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Function $y_1(x)$ and $y_2(x)$ are solutions to the equation $y''+\frac{1}{x}y'+(1-\frac{1}x^2)y=0$ in the open interval $(0,\infty$). It is given that the following conditions are fulfilled:

$y_1(1)=5.5$

$y'_1(1)=3.3$

$y_2(1)=3.5$

$y'_2(1)=2.1$

Please conclude, if possible, whether the functions $y_1(x)$ and $y_2(x)$ are linearly dependent or independent.

I simply looked at the value of the Wronskian, which is positive: because it refers to two functions that are the two solutions to a second order linear homogenous differential equation, I took that to mean that the Wronskian does not equal zero anywhere in the given interval. And therefore the functions are linearly independent.

I am not sure whether this interpretation is correct, as the textbook definitions are not 100% clear to me.

Thank you!

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Hint: check whether the Wronskian is zero or non-zero anywhere in the given interval. If it is zero at any point in the interval, then $y_1$ and $y_2$ are linearly dependent, otherwise they are independent.

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  • $\begingroup$ Thank you Nik! But wasn't there a law that says that if we are talking about two solutions to a second order homogenous differential equation, then the Wronskian is either positive or negative in that entire interval. Am I totally mistaken here? $\endgroup$ – dalta Jul 23 '19 at 19:37
  • $\begingroup$ No, all you need is to check whether the Wronskian is zero or non zero to determine whether the solutions of a given differential equation are dependent or independent. $\endgroup$ – Nik Jul 23 '19 at 19:45
  • $\begingroup$ Thanks again! Then my next question is: how does one check that? It seems nearly impossible to solve this equation, and Wolfram Alpha gives a solution that is so complicated that it does not seem that that is what they intended. I can not obtain any other information about the Wronskian unless I manage to compute $y_1$ and $y_2$, right? Sorry, it's my second week studying Differential Equations, so it is all brand new to me... $\endgroup$ – dalta Jul 23 '19 at 20:01
  • $\begingroup$ You can try solving this differential equation using variation of parameter method. Moreover, there is Abel's formula which may help you to compute Wronskian. $\endgroup$ – Nik Jul 24 '19 at 8:31

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