General Solution for y'+ay=0 | Find the Solution with Step-by-Step Guide

[Math10]

Homework Statement

Find the general solution of y'+ay=0.

Homework Equations

dy/dx=-ay
dy/y=-a dx
∫ dy/y=∫ -a dx
ln(y)=-ax+C
y=e^(-ax)

The Attempt at a Solution

What I want to know is that is y=e^(-ax) the answer or y=ce^(-ax) is the answer?

 

[verty]

Homework Statement

Find the general solution of y'+ay=0.

Homework Equations

dy/dx=-ay
dy/y=-a dx
∫ dy/y=∫ -a dx
ln(y)=-ax+C
y=e^(-ax)

The Attempt at a Solution

What I want to know is that is y=e^(-ax) the answer or y=ce^(-ax) is the answer?

Find which values of c solve the equation. If it is only true for c = 1, THEN y = e^(-ax) will be the answer.

 

[HallsofIvy]

Homework Statement

Find the general solution of y'+ay=0.

Homework Equations

dy/dx=-ay
dy/y=-a dx
∫ dy/y=∫ -a dx
ln(y)=-ax+C

Okay, and now you solve for y by taking the exponential of both sides.

y=e^(-ax)

? What happened to the "C"?

The Attempt at a Solution

What I want to know is that is y=e^(-ax) the answer or y=ce^(-ax) is the answer?

 

[Math10]

Thank you, Hallsoflvy.