Toronto Math Forum

Question:
a) Find the general solution of the given system of equations.
b) Draw a directions field and a few of the trajectories.  In each of these problems, the coefficient matrix has a zero eigenvalue. As a result, the pattern of trajectories is different from those in the examples in the text.
\begin{align*}
x' =
\begin{pmatrix}
3 & 6 \\
-1 & -2 \\
\end{pmatrix}
x \\
\end{align*}

Solution:
\begin{align*}
det
\begin{pmatrix}
3-\lambda & 6 \\
-1 & -2-\lambda \\
\end{pmatrix}
&=0 \\
(3-\lambda)(-2-\lambda)-(-6) &=0 \\
\lambda^2-\lambda &=0 \\
\lambda(\lambda-1) &=0 \\
\lambda = 0 \quad or \quad \lambda = 1 \\
\end{align*}
Case 2: $\lambda$=0
\begin{align*}
\begin{pmatrix}
3 & 6 & \vdots & 0\\
-1 & -2 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
3 & 6 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
\frac{1}{2} & 1 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix} \\
\end{align*}
\begin{align*}
\Rightarrow \qquad
\begin{pmatrix}
x1 \\ x2
\end{pmatrix}
=
\begin{pmatrix}
-2 \\ 1
\end{pmatrix}
t
\end{align*}

Case 1: $\lambda$=1
\begin{align*}
\begin{pmatrix}
3-1 & 6 & \vdots & 0\\
-1 & -2-1 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
2 & 6 & \vdots & 0\\
-1 & -3 & \vdots & 0\\
\end{pmatrix}
\quad \sim \quad
\begin{pmatrix}
\frac{1}{3} & 1 & \vdots & 0\\
0 & 0 & \vdots & 0\\
\end{pmatrix}
\end{align*}
\begin{align*}
\Rightarrow \qquad
\begin{pmatrix}
x1 \\ x2
\end{pmatrix}
=
\begin{pmatrix}
-3 \\ 1
\end{pmatrix}
t
\end{align*}
Therefore, the general solution is
\begin{align*}
    y = c1 *
    \begin{pmatrix}
    -2 \\ 1
    \end{pmatrix}
    + c2 * e^t
    \begin{pmatrix}
    -3 \\ 1
    \end{pmatrix}
\end{align*}

Question: $$y'' + 9y' = 9\sec^2(3t), 0<t<\frac{\pi}{6}$$
We first solve the homogeneous part: $$r^2+9=0$$,
where $$r=\pm3i$$,
so $$y_c(t) = c_1\cos{3t}+c_2\sin{3t}$$
Then we use wronskian to solve non-homoogeneous part:
\begin{align*}
w_1 &=
\begin{vmatrix}
0 & \sin{3t} \\
1 & 3\cos{3t} \\
\end{vmatrix} = -\sin{3t} \end{align*}
\begin{align*}
w_2 &=
\begin{vmatrix}
\cos{3t} & 0 \\
-3\sin{3t} & 1 \\
\end{vmatrix} = \cos{3t}
\end{align*}
\begin{align*}
w &=
\begin{vmatrix}
\cos{3t} & \sin{3t} \\
-3\sin{3t} & 3\cos{3t} \\
\end{vmatrix} = 3\cos^2{3t} + 3\sin^2{3t} = 3
\end{align*}
Subsititues above into formula we get:
 $$\begin{align*}
Y_p(t) &= \cos{3t} \int{\frac{-\sin{3s}\times 9\sec^2{3s}}{3}ds} + \sin{3t} \int{\frac{\cos{3s}\times9\sec^2{3s}}{3}ds} \\
&= \cos{3t} \cdot -3\int{\sin{3s}\frac{1}{\cos^2{3s}}ds} + \sin{3t} \cdot 3\int{\cos{3s}\frac{1}{\cos^2{3s}}ds}\\
&= \cos{3t} \cdot -3\int{\tan{3s}\cdot\sec{3s}ds} + \sin{3t} \cdot 3\int{\sec{3s}ds}\\
&= \cos{3t} \cdot -\sec{3t} + \sin{3t} \cdot \ln{|\sec{3t}+\tan{3t}|} \\
&= -1 + \sin{3t} \ln{|\sec{3t}+\tan{3t}|}
\end{align*}$$
Finally, $$Y(t) = c_1\cos{3t}+c_2\sin{3t} + \sin{3t} \ln{|\sec{3t}+\tan{3t}|} -1 $$

