Question

How do I turn y=Cx^2 back into the original differential equation?

Answer 1

why would you want to?

Answer 2

lets say we have an equation:

3 + x = 5

our solution is x=2

how do we turn x=2 back in to 3+x = 5 .... it doest make a lot of sense to me.

Answer 3

we can subsitute it back in to check of its a valid solution ... but then you would already have your equation to begin with.

Answer 4

This seems more like a concept-check sort of question to me, asking something more along the lines of "Do you understand what a differential equation is and how a solution relates to the original equation?"

As far as finding the original DE goes... We can do something like the following:
\[y=Cx^2=e^{C}e^{\ln x^2}=e^{\ln x^2+C}~~\iff~~\ln y=\ln x^2+C\]
Differentiating would give us
\[\frac{dy}{y}=\dfrac{2\,dx}{x}~~\iff~~\frac{dy}{dx}=\frac{2y}{x}\]
As for whether \(y=Cx^2\) is a solution unique to this equation, I'm not entirely sure...