Answer to Question #178123 in Differential Equations for Rajan Jogani

Question

Find the general solutions of the following system

1. dx/dt = -4x+2y , dy/dt = 5x/2 + 2y

2. X' = ( -1 3

- 3 5) X

3. X' = ( 4 -5

5 -4)X

Answer 1

Consider a general system of linear ODE with two variables.

$\dot{X}(t)=AX(t)$, $X\in R^2$, $A\in M(2,R)$

The solution can be expressed via exponent of the matrix $(At)$ :

$X(t)=e^{At}X(0)$, where $e^{At}=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}A^n$

2. Let $A=\begin{pmatrix}-1 & 3 \\ -3 & 5\end{pmatrix}$ . The characteristic polinomial of A is $\lambda^2-4\lambda+4=(\lambda-2)^2$ .

By the Cayley–Hamilton theorem we have $(A-2I)^2=0$ . Then

$e^{At}=e^{2tI}e^{(A-2I)t}=e^{2t}\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}(A-2I)^n=e^{2t}(I+t(A-2I))$

$e^{At}=e^{2t}\begin{pmatrix}1-3t & 3t \\ -3t & 1+3t\end{pmatrix}$

Answer. $X(t)=e^{At}X(0)=e^{2t}\begin{pmatrix}1-3t & 3t \\ -3t & 1+3t\end{pmatrix}X(0)$ .

3. Let $A=\begin{pmatrix}4 & -5 \\ 5 & -4\end{pmatrix}$. The characteristic polinomial of A is $\lambda^2+9=0.$

By the Cayley–Hamilton theorem we have $A^2+9I=0$, i.e. $A^2=-9I$ . Then

$e^{At}=\sum\limits_{n=0}^{+\infty}\frac{t^{2n}}{(2n)!}A^{2n}+\sum\limits_{n=0}^{+\infty}\frac{t^{2n+1}}{(2n+1)!}A^{2n+1}=\sum\limits_{n=0}^{+\infty}\frac{(-9)^n t^{2n}}{(2n)!}I+\sum\limits_{n=0}^{+\infty}\frac{(-9)^nt^{2n+1}}{(2n+1)!}A=\sum\limits_{n=0}^{+\infty}(-1)^n \frac{(3t)^{2n}}{(2n)!}I+\frac{1}{3}\sum\limits_{n=0}^{+\infty}(-1)^n\frac{(3t)^{2n+1}}{(2n+1)!}A=\cos(3t)I+\frac{1}{3}\sin(3t)A$

$e^{At}=\frac{1}{3}\begin{pmatrix}3\cos 3t +4\sin 3t & -5\sin 3t \\ 5\sin 3t & 3\cos 3t -4\sin 3t\end{pmatrix}$

Answer. $X(t)=e^{At}X(0)=\frac{1}{3}\begin{pmatrix}3\cos 3t +4\sin 3t & -5\sin 3t \\ 5\sin 3t & 3\cos 3t -4\sin 3t\end{pmatrix}X(0)$

1. Assume that the matrix A has two different real roots $\lambda_1$ and $\lambda_2$ . The characteristic polinomial of A is $\lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2=0$. By the Cayley–Hamilton theorem we have $A^2-(\lambda_1+\lambda_2)A+\lambda_1\lambda_2I=0$. Let $X_1$ and $X_2$ be the eigenvectors of the matrix A, corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$ respectively. They form a basis of the vector space $R^2$, hence $X(0)=c_1X_1+c_2X_2$ with some unknown coefficients $c_1$ and $c_2$. Then

$e^{At}X_i=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}A^nX_i=\sum\limits_{n=0}^{+\infty}\frac{t^n}{n!}\lambda_i^nX_i=e^{\lambda_it}X_i$. So

$X(t)=e^{At}X(0)=e^{At}(c_1X_1+c_2X_2)=c_1e^{\lambda_1t}X_1+c_2e^{\lambda_2t}X_2$

Notice that

$(\lambda_1 I-A)X(0)=(\lambda_1 I-A)(c_1 X_1+c_2 X_2)=c_2(\lambda_1-\lambda_2) X_2$

$(\lambda_2 I-A)X(0)=(\lambda_2 I-A)(c_1 X_1+c_2 X_2)=c_1(\lambda_2-\lambda_1) X_1$

therefore,

$c_2X_2=\frac{1}{\lambda_1-\lambda_2}(\lambda_1 I-A)X(0)$,

$c_1X_1=\frac{1}{\lambda_2-\lambda_1}(\lambda_2 I-A)X(0)$ and

$X(t)=c_1e^{\lambda_1t}X_1+c_2e^{\lambda_2t}X_2=e^{\lambda_1t}\frac{1}{\lambda_1-\lambda_2}(\lambda_1 I-A)X(0)+e^{\lambda_2t}\frac{1}{\lambda_2-\lambda_1}(\lambda_2 I-A)X(0)$

If we take $A=\begin{pmatrix}-4 & 2 \\ 5/2 & 2\end{pmatrix}$, then the characteristic polynomial equals to $\lambda^2+2\lambda-13$, its roots are $\lambda_1=-1+\sqrt{14}$ and $\lambda_2=-1-\sqrt{14}$ .

$X(t)=e^{\lambda_1t}\frac{1}{\lambda_1-\lambda_2}(\lambda_1 I-A)X(0)+e^{\lambda_2t}\frac{1}{\lambda_2-\lambda_1}(\lambda_2 I-A)X(0)$ implies

$X(t)=\frac{e^{(-1+\sqrt{14})t}}{2\sqrt{14}}((-1+\sqrt{14}) I-A)X(0)-\frac{e^{(-1-\sqrt{14})t}}{2\sqrt{14}}((-1-\sqrt{14}) I-A)X(0)$

$X(t)=\frac{e^{-t}}{\sqrt{14}}((\sqrt{14}\cosh\sqrt{14}t) I-(I+A)\sinh \sqrt{14}t)X(0)$

$X(t)=e^{-t}((\cosh\sqrt{14}t) I-\frac{\sinh \sqrt{14}t}{\sqrt{14}}(I+A))X(0)$

Answer. $X(t)=e^{-t}((\cosh\sqrt{14}t) I-\frac{\sinh \sqrt{14}t}{\sqrt{14}}(I+A))X(0)$