Differential Equation Word Problem involving y=Ce^(xk) (y=y')

Question

"The rate of change of y is proportional to y. Write and solve the differential equation that models the verbal statement."

This part of the problem is easy. My work is such:

$y'=ky$

$\frac{dy}{dx} = ky$

$dy = ky dx$

$\frac{dy}{y} = k dx$

Then by integrating, I get:

$y = e^{kx + c}$, where $c$ is some constant

Which can also be written as:

$y = e^{kx} e^c$

$y = ce^{kx}$

The next part is rather confusing though:

"Use this statement to find the value of y when x=10, if it satisfies when x=0, y=2, and when x=4, y=8"

How can you find a particular solution of this differential equation when you are missing information about c, k, and y? If it helps, I just started Calculus 2.

Thank you for your time in advance.

Answer 1

We are told that when $x=0, y=2$ so $2=c e^{0} \Rightarrow c=2$. In addition, we are given that when $x=4, y=8$, so we have $8=2e^{4k} \Rightarrow k=(\ln 4)/4$. The $y$ value you want when $x=10$ can now be figured out:

$$y=2 e^{10\frac{ \ln 4}{4}}=64$$

Answer 2

Seeing as you have the data points (0,2) and (4,8) your only real unknowns are c and k. The fundamental theorem of algebra says that we need two equations in order to solve for these 2 unknowns. Even Without knowing that you can see:

2 = ce^{k * 0}

2 = c

plugging in this as well as (4,8) to the equation yields:

8 = 2*e^4k

e^4k = 4

ln4 = 4k

k ≈ 0.345

from here now that all of our values are solved for plug in (10,y)

y = 2e^{0.345 * 10}

y ≈ 63

Done!

EDIT: I want to mention that the points (0,2) and (4,8) are conditions for the equation and not something the question wants you to show. Without them you wouldn 't have enough information to solve