(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

dy/dt=y((3t^2)-1), y(1)=-2

2. Relevant equations

Basic integrals

3. The attempt at a solution

integrate on both sides: dy/y=dt((3t^2)-1)

========>ln(y)=(t^3)-t+c

========>y=e^((t^3)-t+c)

========>y=e^((t^3)-t)e^(c)

I am not sure if its some e rule that I forgot but how do you get -2 from y(1), since e^C is always positive?

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# First order linear differential equations

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