I can't afford to take the class so I'm working my way though Tenenbaum & Pollard.

Anyway I can't figure out the process for this problem.

Find a differential equation whose solution is the given

15. (y-c)^2 = cx

Answer is given as
4x(y')^2 + 2xy' - y = 0

Solve for y rest is history badabing badaboom

first solve for y by taking the square root of each side then adding c
you get:

y = (cx)^(1/2) + c

because it will be needed, find y' now

y' = (1/2)c(cx)^(-1/2)

Now it is just a matter of verifying that 4x(y')^2 +2xy' - y = 0 is true for the values of y and y' that we have found.
4x[(1/4)c^2(cx)^-1] + 2x[(1/2)c(cx)^(-1/2)] -(cx)^1/2 - c = 0
simplifying terms, we find this to say
c + (xc)^(1/2) - (xc)^(1/2) - c = 0
0 = 0
seems legit

I'm not sure what exactly your question was asking or what info you were given to start with...it seems like a weird question for differential equations.
BUT this is showing that the given y (which had to be solved for) is a solution to the differential equation 4xy'^2 +2xy' -y = 0

Just read your question, don't think I answered it at all. I guess this is a sort of trial and error one.
You'd want to start by seeing if you can find something to multiply a y' and a y term by to produce a zero.
If that doesn't work, I guess you could introduce a y'^2 term to the mix and try to find coefficients as a function of x that would produce the zero.
If you're diligent enough (I don't know why you would need to solve a problem like this, though), you could eventually come up with the 4x and 2x and 1 coefficients that they got for y'^2, y', and y.
It seems like a pretty silly problem, I don't know why you would want to start at the solution and look for the differential equation, other than just as a way to really think about what's going on in a differential equation.

This post was edited by ringo794 on May 26 2016 08:13pm

Wrong

Basically derive the differential equation from the solution. I'm not given the differential equation, just the solution so they want you to eliminate c by using replacements and derivitives.

I can't figure out the process for this one. I entered the answer into wolfram and it didn't give me the function provided.

Differential equations are applicable in changing systems like thermodynamics, waves, etc.

Meteorologists also use it and I used PDE in biomathematics.

I pretty clearly said that in my post.