Find solution of differential equation $y'(t)=-2y(t)+1$

Question

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 $$y'(t)=1,\quad y(0)=1?$$

Answer 1

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!)

Answer 2

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 $$y(t)=\frac{1}{2}(1+e^{-2t})$$

Answer 3

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 $y(t)=t+1$.