Integration question

Question

A solution y' = ky is the function P(x) = Ce^kx

If P'(x) = 1/5P(x) and P(0) = 20

Find C and K ?

Answer 1

Indeed,

if y = P(x) = C*e^(kx)

then

y' = (Ck) * e^(kx) = k * (C*e^(kx) = k* P(x) = k*y

y' = P'(x) = (1/5)*P(x)

implies that

y = P(x) = C*e^(kx)

y' = (Ck)*e^(kx)

(1/5)*P(x) = (Ck)*e^(kx)

(1/5)*C*e^(kx) = (Ck)*e^(kx)

1/5 = Ck

Since P(0) = 20 ---> 20 = C*e^(k*0) = C*e^(0) = C*1 = C

So C=20

1/5 = 20*k

k = 1/5 divided by 20 = 1/100

Checking....

The function is P(x) = 20 * e^(x/100) where C=20 and k=1/100

:P'(x) = 20/100 * e^(x/100) = 1/5*e^(x/100) = 1/5*e^(x/100)