Toronto Math Forum

a)For critical points, we just set $x'=0$ and $y'=0$ which will yield four critical points: $P1=(0,0), P2=(-1,-1), P3=(2,2), P4=(-2,0)$. For linearization, we have to compute the Jacobian i.e the matrix of first derivatives of the functions. If $F=(2+x)(y-x), G=y(2+x-x^2)$, then $F_x=y-2x-2, F_y=2+x, G_x=y(1-2x), G_y=2+x-x^2$. Now we have got everything to find corresponding linearized systems, just use equation 13 from page 522 (book).

b) To determine the nature of solutions at each critical point, we just have to find the eigenvalues of the matrices we get from evaluating the Jacobian at the respective critical points.

For the point $P1$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-2$, so the the critical point is a saddle (look picture from handout), this is unstable.

For the point $P2$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+r+3=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P3$, we evaluate the Jacobian to find that the resulting matrix has complex eigenvalues (the characteristic equation is $r^2+4r+24=0$), so the the critical point is a spiral point, this is (asymptotically) stable as the real part of the eigenvalues is negative.

For the point $P4$, we evaluate the Jacobian to find that the resulting matrix has eigenvalues $r_1=2, r_2=-4$, so the the critical point is a saddle, this is unstable.

For visualization of solutions, go to the course homepage and look at the section of learning resources. Pick the link to math.rice, select PPLANE.2005.10, type in the equations and enjoy the picture! You can also see the answer to the bonus question from the picture.

We use Milman's method:

$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m-1}}{(m-1)!}+Q''(r)\frac{x^{m-2}}{2!(m-2)!}+...  \right)$$
In this case $Q=r^3-2r^2+4r-8$. As we see an exponent in the power of three, let $r=3$. We need to evaluate $Q(3)=13$. As there are no polynomial terms, let $m=0$.

Then $L(Ae^{3x})=13Ae^{3x}$ which implies $A=\frac{1}{13}$.

Solution is $Y_p=\frac{1}{13}e^{3x}$.

We use Milman's method to find a particular solution:
$$L\left[A\frac{x^m}{m!}e^{rx}\right]=Ae^{rx}\left(Q(r)\frac{x^m}{m!}+Q'(r)\frac{x^{m-1}}{(m-1)!}+Q''(r)\frac{x^{m-2}}{2!(m-2)!}+...  \right)$$
In this case
$Q=r^2-3r+2, Q'=2r-3, Q''=2$. We want to have $r=3$. So we evaluate that $Q(3)=2, Q'(3)=3, Q''(3)=2$.
Also, let $m=0$. Then $L(Ae^{3x})=2Ae^{3x}$. This implies that $A=1$.
In conclusion, solution is $Y_p=e^{3x}$.

For general solution to homogeneous equation, solve $r^2-3r+2=0$ yielding $r_1=2$, $r_2=1$.
So, $Y_{gen.hom}=c_1e^{2x}+c_2e^{x}$

General solution for the whole system is $Y_G=Y_{gen.hom}+Y_p=c_1e^{2x}+c_2e^{x}+e^{3x}$.
Condition $Z(0)=1$ yields that $c_1+c_2+1=1$. Condition $Z'(0)=0$ yields that $2c_1+c_2+3=0$. Solving these gives $c_1=-3, c_2=3$.

In conclusion we get that: $y=-3e^{2x}+3e^x+e^{3x}$.

We first solve the characteristic polynomial: $r^3+r=0$ which yields $r_1=0, r_{2,3}=+- i$. So, solution to the general homogeneous equation is
$$y_g=u_1+u_2\cos(t)+u_3 \sin(t)$$
Idea in method of parameters is to let $u_1,u_2,u_3$ to be functions and then plot the $y_g$ into the equation (in this case compute it's first and third derivative), solving for $u_1,u_2,u_3$.
This can be done, but textbook page 242 gives you a nice final formula: $$u'_m=\frac{gW_m}{W}$$

So, we use this - maybe this was the reason why I finished 7 minutes earlier.

It turns out that $W(1,\cos(t), \sin(t))=1$. $W_m$ is just the same wronskian but the m-th column is replaced by the column $(0,0,1)$. With this definition
$$W_1=1, W_2=-\cos(t), W_3=-\sin(t)$$
Using the previously stated formula and the fact that in this case $g=\tan(t)$:

$$u_1'=\tan(t) \implies u_1=-\ln(\cos(t))$$
$$u_2'=-\tan(t)\cos(t)=-\sin(t)\implies u_2=\cos(t)$$
$$u_3'=-\tan(t)\sin(t)=-\left(\frac{1-\cos^2(t)}{\cos(t)} \right)=-\left(\frac{1}{\cos(t)}-\cos(t)\right)\implies
 u_3=-\ln(\sec(t)+\tan(t))+\sin(t)$$
The third integral was the hardest. But in the homework problems you also had to compute $\int \frac{dt}{\cos(t)}$ which is surprisingly difficult. So, I had worked on the integral before and remembered the solution.

Now, we just have to plug $u_1,u_2,u_3$ back to the general solution of homogeneous equation $y_g$ and we obtain that the particular solution is:
$$Y_p=-\ln(\cos(t))+\cos^2(t)+\sin^2(t)-\sin(t)\ln(\sec(t)+tan(t))=-\ln(\cos(t))+1-\sin(t)\ln(\sec(t)+\tan(t))$$
General solution is therefore
$$Y_G=Y_p+c_1+c_2\cos(t)+c_3 \sin(t)$$