Differential Equations: Third Test

Problem 1. Find the solution to the IVP

$$\left\{\begin{array}{ccccc}x'&=&2x+y\\ y' &=&-2x+2y\end{array... ...h}\;\;\; Y(0) = \left(\begin{array}{c}-1\\ 1 \end{array}\right)$$

Answer. The coefficient matrix is

$$A = \left(\begin{array}{rrr} 2 & 1\\ -2 &2 \end{array}\right)$$

So the characteristic equation is

$$\lambda^2 -(2+2) \lambda + 4+2 = 0$$

or

$$\lambda^2 - 4\lambda + 6 = 0$$

Its roots are

$$\lambda = \frac{4 \pm \sqrt{-8}}{2} = 2 \pm i \sqrt{2}$$

The general solution is

$$Y = c_1 e^{2t} \left(\begin{array}{rr} \cos(\sqrt{2} t)\\ -\... ...\sin(\sqrt{2} t)\\ \sqrt{2} \cos(\sqrt{2}t) \end{array}\right)$$

The initial conditions imply

$$Y(0) = \left(\begin{array}{rr} -1\\ 1 \end{array}\right) = c... ...)+ c_2 \left(\begin{array}{rr} 0\\ \sqrt{2} \end{array}\right)$$

which implies

$$c_1 = -1 \;\;\mbox{and}\;\; c_2 = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

So the solution is

$$Y = -e^{2t} \left(\begin{array}{rr} \cos(\sqrt{2} t)\\ -\sqr... ...\sin(\sqrt{2} t)\\ \sqrt{2} \cos(\sqrt{2}t) \end{array}\right)$$

Problem 2. Find the general solution to

$$\left\{\begin{array}{ccccc}x'&=&5x+2y\\ y' &=&-2x+y\end{array}\right.$$

Answer. The coefficient matrix is

$$A = \left(\begin{array}{rrr} 5 & 2\\ -2 & 1 \end{array}\right)$$

So the characteristic equation is

$$\lambda^2 -(5+1) \lambda + 5+4 = 0$$

or

$$\lambda^2 - 6\lambda + 9 = 0$$

Its roots are

$$\lambda = \frac{6 \pm \sqrt{0}}{2} = 3$$

One solution is

$$Y_1 = e^{3t} \left(\begin{array}{rr} 2\\ 3-5 \end{array}\right)= e^{3t} \left(\begin{array}{rr} 2\\ -2 \end{array}\right)$$

Since the roots are double the second solution is given by

$$Y_2 = e^{3t} \left(\begin{array}{rr} p\\ q \end{array}\right) + t e^{3t} \left(\begin{array}{rr} 2\\ -2 \end{array}\right)$$

The $p$ and $q$ are given by the system

$$\left\{ \begin{array}{llll} (5-3)p + 2 q &=& 2\\ -2 p + (1-3)q &=& -2 \end{array}\right.$$

this reduces to one equation

$$p + q = 1$$

Set $p = 0$ we get $q=1$ or

$$Y_2 = e^{3t} \left(\begin{array}{rr} 0\\ 1 \end{array}\right) + t e^{3t} \left(\begin{array}{rr} 2\\ -2 \end{array}\right)$$

The general solution is then

$$Y = c_1 e^{3t} \left(\begin{array}{rr} 2\\ -2 \end{array}\ri... ...c_2 t e^{3t} \left(\begin{array}{rr} 2\\ -2 \end{array}\right)$$

Problem 3. Consider the first order system

$$\left\{\begin{array}{ccccc}x'&=&xy^2+x^2\\ y' &=&y^2 +y\end{array}\right.$$

(1)

Find all the equilibrium points.

(2)

Use the Jacobian technique to classify the equilibrium points (source, sink, etc....).

Answer. The equilibrium points are

$$(0,0)\;\; (0,-1)\;\; (-1,-1)$$

The jacobian is

$$J(x,y) = \left(\begin{array}{cc} 2x+y^2 & 2xy \\ 0 & 2y+1 \end{array}\right)$$

So for $(0,0)$ we have

$$J(0,0) = \left(\begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right)$$

Its eigenvalues are 0 and 1. So no conclusion.

For $(0,-1)$ we have

$$J(0,-1) = \left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right)$$

Its eigenvalues are 1 and -1. So we have a saddle.

For $(-1,-1)$ we have

$$J(-1,-1) = \left(\begin{array}{rr} -1 & 2 \\ 0 & -1 \end{array}\right)$$

Its eigenvalues are -1 and -1. So we have a sink.

Problem 4. Consider the linear system

$$\left\{\begin{array}{llll}x'&=&4x+\alpha y\\ y' &=&x\end{array}\right.$$

(1)

For which values of $\alpha$, the origin is the only equilibrium point?

(2)

When the origin is the only equilibrium point, classify it as source, sink, etc...

Answer. The coefficient matrix is

$$A = \left(\begin{array}{rr} 4 & \alpha \\ 1 & 0 \end{array}\right)$$

First $(0,0)$ is the only equilibrium point if and only if the determinant of $A$ is not 0. That is we must have

$$- \alpha \neq 0 \;\;\mbox{or}\;\; \alpha \neq 0$$

So let us take $\alpha \neq 0$. In this case the characteristic equation is

$$\lambda^2 -(4+0) \lambda - \alpha = 0$$

or

$$\lambda^2 - 4\lambda -\alpha = 0$$

Its roots are

$$\lambda = \frac{4 \pm \sqrt{16 + 4 \alpha}}{2} = 2 \pm \sqrt{\alpha + 4}$$

The classification of the equilibrium point depends on these roots. So we have

(1)

If $\alpha + 4 < 0$, we have complex roots with real part being positive. So $(0,0)$ is a spiral source.

(2)

If $\alpha + 4 = 0$, we have double roots equal to 2. So $(0,0)$ is a source.

(3)

If $\alpha + 4 > 0$, we have two different real roots. One is obviously positive. Let us check the sign of the other one that is

$$\lambda = 2 - \sqrt{\alpha + 4}$$

It is easy to see that this roots is positive if and only if $\alpha < 0$. So we have:

(3.1)

if $-4 < \alpha < 0$, then $(0,0)$ is a source.

(3.1)

if $0 < \alpha$, then $(0,0)$ is a saddle.

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