calculus II separable equations

Question

Find the solution to dy/dx = 5y satisfying y(3) = 1, what does y=

Answer 1

dy/dx = 5y

cross multiplies:

dy = 5y dx

separates:

(1/5)(dy/y) = dx

integrates:

(1/5) ln |y| = x+C

solves for y:

ln |y| = 5x + 5c

y = exp(5x + 5c)

y(3)=1 --> 1 = exp(15+5c)

ln 1 = 15 +5c

0 = 15+5c

-15 = 5c

c = -3

y = exp(5x-15)

check by differentiation:

dy/dx = 5 exp(5x-15) = 5y, so yes it checks

Answer 2

Move the equation around. dy/y=5dx. Now we can integrate both sides. Ln(y/y0)=5(x-x0) where (x0,y0) is a solution. Take the exponential of both sides and find y/y0=exp(5(x-x0)). That leaves y=y0*exp(5(x-x0)). Verify that x0=3 and y0=1 solves it: 1=1*exp(5*(3-3))=1*exp(0)=1. The answer is y=exp(5*(x-3)).