Linearly dependent solution of $y'' + 9y =0$

Question

Which of following pair of function is linearly dependent pair of solution of $$y'' + 9y =0$$

1. $\sin 3x, \sin 3x - \cos 3x$
2. $\sin 3x + \cos 3x, 3\sin x-4\sin^3x $
3. $\sin3x, \sin(3x)\cos(3x)$
4. $\sin 3x+\cos 3x, 4\cos^3x -3\cos x$

I tried computing wronskian to see if it zero. My text book says answer is option $3$, but Wronskian of option $3$ is $-3\sin^33x $ which looks nonzero. Also option $1$ looks linearly dependent to me.

Answer 1

Compute the Wronskian as follows: $$ \begin{vmatrix} \sin 3x & \sin(3x)\cos(3x)\\ 3 \cos 3x & 3[\cos^2(3x) - \sin^2(3x)] \end{vmatrix} = \\ 3\sin(3x) \cos^2(3x) - 3 \sin^3(3x) - 3 \sin(3x) \cos^2(3x) = \\ -3 \sin^3(3x) $$ So it seems that you were indeed correct in your prediction that the functions, as given, are linearly independent.

Notice, however, that $y = \sin(3x)\cos(3x) = \frac 12 \sin(6x)$ is not actually a solution to the equation $y'' + 9y = 0$. This leads me to believe that one of the following occurred:

• This is a strangely worded "trick" question, and choice 3 was the answer because one of the functions was not a solution
• there was a typo (perhaps they meant $y = \sin(\frac 32 x)\cos(\frac 32 x)$).

Answer 2

I see your concern. Well, option 1) is a linearly independent set of solutions of the ODE. To see this write the relation $A\sin 3x+ B(\sin 3x -\cos 3x) =0$ and differentiate it. You get two equations in the two unknowns $A,B$. It is easy to see that this has only the trivial solution, $A=B=0$. Thus, independence.

Option 3) (the answer?) is not possible either. Again write $A\sin 3x+ B\sin 3x \cos 3x =0$ and differentiate it. You get two equations in the two unknowns $A,B$.Once again solving for B it is easy to see that $B=0$ and so it is also an independent set. Furthermore, $\sin 3x\cos3x$ is NOT EVEN a solution of the ODE.

Let's try 2). To see this note that $3\sin x - 4 \sin^3x$ is ACTUALLY a solution of the ODE (you need to play with trig identities here). So, setting $A(\sin 3x+\cos 3x)+B(3\sin x - 4 \sin^3x)=0$ and differentiating as before, then solving for $B$ you get $B=0$ as the determinant of the system is actually identically equal to $1$ (again use identities).

The last one, 4) should therefore be the answer: As before $4\cos^3x-3\cos x$ is a solution of the ODE. But proceeding as in 2) above and expanding we see that the determinant in this case is identically $-3$. Hence another independent set. So, unless I didn't have enough coffee yet, the answer should be "None of these".