# Find solution of y′(t) = ky(t) with y(1) = 5 y′(1) = 4. y = Ce^(kt)

Find solution of $y\prime \left(t\right)=ky\left(t\right)$ with $y\left(1\right)=5y\prime \left(1\right)=4$. $y=C{e}^{kt}$

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Szeteib

Write as $\frac{dy}{dt}=ky$.
$\int \frac{dy}{y}=k\int dt+c.$
$\mathrm{ln}\left(y\right)=kt+c.$
Therefore $k=\frac{\mathrm{ln}\left(y\right)-c}{t}.$ When $y=5,t=1,\mathrm{ln}\left(5\right)=k+c.k=\mathrm{ln}\left(5\right)-c.$
$y={e}^{kt+c}={e}^{t\left(\mathrm{ln}\left(5\right)-c\right)+c}.\frac{dy}{dt}=\left(\mathrm{ln}\left(5\right)-c\right){e}^{t\left(\mathrm{ln}\left(5\right)-c\right)+c}.$
When $\frac{dy}{dt}=4,t=1:4=5\left(\mathrm{ln}\left(5\right)-c\right).\mathrm{ln}\left(5\right)-c=0.8$, so $c=\mathrm{ln}\left(5\right)-0.8.$
Therefore k=0.8.
So $y={e}^{0.8t+\mathrm{ln}\left(5\right)-0.8}=5{e}^{0.8t}\left(0.4493\right)=2.2466{e}^{0.8t}$ approx.