a function has a derivative of dy/dx=3x^2y. if f(2)=3e, what is f(3)?
please show all steps and reasoning, thank you so much!
please show all steps and reasoning, thank you so much!
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y' = 3x^2y
dy/y = 3x^2 dx
Integrate both sides.
ln(y) = x^3 + c
y = e^(x^3) + e^c
y = ce^(x^3)
f(2) = 3e Plug in 2 for x, and 3e for y.
3e = ce^8 Divide by e^8.
c = 3e^-7.
y = 3e^(x^3 - 7)
So at f(3). = 3e^20.
dy/y = 3x^2 dx
Integrate both sides.
ln(y) = x^3 + c
y = e^(x^3) + e^c
y = ce^(x^3)
f(2) = 3e Plug in 2 for x, and 3e for y.
3e = ce^8 Divide by e^8.
c = 3e^-7.
y = 3e^(x^3 - 7)
So at f(3). = 3e^20.
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Sorry that should be ln(y) = x^3 + c, then it becomes y = e^(x^3 + c). Then you have something that looks like e^(a + b) = e^a * e^b. So this becomes e^(x^3) * e^c. e^c is a constant, so that just becomes ce^(x^3).
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