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.