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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 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?2011-05-25
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    @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.2011-05-25
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    ok so with boundaries we have for example y(0) = 32011-05-25
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    @Mario: for a second order equation you need two boundary conditions. To make $y(0)=3$ in your original equation you need $3sin v=3$.2011-05-25
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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?2011-05-25
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    @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.2011-05-25
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    yes so the solution for my question which is: Y = 3 sin (4x + v) y' = 12 cos 4x + v) y'' = - 48 sin (4x + v)2011-05-25
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    @ja: No problem! I was worried briefly. And your question certainly has the same solution as mine, and is easier!2011-05-25
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    @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!2011-05-25
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    This makes me confused... hate when I don't get a concrete question. Guess this is what real life and problems are all about... So I could write something like this? 16y + y'' = 0 is this correct?2011-05-25
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    @Mario: Yep! ${}{}{}{}$2011-05-25
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    So during exams don't they require certain steps and so on for the solution? or should I just write down the second and first derivate and then my final soulution?2011-05-25
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    @Mario: This sort of question will never be asked of you on an exam - as it's altogether too simple. And if it is, it's very easy to check. They would just plug the solution in and see if it's correct. This question seems to be aimed solely at determining if you know what a differential equation is, and to get you oriented to solving for functions instead of constants. This is akin to how you will never be asked: "Find a problem whose solution is x = 2". It's too simple.2011-05-25
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    I understand... but if we take boundaries in account how would it be then?2011-05-25
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    Would it be something like this for example: y(0) = 52011-05-25
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    @Mario: Yep! ${}{}{}$2011-05-25
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    @Mixedmath, many thanks for your help! I really learned alot ;-)2011-05-25