Solving Integral Equations: Placing Arbitrary Constants in ln/e Solutions

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The discussion centers on solving the differential equation y' = 3y and the placement of arbitrary constants in the solution. One participant derives the solution as y = e^{3x} + e^{3c}, while the textbook states it should be y = ce^{3x}. The confusion arises from the proper handling of constants in logarithmic and exponential forms. A clarification is made regarding the properties of exponents, emphasizing that e^{x+y} equals e^x * e^y, not e^x + e^y. The conversation concludes with acknowledgment of the mistake and appreciation for the correction.
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This is a question...

For the following question:
y^{'}=\frac{dy}{dx}=3y

I get the solution...
\int \frac{1}{3y} dy = \int dx
\frac{1}{3}ln y = x + c
y = e^{3x}+e^{3c}

However the textbook example says the solution is...
y = ce^{3x}

My question is would my answer be incorrect? How should the arbitrary constant be placed in ln and e integral solutions?
 
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e^{x+y}=e^xe^y and not e^{x+y}=e^x+e^y. You made this mistake in your last line.
 
You have to take into account that
e^{x+y}=e^x \cdot e^y \neq e^x + e^y
To get a feeling for that relation, take for example
2^{3+4}=(2 \cdot 2 \cdot 2 )\cdot (2 \cdot 2 \cdot 2 \cdot 2 )=2^3 \cdot 2^4
 
Oh yes, that is correct. Silly mistake. Thanks.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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