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

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?



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?

• $$x(t) = -2e^{-t} + 6 e^t + \sin t -4 \\ y(t) = -6 e^{-t} + 6 e^t + 2 \sin t - \cos t -6$$
– Moo
Oct 5, 2020 at 19:25
• So I must have done something wrong.right? @Moo Oct 5, 2020 at 19:27
• You can always check your answers (a great attribute of DEQs) by substituting your result into each original ODE and verifying equality and see. I think something is wrong - likely a simple algebra or two issue I am guessing. Your eigenvalues and eigenvectors are correct.
– Moo
Oct 5, 2020 at 19:29
• The equation $b_2=2a_2$ is wrong, as $(a_2,b_2)$ has to be a right eigenvector to $k=-1$. If you check again you will find that $b_2=3a_2$ follows also from insertion of the basis solutions. Oct 5, 2020 at 20:08

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

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