# [SOLVED]1.2.2a initial value problem

#### karush

##### Well-known member
\begin{align*}\displaystyle 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 ???

#### MarkFL

Staff member
\begin{align*}\displaystyle 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 ???
Note that:

$$\displaystyle -5\int e^{-u}\,du=5e^{-u}+C$$

And so after integrating, you should have:

$$\displaystyle e^{-t}y=5e^{-t}+C$$

Hence:

$$\displaystyle y(t)=5+Ce^{t}$$

Now, we are given:

$$\displaystyle y(0)=5+C=y_0\implies C=y_0-5$$

And so:

$$\displaystyle y(t)=5+(y_0-5)e^{t}$$

Does that make sense?

#### karush

##### Well-known member
Ok

I had the C thing not happening

I noticed we get a lot of views on these!