Find solution of differential equation $y'(t)=-2y(t)+1$ Could you help me explain how to find the solution of the differential equation
$$
y'(t)=-2y(t)+1, 
$$
with 
$$y(0)=1.$$
I know that the solution is 
$$y(t)=\frac12 (1+e^{-2t}).$$
How about the IVP
$$y'(t)=1,\quad y(0)=1?$$
 A: So we have $$\frac{dy}{dt} = -2y + 1 \implies \frac{1}{-2y+1}dy = dt \implies \int \frac{1}{-2y+1}dy = \int dt$$ can you carry on from here? (Hint: natural logs. Also, remember the constant of integration!)
A: Homogenous solution
$y'(t)=-2y(t)\iff \frac{y'(t)}{y(t)}=-2\iff\frac{dy}{y}=-2dt\iff \ln(y(t))=-2t+C\iff y(t)=De^{-2t}$
Particular solution
It's of the form $w(t)=c$. You remplace in the equation and you get
$-2c+1=0\iff c=\frac{1}{2}$
Finally, the solution is given by $$y(t)=De^{-2t}+\frac{1}{2}$$
And with the condition $y(0)=1$, we get
$$D+\frac{1}{2}=1\iff D=\frac{1}{2}$$
and so the solution is $$y(t)=\frac{1}{2}(1+e^{-2t})$$
A: This initial value problem has exactly one solution: If $y=y(t)$ is the solution, then
$$
\mathrm{e}^{2t}y'(t)=-2\mathrm{e}^{2t}y(t)+\mathrm{e}^{2t},
$$
or
$$
\big(\mathrm{e}^{2t}y(t)\big)'=\mathrm{e}^{2t}y'(t)+2\mathrm{e}^{2t}y(t)=\mathrm{e}^{2t}
=\left(\frac{1}{2}\mathrm{e}^{2t}\right)',
$$
or equivalently
$$
\mathrm{e}^{2t}y(t)=\frac{1}{2}\mathrm{e}^{2t}+c,
$$
where $c$ is some constant, and due to the fact that $y(0)=1$, the value of $c$ has to be $c=1/2$. Thus
$$
y(t)=\frac{1}{2}+\frac{1}{2}\mathrm{e}^{-2t}.
$$
Note. The procedure followed above establishes uniqueness as well, for if $z=z(t)$ were another solution, then following for $z$ the same procedure we would reach to the same answer.
Edit. For the IVP
$$
y'=1,\quad y(0)=1,
$$
the unique solution is $y(t)=t+1$.
