two differential equation and substitute in the second one. Then you have a second order linear differential equation that you can easily solve.