Proof that if Wronskian is nonzero

Question

Let $y_1$ and $y_2$ be real functions whose Wronskian is nonzero. Suppose that $Ay_1 + By_2= 0$. Prove that $A=B=0$

I have proved that $y_1, y_2 \neq 0$, if $A = 0 \rightarrow B =0$, and if $B = 0 \rightarrow A = 0$. However, how do I complete the proof by showing that both $A,B$ have to be zero. (i.e. can't be non-zero?)

Take a derivative of your equality. Get a system of linear equations with unknown $A, B$. Use Kramer rule to solve it. — markvs, Nov 09 '21 at 08:31

Kavi Rama Murthy · Accepted Answer · 2021-11-09T08:41:17.413

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If $A\neq 0$ then $y_1=-\frac B A y_2$ and you can see (from the definition of Wronskian) ) that the Wronskian) is $0$ in this case. [The first column is $-\frac B A$ times the second column, so the determinant is $0$]. Hence, $A=0$. Similarly, $B=0$.

edited Nov 09 '21 at 08:41

answered Nov 09 '21 at 08:30

Kavi Rama Murthy

311,013

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so are you plugging in $(-B/A)y_2$ and $(-A/B)y_1$ into the wronskian – rabito Nov 09 '21 at 08:40
Put $y_1=-\frac B A y_2$ and $y'_1=-\frac B A y'_2$ in the definition of the Wronskian @rabito – Kavi Rama Murthy Nov 09 '21 at 08:42
when computing I get $y_2y_1' - y_2'y_1$ which doesn't equal zero – rabito Nov 09 '21 at 09:06
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You did not substitute for $y_1$ in terms of $y_2$. $\det \left[\begin{array}{llll} {-\frac B A y_2} & {y_2} \ {-\frac B A y'_2} & {y'_2} \end{array}\right]=0$ @rabito – Kavi Rama Murthy Nov 09 '21 at 09:09

Proof that if Wronskian is nonzero

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