Find the values of the parameter k for which the function y = e ^ kx is a solution to the differential equation: y ‘= y.

Let’s find the derivative of our given function: f (x) = (e ^ x) * (x ^ 3).

Using the basic formulas and rules of differentiation:

(x ^ n) ‘= n * x ^ (n-1).

(e ^ x) ‘= e ^ x.

(c) ‘= 0, where c is const.

(c * u) ’= c * u’, where c is const.

(uv) ‘= u’v + uv’.

y = f (g (x)), y ‘= f’u (u) * g’x (x), where u = g (x).

Thus, the derivative of our given function will be as follows:

f (x) ‘= ((e ^ x) * (x ^ 3))’ = (e ^ x) ‘* (x ^ 3) + (e ^ x) * (x ^ 3)’ = (e ^ x) * (x ^ 3) + (e ^ x) * 3 * x ^ 2 = (e ^ x) * (x ^ 3) + (e ^ x) * 3x ^ 2.

Answer: The derivative of our given function will be equal to f (x) ‘= (e ^ x) * (x ^ 3) + (e ^ x) * 3x ^ 2.