Diff. EQ. Inhomogeneous Linear System (Double check please)

[VinnyCee]

Here is the problem:

$$X'\,=\,\left(\begin{array}{cc} 0 & 2 \\ -1 & 3 \end{array}\right)\,X\,+\left(\begin{array}{c} 2 \\ e^{-3t}\end{array}\right)$$

Here is what I have:

$$r_1\,=\,2,\,\,\,\,\,\,r_2\,=\,1$$

$$u_1'\,=\,-2\,e^{-2t}\,+2\,e^{-5t}$$

$$u_2'\,=\,2\,e^{-t}\,-\,e^{-4t}$$

$$\psi\,=\,\left(\begin{array}{cc} e^{2t} & 2\,e^t \\ e^{2\,t} & e^t \end{array}\right)$$

After integrating the $u_1$ and $u_2$ terms and multiplying the answer by $\psi$, I get:

$$X\,=\,\left(\begin{array}{c} C_1\,e^{2t}\,+\,2\,C_2\,e^t\,+\,\frac{1}{10}\,e^{-3t}\,-\,3 \\ C_1\,e^{2t}\,+\,C_2\,e^t\,-\,\frac{3}{20}\,e^{-3t}\,-\,1 \end{array}\right)$$

Then seperating and combining terms:

$$X\,=\,C_1\,e^{2t}\,\left(\begin{array}{c} 1 \\ 1 \end{array}\right)\,+\,C_2\,e^t\,\left(\begin{array}{c} 2 \\ 1 \end{array}\right)\,+\,\frac{1}{20}\,\left(\begin{array}{cc} 2\,e^{-3t}\,-\,60 \\ 3\,e^{-3t}\,-\,20 \end{array}\right)$$

Does this look correct? Thanks.

 

[VinnyCee]

Any ideas?

Anyone have the same thing when solving this problem?

 

[saltydog]

Anyone have the same thing when solving this problem?

Hello VinnyCee. Takes a while. It's a challenging problem. Your answer is NOT correct: Missing a minus sign for the second equation. You know, you can back-substitute the answer into the original ODE to verify the answer or if you're lazy like me, just input the results into Mathematica and have it back-substitute and compare the results.

I get:

$$x_p(t)=-3+1/10 e^{-3t}$$

$$y_p(t)=-1-3/20e^{-3t}$$

Hope you didn't get my first (unedited) post where I said it was correct. Didn't notice the minus sign missing.

 

[VinnyCee]

Could you show me that?

This backsubstitution that you mention. How do you type this? Can you show me the routine you did in Mathematica, so I can double check others also?

 

[saltydog]

This backsubstitution that you mention. How do you type this? Can you show me the routine you did in Mathematica, so I can double check others also?

x[t_]:=-3+1/10 e^(-3 t);
y[t_]:=-1-3/20 e^(-3 t);
D[x[t],t]
Simplify[2 y[t]+2]
D[y[t],t]
Simplify[-x[t]+3 y[t]+e^(-3 t)

Just define the solutions (x[t] and y[t]), calculate the derivatives, then insert the x[t] and y[t] into the right hand side and compare the resuts.