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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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    @Srivatsan: Yours seems good enough. No worries.2011-08-31

1 Answers 1

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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.

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    @alok Simplify the final equation in the answer by evaluating each determinant. Then you should be able to cancel off a $e^{3x} \cdot e^{2x} = e^{5x}$ factor from the equation to get exactly what you wanted.2011-08-31