Let y'=4 x. find all values of r such that y = rx^{2} satisfies the differential equation. if there is more than one correct answer, enter your answers as a comma separated list.

First we solve the differential equation:

y ' = 4 x

dy / dx = 4 x

dy = 4x * dx

Integrating both sides we have

int (dy) = int (4x * dx)

y = 4 (x^2/2)

y = 2x^2

Therefore, comparing both functions:

y = 2x ^ 2

y = rx ^ 2

We conclude that

r = 2

answer

The value of r that satisfies the differential equation is

r = 2

Y' = dy/dx = 4x

To obtain y we integrate wrt x, so y = 4 int (x)

y = 4 x^2/2 = 2x^2

But y = rx^2

So 2x^2 = rx^2

Comparing coefficients we find that r = 2