- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

y'&=y-5\quad y(0)=y_0\tag{given}\\

y'-y&=-5\\

u(x)&=\exp\int-1\, dx = e^{-t}\\

(e^{-t}y)&=-5e^{-t}\\

e^{-t}y&=-5\int e^{-t} dt = -5e^{-t}+c\\

&y=-5\frac{e^{-t}}{e^{-t}}+\frac{c}{e^{-t}}\\

y&=\color{red}{5+(y_0-5)e^t}

\end{align*}$

this is similiar to one I posted before but can't seem to get the book answer (red)

the $y_0$ is ???