a)
$$(r-2)(r-3)=0$$
$$r=2, r=3$$
$$y_c = c_1e^{2x}+c_2e^{3x}$$
$$y_{p}=A\sin{2x}+B\cos{2x}$$
$$y'_{p}=2A\cos{2x}-2B\sin{2x}$$
$$y''_{p2}=-4A\sin{2x}-4B\cos{2x}$$
Therefore, $$-4A\sin{2x}-4B\cos{2x}-10A\cos{2x}+10B\sin{2x}+6A\sin{2x}+6B\cos{2x}=52\cos{2x}$$
$$A=-5, B=1$$
$$y_{p}=c_1e^{-x}+c_2e^{6x}$$
$$y=c_1e^{2x}+c_2e^{3x}+\cos{2x}-5\sin{2x}$$
b)
$$y'=2c_1e^{2x}+3c_2e^{3x}-2\sin{2x}-10\cos{2x}$$
substitutes $$y(0)=0, y'(0)=0$$
$$c_1+c_2+1=0$$
$$2c_1+3c_2-10=0$$
$$c_2=12, c_1=-13$$
$$y=-13e^{2x}+12e^{3x}+\cos{2x}-5\sin{2x}$$

Find the general solution of y'' - y' - 2y = cosh(2t):
The associated homogenous equation is: $$\begin{align*} y'' - y' - 2y &= 0 \\ (r-2)(r+1) &= 0
 \end{align*}$$
$$ r=2, r=-1$$
Therefore, the complementary function is: $$y_c(t) = c_1e^{-t} + c_2e^{2t}$$
Since $$cosh(t) = \frac{1}{2}(e^t+e^{-t})$$ $$cosh(2t) = \frac{1}{2}(e^{2t}+e^{-2t}) $$
we could rewrite the funtion as: $$ y'' - y' - 2y = \frac{1}{2}(e^{2t}+e^{-2t})$$
Then we use the method of undetermined coefficients to find the solution for the non-homogeneous function above:
Let $$y_p(t) = Ae^{2t} + Be^{-2t}$$
Then we find $$ Ae^{2t}, c_2e^{2t} $$ has the same format, so we times t to get a new equation:
$$y_p(t) = Ate^{2t} + Be^{-2t}$$
Then, $$y_p'(t) = Ae^{2t} + 2Ate^{2t} - 2Be^{-2t}$$ $$y_p''(t) = 2Ae^{2t} + 4Ate^{2t} + 2Ae^{2t} + 4Be^{-2t}$$
Subsititutes back to the equation:
$$ 2Ae^{2t} + 4Ate^{2t} + 2Ae^{2t} + 4Be^{-2t} -Ae^{2t} - 2Ate^{2t} + 2Be^{-2t} -2Ate^{2t} - 2Be^{-2t} = \frac{1}{2}(e^{2t}+e^{-2t}) $$
Then we have: $$\begin{align*} 2A + 2A - A &= \frac{1}{2} \\ A &= \frac{1}{6} \\ 4B+2B-2B &= \frac{1}{2} \\ B &=\frac{1}{8}\end{align*}$$
Hence, the particular solution is:
$$y_p(t) = \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t} $$
Finally, $$\begin{align*} y(t) &= y_c(t) + y_p(t) \\ &= c_1e^{-t} + c_2e^{2t} + \frac{1}{6}te^{2t} + \frac{1}{8}e^{-2t} \end{align*}$$

Question: $$ \frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} + \frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \cdot \frac{dy}{dx} = 0 $$
First, we need to find $$ M_y(x,y) $$
Where, $$\begin{align*}
 M_y(x,y) &= \frac{\partial}{\partial y}M(x,y) \\
&= \frac{\partial}{\partial y}\frac{x}{{(x^2+y^2)}^{\frac{3}{2}}} \\
&= x \cdot \frac{\partial}{\partial y}(x^2+y^2)^{-\frac{3}{2}} \\
&= x \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2y \\
&= -\frac{3xy}{(x^2+y^2)^{\frac{5}{2}}}
 \end{align*}$$
Then for the $$N_x(x,y)$$
Where, $$\begin{align*}
N_x(x,y) &= \frac{\partial}{\partial x} \\
&= \frac{\partial}{\partial x}\frac{y}{{(x^2+y^2)}^{\frac{3}{2}}} \\
&= y \cdot \frac{\partial}{\partial x}(x^2+y^2)^{-\frac{3}{2}} \\
&= y \cdot ({-\frac{3}{2}}) \cdot (x^2+y^2)^{-\frac{5}{2}} \cdot 2x \\
&= -\frac{3yx}{(x^2+y^2)^{\frac{5}{2}}}
\end{align*}$$
Since $$M_y(x,y) = N_x(x,y)$$ are exact, there exist a Φ, where Φx = M, Φy = N.
Therefore, $$\begin{align*} \phi &= \int \frac{x}{{(x^2+y^2)}^{\frac{2}{3}}} dx \\
&=\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} dx + g(y)
 \end{align*}$$
$$\phi_y = \frac{y}{{(x^2+y^2)}^{\frac{2}{3}}} + g'(y) $$
Compare it to N(x,y), we know that g'(y) = 0, so g(y) = constant.
Therefore, $$\phi = \frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} + C $$
General Solution: $$\frac{-1}{{(x^2+y^2)}^{\frac{1}{2}}} = C$$