let $$y''+p(t)y'+q(t)y=g(t)$$ then By the Abel theorem wronskian equals to :
$$W= c e^{\int p(t)dt}$$
Now We want to generalise this to a 3rd order differential equation where :
$$y'''+p_1(t)y''+p_2(t)y'+p_3(t)y=0$$
And The book follows the procedure below, I had no idea what is meant would be grateful if someone manages to explain :
First write down $W'$. Then, substitute for $y_1''',y_2''' \ and \ y_3'''$ from the differential equation; multiply the first row by $p_3$, multiply the second row by $p_2$ and add these to the last row to obtain : $$W'= -p_1(t)W$$
After this point I believe I can show the rest. However I didn't understand any word from the sentence above. Could someone help me out? Thanks!
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$\begingroup$The Wronskian of functions $f_1(x), f_2(x), \cdots, f_n(x)$ is defined as $$ W=\begin{vmatrix} f_1(x) & f_2(x) & \cdots &f_n(x)\\ f_1'(x) & f_2'(x) & \cdots &f_n'(x)\\ f_1''(x) & f_2''(x) & \cdots &f_n''(x)\\ \vdots\\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & \cdots &f_n^{(n-1)}(x) \end{vmatrix} $$ Now suppose you have a differential equation $$ y^{(n)}=p_1(x) y^{(n-1)}+\cdots + p_{n-1}(x)y+p_n(x)=\sum_{m=0}^{n-1}p_{n-m}(x)y^{(m)} $$ with the convention that $y^{(0)}=1$. Abel's theorem is all about calculating the Wronskian of the $n$ solutions of the above ODE, say $y_1,\cdots, y_n$. So let's just do that... First note that (prove it) $$W'=\begin{vmatrix} y_1 & y_2 & \cdots &y_n\\ y_1' & y_2' & \cdots &y_n'\\ y_1'' & y_2'' & \cdots &y_n''\\ \vdots\\ y_1^{(n-2)} & y_2^{(n-2)} & \cdots &y_n^{(n-2)}\\ y_1^{(n)} & y_2^{(n)} & \cdots &y_n^{(n)} \end{vmatrix}=\begin{vmatrix} y_1 & y_2 & \cdots &y_n\\ y_1' & y_2' & \cdots &y_n'\\ y_1'' & y_2'' & \cdots &y_n''\\ \vdots\\ y_1^{(n-2)} & y_2^{(n-2)} & \cdots &y_n^{(n-2)}\\ \sum_{m=0}^{n-1}p_{n-m}y_1^{(m)} & \sum_{m=0}^{n-1}p_{n-m}y_2^{(m)} & \cdots &\sum_{m=0}^{n-1}p_{n-m}y_n^{(m)} \end{vmatrix}$$ where the second equality comes from the differential equation. Now by using the properties of determinant (specifically its multi-linearily) $$ W'=\sum_{m=0}^{n-1}p_{n-m} \begin{vmatrix} y_1 & y_2 & \cdots &y_n\\ y_1' & y_2' & \cdots &y_n'\\ y_1'' & y_2'' & \cdots &y_n''\\ \vdots\\ y_1^{(n-2)} & y_2^{(n-2)} & \cdots &y_n^{(n-2)} \\ y_1^{(m)} & y_2^{(m)} & \cdots &y_n^{(m)} \end{vmatrix}= p_{1}W $$ I leave it to you to figure out how each equality in the above identity came to be. Now we have found that $W'=p_1(x)W$ which has a easy solution $W=c\int p_1(t)dt$. QED.
If this proof seems abstract to you, or if you are not familiar with properties of determinants, try proving each step in the above proof with the properties of $3\times 3$ determinants that you know.
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