Linear Differential Equation - Initial Value Problem

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The discussion revolves around solving a linear differential equation as an initial value problem. The user initially struggles with the integration factor and the resulting equation after multiplication. They express confusion about the disappearance of a term in their calculations. Ultimately, they realize that the left side of the equation represents the derivative of the product of the integration factor and the function y, allowing for simplification. This realization resolves their initial confusion regarding the solution.
thaalescosta
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Hello, I'm struggling with a simple problem here.

It asks me to solve the following initial value problem:
fSfRWNn.png


So far I've calculated the integration factor μ(x) = ex-x2 and I multiplied both sides of the equation by it and got this:
http://www4c.wolframalpha.com/Calculate/MSP/MSP36171f59839bdf0g6cbd000044defig6e40h0ei0?MSPStoreType=image/gif&s=64&w=256.&h=47.

The problem I'm having is that the equation I got is different than the equation in the first line of the solution:
2zu70dx.png


I don't understand how the ex-x2⋅(1 - 2x)⋅y disappearedEDIT: Actually, nevermind. My brain was too lazy to realize that the left part of the equation is equal to the derivative of ex-x2⋅y so I could just simplify it
 
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Well done.
 
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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