Find the solution $(x(t), y(t))$ of Cauchy problem

Question

Find the solution $(x(t), y(t))$ of Cauchy problem \begin{align*}&\frac{dx}{dt}=2x-y+2 , \ x(0)=0 \\ &\frac{dy}{dt}=3x-2y+2\sin t, \ y(0)=-7\end{align*} Is the solution unique?

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I have done the following:

Firstly, we consider the system \begin{align*}&\frac{dx}{dt}=2x-y \\ &\frac{dy}{dt}=3x-2y\end{align*} Using the characteristic polynomial we find $(x_1, y_1)=(a_1e^t, b_1e^t)$ and $(x_2, y_2)=(a_2e^{-t}, b_2e^{-t})$.

Replacing these in the system we get $b_1=a_1$ and $b_2=2a_2$.

For $a_1=a_2=1$, we get $a_1=b_a=1$ and $a_2=1, b_2=2$.

So the general solution of the homogeneous system is $$(x,y)=c_1(e^t, e^t)+c_2(e^{-t}, 2e^{-t})$$

We have the matrix $A=\begin{pmatrix}2 & -1 \\ 3 & -2\end{pmatrix}$.

For the eigenvalus $k=1$ we get the eigenvector $(1,1)$.

For the eigenvalus $k=-1$ we get the eigenvector $(1,3)$.

We are looking for a partial solution $\phi(t)=(\phi_1(t), \phi_2(t))$ of the non-homogeneous system in the form $$\phi_1(t)=c_1(t)e^t+c_2(t)e^{-t} \\ \phi_2(t)=c_1(t)e^t+c_2(t)2e^{-t}$$

Replacing $\phi$in the initial system we get $$c_1'(t)e^t+c_2'(t)e^{-t}=2 \\ c_1'(t)e^t+c_2'(t)2e^{-t}=2\sin t$$ Solving this system we get $$c_1(t)=-4e^{-t}+e^{-t}\sin t+e^{-t}\cos t$$ and $$c_2(t)=e^{t}\sin t-e^{t}\cos t-2e^t$$

Therefore we get $$(x,y)=c_1(e^t, e^t)+c_2(e^{-t}, 2e^{-t})+(\phi_1, \phi_2)$$ with $$\phi_1=-6+2\sin t \ \text{ and } \ \phi_2=-8+3\sin t-\cos t$$

Is everythinng correct?

Answer 1

I will use a different approach than the answer I gave above to see if it aligns.

From the first equation, we have

$$y = 2x - x' + 2 \implies y' = 2 x' - x''$$

We can find

$$x'(0) = 2 x(0) - y(0) + 2 = 9$$

Substituting the above into the second equation

$$x'' - x - 4 + 2 \sin t = 0, x(0) = 0, x'(0) = 9$$

Solving

$$x(t) = -2 e^{-t} + 6 e^t + \sin t - 4$$

We can use this $x(t)$ to find $y(t)$

$$y = 2 x - x' +2$$

This results in

$$y(t) = -6 e^{-t} + 6 e^t + 2 \sin t - \cos t -6$$

Alternate Method Using Example 2 (many other methods are also possible)

$$x_c(t) = c_1 e^{-t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} + c_2 e^{t} \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

This gives

$$X = \begin{pmatrix} e^{-t} & e^t\\ 3e^{-t} & e^t \end{pmatrix},~~~~ g = \begin{pmatrix} 2 \\ 2 \sin t \end{pmatrix}$$

From this we have

$$x_p(t) = X \int X^{-1} g~dt$$