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In the paper of Dalgaard and Strulik (2014) titled "Optimal Aging and Death" they use following differential equation: \begin{equation} \dot{D}(t) = \mu (D(t) - E) \end{equation} They ingerate the above equation and get the result like this: \begin{equation} D(t) = D_{0} e^{\mu t} - E e^{\mu t} + E \end{equation} I want to know the steps of how they came to this solution?

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    Is $\mu$ a constant or a function? Taking the derivative of both sides of the second equation gets $D'(t)=\mu e^{\mu t}(D_0-E)$ but I am not sure how helpful this is to you, since there is a factor of $e^{\mu t}$ that isn't present in the first equation. – 2017-02-05
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    μ is a constant. and Can you derive the second equation from the first? I need to derive second equation from the first by integration. – 2017-02-05

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I read the paper you referred to and saw that you forgot to provide one piece of information: $$D(0)=D_0$$ Your ODE is first order and autonomous.

Therefore, you can separate the variables as follows: $$\frac{dD}{dt}=\mu(D-E)$$ $$\Rightarrow \int \frac{1}{D-E}~dD=\int \mu~dt$$ Integrating both sides gives: $$\ln(D-E)=\mu t+C$$ Exponentiate both sides: $$D-E=e^{\mu t+C}$$ $$D-E=e^{\mu t}\cdot e^C$$ Let $e^C=k$: $$D-E=ke^{\mu t}$$ Therefore, we have the following general solution: $$D(t)=ke^{\mu t}+E \tag{1}$$ Now, consider your initial condition $D(0)=D_0$. Substitute this into equation $(1)$ to find $k$, and then substitute the value of $k$ you obtained into $(1)$.

If done correctly, you should get the solution you've provided.

If you don't get to the solution, please do not hesitate to ask.

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They don't integrate the equation, they solve it. It goes like this:

Write $\dot{D}(t)$ as $\frac{dD}{dt}$. Then you separate the variables:

$$dD = \mu (D- E) dt $$ $$\Rightarrow \frac{dD}{\mu (D- E)} = dt $$

And now you integrate:

$$\int_{D_0}^D \frac{dD'}{ (D'- E)} = \mu t$$ $$\Rightarrow \log{\left(\frac{D-E}{D_0 -E}\right)} =\mu t $$

and solve for $D$:

$$\log{(D-E)} = \mu t + \log{(D_0 -E)} $$ $$\Rightarrow D-E = e^{\mu t}(D_0 -E) $$ $$\Rightarrow D = e^{\mu t}(D_0 -E) +E $$

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    In the context of differential equations, *integration* is synonym of resolution; and a solution is also called an *integral*. – 2017-02-05
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    Interesting, I have never heard anyone say that. But it seems to my like a not *so* good choice of words... – 2017-02-05
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    What you use to call integration and integral just correspond to the particular case of the differential equation $y'=f(x)$. – 2017-02-05
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Here is my solution:

\begin{align} \frac{dD}{dt} &= \mu D - \mu E \\ \frac{1}{D-E}dD &= \mu dt \\ \log\left(\frac{D(t)-E}{D_0-E}\right) &= \mu t \\ \frac{D(t)-E}{D_0-E} &= e^{\mu t}\\ D(t) &= D_0 e^{\mu t} - E e^{\mu t} + E \end{align}

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With $G(t):=D(t)-E$, the equation simplifies to

$$\dot G(t)=\mu G(t)$$ which is a very classical one (first order linear with constant coefficients, both homogenous and separable). The solution is known to be

$$G(t)=G_0e^{\mu t}$$ where $G_0:=G(0)$. (You easily find this result by integrating $dG/G=\mu\,dt$.)

Then

$$D(t)=(D_0-E)e^{\mu t}+E.$$