# Given a function, how to obtain a differential equation?

I could obtain a differential equation upon eliminating arbitrary constants from this equation $y = e^x(A \cos x + B \sin x)$. Here are the steps. $$\frac{dy}{dx} = e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x) = y +e^x(-A \sin x + B \cos x)$$

$$\frac{d^2 y}{dx^2} = \frac{dy}{dx} + e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x) = \frac{dy}{dx}+\left(\frac{dy}{dx} - y\right) - y ,$$

or $$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$$ which is the required differential equation.

Now how do I obtain a differential equation for this one: $y = cx + c^2$? I am assuming $c$ is some arbitrary constant. Here are the steps that I've tried. $$\frac{dy}{dx} = c$$

$$\frac{d^2y}{dx^2} = 0$$ and I am stuck.

I am having trouble with this one too: $y = Ae^{3x} + Be^{2x}$ where A and B are constants.Let me edit this.For the above function differentiating w.r.t x first time gives $$\frac{dy}{dx} = 3Ae^{3x} + 2Be^{2x}$$ Differentiating again w.r.t x gives $$\frac{d^2y}{dx^2} = 9Ae^{3x} + 4Be^{2x}$$ What next?

All Right, I've edited as per request. And also I corrected the pointed mistake. Sorry for the inconvenience.

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The derivative of $cx+c^2$ with respect to $x$ is just $c$... the exponential solution with two constants requires you to differentiate twice. – J. M. Aug 31 '11 at 18:24
As for the question in the title: for general functions, it can be very hard to do... – J. M. Aug 31 '11 at 18:25
@alok, why don't you try typesetting the math in TeX (and I don't mean this in a negative way)? It's quite simple: the first step is simply to enclose the math expressions inside dollar (\$) signs. E.g., \$ e^x (A cos x + sin x) \$. This wouldn't be perfect, but would be a good first step; you can always improve once you're more comfortable. – Srivatsan Aug 31 '11 at 18:26 @Arturo There seems to have been too many simultaneous edits to the question, and I think I should take the blame. Also, one of your edits perhaps just disappeared because of mine. I apologize for this :-) (I hadn't noticed you made the changes until after I submitted mine.) – Srivatsan Aug 31 '11 at 19:05 @Srivatsan: Yours seems good enough. No worries. – Arturo Magidin Aug 31 '11 at 19:25 ## 1 Answer As for your first question, you just need to substitute$c$in your first equation: $$y = y'x + (y')^2$$ and you already have a differential equation whose general solution is your function$y = cx + c^2$. (Check this!) As for the second one, since it depends on two parameters,$A$and$B, it's a solution of a second order differential equation. So you should derive two times: \begin{aligned} y &= Ae^{3x} + B e^{2x} \\ y' &= 3Ae^{3x} + 2B e^{2x} \\ y'' &= 9Ae^{3x} + 4B e^{2x} \end{aligned} Now, you look at these equalities as one between vectors in\mathbb{R}^3$: $$\begin{pmatrix} y \\ y' \\ y'' \end{pmatrix} = A \begin{pmatrix} e^{3x} \\ 3e^{3x} \\ 9e^{3x} \end{pmatrix} + B \begin{pmatrix} e^{2x} \\ 2e^{2x} \\ 4e^{4x} \end{pmatrix}$$ This way, they say that you have three linearly dependent vectors in$\mathbb{R}^3$. So their determinant must be zero: $$\begin{vmatrix} y & e^{3x} & e^{2x} \\ y' & 3e^{3x} & 2e^{2x} \\ y'' & 9e^{3x} & 4e^{2x} \end{vmatrix} = 0$$ And this is your second order differential equation. EDIT. More explicitly, computing this determinant, we get: $$0 = y \begin{vmatrix} 3e^{3x} & 2e^{2x} \\ 9e^{3x} & 4e^{2x} \end{vmatrix} -y' \begin{vmatrix} e^{3x} & e^{2x} \\ 9e^{3x} & 4e^{2x} \end{vmatrix} +y'' \begin{vmatrix} e^{3x} & e^{2x} \\ 3e^{3x} & 2e^{2x} \end{vmatrix}$$ Etc. - Oh yaa..Silly me.. – alok Aug 31 '11 at 19:26 @alok: Another possibility would be the differential equation$y''=0\$. This has more solutions than the required one, though. – Rasmus Aug 31 '11 at 19:34
I have no idea as to what to do.All i'am able to do is differentiate the same function twice and leave it as it is. – alok Aug 31 '11 at 19:37
@Agusti:The back of my text gives d^2y/dx^2 - 5dy/dx + 6y = 0 How do i obtain this? – alok Aug 31 '11 at 20:36
Thinking a little bit. :-) – a.r. Aug 31 '11 at 23:53