1
$\begingroup$

I have this system of ODEs : \begin{align} &\mathbf{x}'= \begin{pmatrix}0 & 1 \\-\frac{1}{t^2} & \frac{1}{t}\end{pmatrix}\mathbf{x}+ \begin{pmatrix}0\\\frac{1}{t^2}\end{pmatrix}\\ \Rightarrow & x_1'=x_2,\\&x_2'=-\frac{1}{t^2}x_1+\frac{1}{t}x_2+\frac{1}{t^2}\\ \end{align} So I tried to solve this by converting this system into a 2nd order linear ODE. I used 2 approaches,

$1)$ Converting the system into a 2nd order linear ODE in terms of $x_1$, I eventually get$$\mathbf{x}=c_1\binom{t}{1}+c_2\binom{tln|t|}{1+ln|t|}+\binom{1}{0}$$ $2)$ Converting the system into a 2nd order linear ODE in terms of $x_2$, I eventually get$$\mathbf{x}=c_1\binom{tln|t|-t}{ln|t|}+c_2\binom{t}{1}+\binom{1}{0}$$ Everything is the same except for that one term. I thought they could be constant multiple of each other but I could not show that.

Can someone please enlighten me?

$\endgroup$

1 Answer 1

1
$\begingroup$

Your two answers are equivalent: $$\begin{eqnarray} \mathbf{x}&=&c_1\binom{t}{1}+c_2\binom{tln|t|}{1+ln|t|}+\binom{1}{0}\\ &=&c_1\binom{t}{1}+c_2\binom{tln|t|}{ln|t|}+c_2\binom{0}{1}+\binom{1}{0}\\ &=&c_1\binom{t}{1}+c_2\binom{tln|t|-t}{ln|t|}+c_2\binom{t}{1}+\binom{1}{0}\\ &=&c_2\binom{tln|t|-t}{ln|t|}+(c_1+c_2)\binom{t}{1}+\binom{1}{0}\\ &=&k_1\binom{tln|t|-t}{ln|t|}+k_2\binom{t}{1}+\binom{1}{0} \end{eqnarray}$$ where $k_1=c_2$ and $k_2=c_1+c_2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .