Question

General Gompertz solution Solve the initial value problem $$M^{\prime}(t)=-r M \ln \left(\frac{M}{K}\right), M(0)=M_{0}$$ with arbitrary positive values of \(r, K,\) and \(M_{0^{\prime}}\)

Short answer

Question: Solve the initial value problem for the given differential equation and initial condition using the General Gompertz model: $$M'(t) = -rM \ln \left(\frac{M}{K}\right), \quad M(0) = M_0$$ Answer: The explicit solution for the General Gompertz model is given by: $$M(t) = M_0e^{-\frac{r}{K}t}$$

Step-by-step solution

Identify the differential equation and initial condition

The given differential equation is: $$M'(t) = -rM \ln \left(\frac{M}{K}\right)$$ with an initial condition: \(M(0) = M_0\).

Separate the variables

We want to separate the variables to solve the equation. Multiply both sides of the equation by \(\frac{1}{M\ln\left(\frac{M}{K}\right)}\): $$\frac{dM}{M\ln \left(\frac{M}{K}\right)} = -r \, dt$$

Integrate both sides

Now, integrate both sides of the equation with respect to their respective variables: $$\int \frac{dM}{M\ln \left(\frac{M}{K}\right)} = -r \int dt$$

Use substitution on the left side of the equation

Let's substitute \(u = \ln\left(\frac{M}{K}\right)\), which implies \(\frac{du}{dM} = \frac{1}{MK} \Rightarrow dM = MK\, du\). The left side of the integral becomes: $$\int \frac{MK \, du}{M u} = \int K \, du$$

Solve the integrals

Now solve the integrals: $$K\int du = -r\int dt \Rightarrow Ku = -rt + C$$ where \(C\) is the integration constant.

Substitute back and solve for M(t)

Substitute back \(u = \ln\left(\frac{M}{K}\right)\): $$K\ln \left(\frac{M}{K}\right) = -rt + C$$ Now, solve for \(M(t)\), first, divide both sides by \(K\): $$\ln \left(\frac{M}{K}\right) = -\frac{r}{K}t + \frac{C}{K}$$ Take exponent on both sides: $$\frac{M(t)}{K} = e^{-\frac{r}{K}t + \frac{C}{K}}$$ Finally, multiply both sides by \(K\): $$M(t) = Ke^{-\frac{r}{K}t + \frac{C}{K}}$$

Find the integration constant using the initial condition

Use the initial condition \(M(0) = M_0\): $$M_0 = K e^{-\frac{r}{K}\cdot 0 + \frac{C}{K}} = K e^{\frac{C}{K}}$$ Solve for \(C\): $$C = K \ln\left(\frac{M_0}{K}\right)$$

Write the final solution

Substitute the value of \(C\) back into the solution of \(M(t)\): $$M(t) = Ke^{-\frac{r}{K}t + \ln\left(\frac{M_0}{K}\right)}$$ You can simplify the expression inside the exponent as follows: $$M(t) = Ke^{-\frac{r}{K}t + \ln\left(\frac{M_0}{K}\right)} = K\frac{M_0}{K}e^{-\frac{r}{K}t}$$ Finally, the explicit solution for the General Gompertz model is: $$M(t) = M_0e^{-\frac{r}{K}t}$$

Key concepts

Differential Equation

In mathematics, a differential equation is an equation that involves an unknown function and its derivatives. These equations are powerful tools for modeling how things change over time or space. For example, the equation \( M'(t) = -rM \ln \left(\frac{M}{K}\right) \) captures the Gompertz growth model. This model describes how quantities like populations or cells grow in a sigmoidal manner, often used in biology or demography.

The main goal when working with differential equations is to find a function \( M(t) \) that satisfies the given equation for all values of \( t \). The equation provides a relationship between \( M(t) \) and its rate of change \( M'(t) \), which helps to predict how \( M \) changes over time.

Initial Value Problem

An initial value problem gives us a differential equation along with an initial condition. This initial condition specifies the value of the unknown function at a particular point, providing crucial information required to solve the differential equation uniquely.

In our exercise, the initial condition is given by \( M(0) = M_0 \). This means that at time \( t = 0 \), the function \( M(t) \) has the value \( M_0 \). Using an initial condition like this helps us determine any constants of integration, ensuring that our solution to the differential equation is specific and matches real-world observations.

• This type of problem ensures the solution is not just any function, but the exact one that fits a particular real-world scenario.
• It aids in clearly specifying dynamics modeled by the equation, especially at the beginning of the time span considered.

Separation of Variables

Separation of variables is a technique used to solve differential equations, which simplifies the problem by rearranging terms involving different variables. This technique allows us to express each side of the equation as a function of a single variable.

In this exercise, after recognizing the differential equation, you separate variables by rearranging it as follows:

• Move the terms involving \( M \) to one side: \( \frac{dM}{M\ln \left(\frac{M}{K}\right)} \)
• Move terms involving \( t \) to the other side: \( -r \, dt \)

By doing this, each side of the equation only depends on one variable. This makes it possible to integrate both sides separately and find a solution that includes the integration constant.

Integration Constant

The integration constant, denoted usually as \( C \), appears when we integrate during the process of solving a differential equation. When you integrate both sides of the separated equation in our problem to solve for \( M \), you introduce an arbitrary constant \( C \).

Integration is essentially finding all of the possible antiderivatives of a function, which includes a constant because the derivative of a constant is zero. Without an initial condition, any number could be added as a constant, resulting in infinitely many solutions. Therefore, \( C \) might initially seem undefined, but it is crucial because it allows us to match the specific situation defined by an initial value problem.

• For example, with the initial condition \( M(0) = M_0 \), we can solve for \( C \), ensuring our solution is unique and accurate.
• In our exercise, we find \( C \) such that the function \( M(t) \) passes through the point \( (0, M_0) \).

This tailor-fits the general solution to our specific problem scenario.