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.
Best 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.