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The fundamental set of solution $$y^{(4)} - 16y = 0$$

I worked this problem out but I was under the impression that I can apply the general method of characteristic equation to solve :

$$ r^4 - 16 = 0$$

$r = \pm 2 $ so my solutions are $e^{2t} $ and $e^{-2t}$ and then I can use the Wronskian to check if it is fundamental or not. If $$W(e^{2t}, e^{-2t}) \neq 0 $$ then it is fundamental (and also linearly dependent).

This is true. However, the book's solution to this problem is $$e^{2t}, e^{-2t}, cos(2t),sin(2t)$$

But I don't understand why there could be sinusoidal functions in the set of fundamental solutions since the gen. solution to the problem has no imaginary part.

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If I understood your question correctly, then you should look into the fact that $r^{4} - 16$ has FOUR roots, not two. –  AWertheim May 1 '13 at 18:08
    
The OP made a proper, well explained question which made it easy for us to identify his mistake. Why does he only have my upvote? –  Git Gud May 1 '13 at 18:15
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3 Answers

up vote 5 down vote accepted

Your mistake is here:

...$r = \pm 2 $ so my solutions are $e^{2t} $ and $e^{-2t}$...

Note that $r^4-16=(r^2-4)(r^2+4)=(r-2)(r+2)(r-2i)(r+2i)$.

Therefore a basis of solutions is given by $\{t\mapsto e^{-2t}, t\mapsto e^{2t}, t\mapsto \cos (2t), t\mapsto \sin (2t)\}$, on some non-degenerate interval $I$.

Theory gives that the above set is in fact a basis, so you don't need to compute the Wronskian to conclude this.

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@Babak S. Appreciated for the improvement. –  Git Gud May 1 '13 at 18:14
    
Sorry, I don't quite understand why I don't need to use Wronskian to conclude this. In which case then do I need to use the Wronskian to conclude that something is a fundamental set of solution? Thank you –  40Plot May 1 '13 at 18:16
    
@40Plot Usually it is presented in the theory that solutions obtained this way form a basis of solutions. Check this: "Applying this to all roots gives a collection of $n$ distinct and linearly independent functions". But if for some reason you're required to compute the wronskian, then do it. –  Git Gud May 1 '13 at 18:19
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Since $r^4=16$ is a fourth degree equation, it has 4 roots: $\pm2, \pm2i$. That's where $\sin(2t)$ and $\cos(2t)$ come from.

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Two excellent answers have been posted, but your question seems to suggest that you might understand that there are complex roots and object to their inclusion in the fundamental set of solutions. If that is the case, then I'll try to explain that here.

Paul's online math notes do a very nice job of describing how $e^{rt}, r \in \mathbb{C}$ actually "produce" real valued solutions, so I'll post that here first. http://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx

Nevertheless, I think there is another explanation which is really nice, and it comes from the fact that CCLDEs act as linear operators on solutions (CCLDEs involve repeated differentiation, and differentiation is a linear operation) - hopefully you are familiar with what a linear operator is, but if not, it can be explained. If we represent a differential equation as a linear operator $L$, then solutions to that differential equation are functions $y$ such that $L(y) = 0$. Since $L$ is linear, it follows that if $L(af(x) + bg(x)) = 0$ (where $a$ and $b$ are constants), then $aL(f(x)) + bL(g(x)) = 0$. Hence, if $\cos x + i\sin x$ is a solution to our CCLDE, then it follows that $L(\cos x + i \sin x) = 0$, and so $L(\cos x) + iL(\sin x) = 0$. But $L(\cos x)$ and $L(\sin x)$ are real-valued, and so we must have that $L(\cos x) = 0$ and $L(\sin x) = 0$. So $\cos x$ and $\sin x$ are in the fundamental set of solutions.

This might be a bit overkill, but hopefully it helps. (If anyone notices any egregious mistakes in my explanation, please edit!)

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