Can anyone give some clues on how to solve this differential equation:
f '(x) = 3 * f(x)
Let me see if I can help...
If the problem is asking to solve for f(x)...
Then, I'd first ask myself, what function is equal to its derivative.
I only know of one function: e^x
Therefore, the answer must be of be form: e^(kx), where k is some constant.
Does that help?
f'(x) = 3f(x)
lets say y = f(x) for funsies
y' = 3y
(dy/dx) = 3y
dy/(3y) = dx
Integrate with respect to both sides...
right side = X+ C
Left side, , it equals (1/3)*(lny) (technically absolute value of y, but whatevs)
ln(y) = 3x + 3c ... which we can also say 3x + C, since C is an arbitrary constant
ln(y) = 3x+C
y = e^(3x+C)
y = (e^(3x))(e^C)
e^C is a constant, so we can say thats C
y = Ce^(3x)
Thank you guys, got it now.
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