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As the title says, I am asked to specify a differential equation with the solution $y = 3 \sin(4x + v)$; boundary conditions are not required.

I have a question from my book and don't know how to deal with it! How do I do it? Thank you in advance...

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If this is homework, please tag it as such. Also, please include any work you've attempted or ideas you've had on the problem. –  mixedmath May 25 '11 at 14:35
    
oh no... it's not homework! By the way... it should be y = 3sin (4x + v) if it where y = sin (4x + v) i would easily extend it to: sin 4x + cos 4x but this isn't the situation. –  Adam May 25 '11 at 14:46
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2 Answers

If you want a second order differential equation then you can construct like $ y''+C\,y=0 $. Can you find what the constant $C$ is going to be for your case?

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I just did something in that style... Like i wrote in the comment to @mixedmath : So I could write something like this? 16y + y'' = 0 is this correct? Is it the final and only answer... was thinking about my exam that I will have later on. Doesn't they require some steps and so on? –  Adam May 25 '11 at 14:57
    
@mario - It is not the only answer. Just the simplest one. When it comes to diff. equs. typically you guess an answer and try to fit the coefficients to your boundary conditions. In this case your boundary conditions exist in the entire domain with $y(x)$ perscribed. The steps you need to show is how to evaluate $y'$, $y''$, and how they can be combined in one equation. –  ja72 May 25 '11 at 15:01
    
ok so with boundaries we have for example y(0) = 3 –  Adam May 25 '11 at 15:05
    
@Mario: for a second order equation you need two boundary conditions. To make $y(0)=3$ in your original equation you need $3sin v=3$. –  Ross Millikan May 25 '11 at 16:32
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I'll do a related question. Suppose I were asked to give a differential equation with answer $y = e^{2x}$. Then I might note that the second derivative of y is $4e^{2x}$ and the first derivative is $2e^{2x}$. But then y is just a solution to $y'' + 2y' = 8e^{2x}$.

Does that make sense?

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Why wouldn't the solution be $\dot{y}-2 y=0$ which is homogeneous and simpler to solve? –  ja72 May 25 '11 at 14:43
    
@mixedmath - my bad, the commend was supposed to go on the original question and I clicked the wrong button. I could not delete it, so I changed the comment. Sorry for the confusion. –  ja72 May 25 '11 at 14:47
    
yes so the solution for my question which is: Y = 3 sin (4x + v) y' = 12 cos 4x + v) y'' = - 48 sin (4x + v) –  Adam May 25 '11 at 14:48
    
@ja: No problem! I was worried briefly. And your question certainly has the same solution as mine, and is easier! –  mixedmath May 25 '11 at 14:49
    
@Mario: I was just giving an example. You could set up any sort of differential equation you wanted! You could make it of any order, non-linear, etc. Anything at all! –  mixedmath May 25 '11 at 14:50
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