Toronto Math Forum

Therefore, 𝐻(𝑥,𝑦)=1/2x⁴+x²y²-16x²1/2y⁴+4y²=C

$+$ missing

I also think the A for At$e^{t}$ should be 10,so the final solution for non home part should be $10te^{t}-e^{-t}+2cost-6sint$ which is same as Jingyi's

You mean $2\cos(t) + 6\sin(t)$ right?

I also think one of the particular solution should be $10te^{t}$ $L'(1) = 1$, by $AL'(1) = 10$ $A = 10$. In your solution the $e^{t}$ part had an extra $-2A$.

I agree with Jingze's solution. It is obvious that Boyu's answer is not (0, 0) at $t =0$

a)

b) Characteristic equation reads
$r = 1, -1$ where $-1$ is a repeated eigenvalue, hence the solutions are
After some row operations,
This agree with part a) where $c = 4$

c) Since $e^{-t}, te^{-t}$ are solutions to homogeneous equation, the form of particular solution is $At^2e^{-t}$, where
Hence the solution is

characteristic equation is
$$r^3 - r = 0\\
r(r+1)(r-1) = 0\\
r = 0, 1, \text{  or   } -1$$
The solution to homogeneous equations are $c_1e^t, c_2e^{-t}, c_3$
Let $y_1 = e^t, y_2 = e^{-t}, y_3 = 1$
$$W(y_1, y_2, y_3) = y_1'y_2'' - y_2'y_1'' = 1 + 1 = 2$$
Let $W_i$ be the determinant of wronskian with ith column substituted with $(0, 0, 1)$
$$W_1(t) = -e^{-t}\\
$$W_2(t) = e^{t}\\
$$W_3(t) = -2$$
given that $g(t) = \tanh(t)$
$$Y_1 = y_1 \int \dfrac{W_1(s)  g(s)}{W(s)}ds = \dfrac{-e^t}{2}(\int \tanh(s)e^{-t}ds) = -\dfrac{1}{2} + e^t\arctan(e^{-x})\\
Y_2 = y_2 \int \dfrac{W_2(s)  g(s)}{W(s)}ds = \dfrac{e^{-t}}{2}(\int \tanh(s)e^{t}ds) = \dfrac{1}{2} - e^{-t}\arctan(e^{x})\\
Y_3 = \int -tanh(s) ds =  -\ln(\cosh(x))$$
The general solution is therefore
$$y = c_1e^{-t} + c_2e^t + c_3 + e^t\arctan(e^{-x}) - e^{-t}\arctan(e^{x}) - \ln(\cosh(x))$$

Reformat equation into form $y'' + p(t)y' + q(t) = 0$
We can see that
$p(t) = \frac{\sin(t)}{\cos(t)}$
By the equation for wronskian
$$W(t) = \exp \bigl(\int p(t) dt\bigr) = \exp \bigl(\int \frac{\sin(t)}{\cos(t)} dt\bigr) = \exp (-\ln(\cos(t))+\ln (c_0)) = \frac{c_0}{\cos(t)}$$
where $c_0$ depends on choice of $y_2$ and $y_2$