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?
Okay, and now you solve for y by taking the exponential of both sides.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
? What happened to the "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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