Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.