How to solve differential equation $y'=2y$ when $y(7)=6$ ?
The answer is $6e^{2(x-7)}$, but i don't get how come it is $(x-7)$?
I know that it is from formula of $y(x)=Ce^{kx}$, where $C=6$ and $k=2$.
$\endgroup$1 Answer
$\begingroup$This is a Separable Equation
$$\displaystyle \int \dfrac{1}{y}~ dy = \int 2 ~dx \implies y(x) = ce^{2 x}$$
Using the I.C.
$$y(7) = c ~e^{2 \times 7} = c~ e^{14} = 6 \implies c = 6 ~e^{-14}$$
Substitute back
$$y(x) = c~ e^{2x} = 6~ e^{-14} e^{2x} = 6~ e^{-14 + 2 x} = 6 ~e^{2(x-7)}$$
$\endgroup$ 0
