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

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In summary, the conversation is about finding the general solution of y'+ay=0. The equations used are dy/dx=-ay, dy/y=-a dx, ∫ dy/y=∫ -a dx, ln(y)=-ax+C, and y=e^(-ax). The question is whether y=e^(-ax) or y=ce^(-ax) is the correct answer. The final response suggests taking the exponential of both sides to solve for y, but it is unclear what happened to the constant "C" in the solution.
  • #1
Math10
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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?
 
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  • #2
Math10 said:

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.
 
  • #3
Math10 said:

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?
 
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  • #4
Thank you, Hallsoflvy.
 

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