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solve the seperable equation: dx/dt=3xt^2
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cross multiply \[\implies \frac{dx}x = 3t^2 dt\] now integrate both sides does that help?
or do you need more help?
i did that and got ln x = t^3 + C but the back of the book has it as x = Cexp(t^3). How did they get there?
\[\Large \ln x = t^3 + C\] raise both sides by e \[\Large \implies e^{\ln x} = e^{t^3 + C}\] simplify... \[\Large \implies x = e^{t^3 + C}\] according to algebra...this is just \[\Large \implies x = e^{t^3} \times e^C\] constant raised to constant is just a constant \[\Large x = e^{t^3} \times C\] this is also written as \[\Large \implies x = C\text{exp} (t^3)\] did you follow that?
yes thank you for your help
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