Solving this simple DE $y''=y'$ I have $$\int y'' dt = \int y' dt \Rightarrow y' = y + C \Rightarrow \int y' dt= \int (y + C  )dt \Rightarrow y = \frac{y}{2} + Cy + K$$
But not sure how the solution is $y(t)=a+be^{t}$.
EDIT: $$\int y'' dt = \int y' dt \Rightarrow y' = y + C$$
          $$y' - y=C$$
Let the integrating factor, $r(t)=e^{-t}$.
Multiply both sides of the DE by the integrating factor to get
$$e^{-t}y' - e^{-t}y = e^{-t}C$$
So $$(e^{-t}y)' = Ce^{-t}$$
Then $$ \int \frac{\mathrm{d} }{\mathrm{d} t}(e^{-t}y) dt = C\int e^{-t} dt $$.
From this I get the solution $y = Ke^{t} - \frac{1}{t}$.
Is this correct?
 A: Hint: Let $u = y'$ then solve $u'=u$.
A: when you are solving $y^\prime = y + C$ you need to separate the variable like $\frac{dy}{y+C} = dt$ 
the way you were doinf $\int y^\prime dt = \int (y+ C) \, dt Ct + \int y \, dt$ is correct so far, but you cannot write $\int y \, dt = y$ because you don't know how $y$ depends on $t.$ for all we know $y$ could be $t, t^2, \sin t, e^t, \cdots$  
A: For second-order ODEs like this there is a standard method to solve them using the characteristic polynomial. Suppose the solution is of the form $y(t) = e^{rt}$. Then $y' = re^{rt}$ and $y''=r^2e^{rt}$. Multiplying by $e^{rt}$, we have
$$e^{rt}(y'' - y') = r^2e^{rt} - re^{rt} = e^{rt}(r^2 - r) = e^{rt}r(r-1)=0.$$
Now, $|e^z|>0$ for any $z\in\mathbb C$, so this implies that $r(r-1)=0$. So we have two solutions; when $r=0$, $y_1(t) = 1$ and when $r=1$, $y_2(t)=e^t$. Here $a$ and $b$ are arbitrary constants. Since this is a homogeneous equation, any linear combination of $y_1$ and $y_2$ will also be a solution. Suppose $a,b\in\mathbb R$ and $y(t) = a + be^t$. Then $y' = y'' = be^t$, so $y$ is indeed a solution.
