Toronto Math Forum

To find eigenvalues, $det(A-\lambda x)=(1-\lambda)(-3-\lambda)-(-2)(3)=0$,we can get $\lambda = -1-\sqrt{2}$ or $\lambda = -1+\sqrt{2}$
To find eigenvectors, when $\lambda=-1-\sqrt{2}$,
$\begin{pmatrix}
2+\sqrt{2} & 3 \\
-2 & -2+\sqrt{2}
\end{pmatrix}
~
\begin{pmatrix}
2 & 2-\sqrt{2} \\
0 & 0
\end{pmatrix}$
the eigenvector is $\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}$
so $e^{(-1-\sqrt{2})t}\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}
= e^{-t}(\cos\sqrt{2}t-i\sin\sqrt{2}t)\begin{bmatrix}2-\sqrt{2} \\ -2\end{bmatrix}
= e^{-t} (\begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + i\begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix})$
therefore, the general solution is $x(t)=c_1 e^{-t} \begin{bmatrix} 2\cos\sqrt{2}t-\sqrt{2}cos\sqrt{2}t \\ -2cos\sqrt{2}t \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} -2\sin\sqrt{2}t+\sqrt{2}sin\sqrt{2}t \\ 2sin\sqrt{2}t \end{bmatrix}$

(a)
Rewrite the equation:$y''+\frac{2x+2}{x(2x+1)}y'+\frac{2}{x(2x+1)}y=0$
$p(x)=\frac{2x+2}{x(2x+1)}, W=e^{-\int p(x) dx}$
then, by Abel's theorem, $W(y_1,y_2)(x)=ce^{- \int \frac{2x+2}{x(2x+1)}}dx$
We can first calculate the integral $\int \frac{2x+2}{x(2x+1)} dx$
let $\frac{A}{x}+\frac{B}{2x+1}=\frac{2x+2}{x(2x+1)},
then \frac{(2x+1)A+Bx}{x(2x+1)}=\frac{2x+2}{x(2x+1)}$
then we get A=2, B=-2, so the integral $ \int \frac{2x+2}{x(2x+1)} dx = \int 2(\frac{1}{x}-\frac{1}{2x+1}) dx = 2(lnx-\ln(2x+1))=\ln \frac{x^2}{2x+1}$
therefore, $W=ce^{-\ln \frac{x^2}{2x+1}}=c\frac{2x+1}{x^2}$
let c=1, $W=\frac{2x+1}{x^2}$
(b)
$y_1=x+1, y_1'=1, y_1''=0$, put $y_1, y_1'$ and $y_1''$ into the equation, we can get $(2x+1)x \cdot 0+(2x+2) \cdot 1+2(x+1)=0$; therefore; $y_1$ is the solution for ODE
Let another linearly independent solution be $y_2$
Since we know $W=y_1y_2'-y_2y_1'=(x+1)y_2'-y_2$
then we need to solve the equation $(x+1)y_2'-y_2=\frac{2x+1}{x^2}$
Rewrite the equation: $y_2'-\frac{1}{x+1}y_2=\frac{2x+1}{x^2(x+1)}$
$\mu(x)=e^{\int -\frac{1}{x+1} dx}=\frac{1}{x+1}$, multiply on both sides with $\mu(x)$
We can get $\frac{1}{x+1}y_2'-\frac{1}{(x+1)^2}y_2=\frac{2x+1}{x^2(x+1)^2}$
Integrate on both sides, we can get $\frac{1}{1+x}y_2 = -\frac{1}{x(x+1)}+c$
Then $y_2 = -\frac{1}{x}+c(x+1)$, let c=1, $y_2=-\frac{1}{x}+x+1$
(c)
the general solution $y=c_1(x+1)+c_2(-\frac{1}{x}+x+1)$
since y(-1)=1, we can get $c_2=1$
$y'=c_1+c_2+\frac{c_2}{x^2}$, and y'(-1)=0, so $c_1=-2$
Therefore, $y=-(\frac{1}{x}+x+1)$

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If the Wronskian W of f and g is $3e^{4t}$, and if $f(t)=e^{2t}$, find g(t)

Solution: We calculate the Wronskian of f and g:

$W =
\begin{vmatrix}
f(t) & g(t) \\
f'(t) & g'(t) \\
\end{vmatrix}
=
\begin{vmatrix}
e^{2t} & g(t) \\
2e^{2t} & g'(t) \\
\end{vmatrix} $
= $e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
$e^{2t}g'(t)-2e^{2t}g(t)=3e^{4t}$ \newline
Divided by $e^{2t}$ on both sides, we get $g'(t)-2g(t)=3e^{3t}$ \newline
$\mu(t)$ = $e^{\int-2dt}$ = $e^{-2t}$ \newline
multiply $e^{-2t}$ on both sides, $e^{-2t}g'(t)-2e^{-2t}g(t)=3$ \newline
$(e^{-2t}g(t))'=3$ \newline
Integral on both sides, $e^{-2t}g(t)=3t+C$ \newline
g(t) = $3te^2t+Ce^{2t}$

Question: e^x+（e^x  cot⁡(y)+2y csc⁡(y) ） y^'=0, find an integrating factor and solve the given equation.
Solution:
My=0.
Nx=∂/∂x [e^x  cot⁡(y)+2y csc⁡(y) ]=cot⁡(y) e^x.
We know the given equation is not exact, so we need to find μ(t).
R=(My-Nx)/M=(0-cot⁡(y) e^x)/e^x =-cot⁡(y).
μ(t)=e^(-∫Rdy)=e^(-∫cot⁡y dy )=e^ln⁡|sin⁡y | =sin⁡y.
Multiply sin⁡y on both sides.
We get sin⁡y e^x+(e^x  cos⁡y+2y) y^'=0.
There exists ψ(x,y) such that ψ_x (x,y)=sin⁡y e^x.
Integrating on both sides with x, ψ(x,y)=sin⁡y e^x+g(y).
Differentiate on both sides with y, ψ_y (x,y)=cos⁡y e^x+g^' (y).
Since ψ_y (x,y)=N=e^x  cos⁡y+2y, then g^' (y)=2y and so g(y)=y^2+c.
Combine all results above, we get cos⁡y e^x+y^2=c.