Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Show that the variables are not separable in the equation \(y^{\prime}(t)=t+y .\) That is, there are not two functions, \(g(t)\) and \(h(y),\) which for all \(t\) and \(y \quad t+y=g(t) \times h(y)\) A procedure is to assume two such functions, \(g(t)\) and \(h(y)\) exist and then show that the following equations are incompatible. \(\begin{array}{lllll}t=0 & y=0 & t=0 & y=1 & t=1 & y=0\end{array}\) \(g(0) \times h(0)=0+0=0 \quad g(0) \times h(1)=0+1=1 \quad g(1) \times h(0)=1+0=1\) Show that \(g(0) \times h(0)=0, g(0) \times h(1)=1\) and \(g(1) \times h(0)=1\) are incompatible.

Step-by-Step Solution

Answer

The assumption of separability leads to inconsistent equations, proving the variables are not separable.

Step 1: Assume Existence of Separability

Assume that there exist two functions \(g(t)\) and \(h(y)\) such that \(t + y = g(t) \times h(y)\) for all \(t\) and \(y\).

Step 2: Verify Consistency at Specific Points

Evaluate \(g(t)\) and \(h(y)\) at the given points: \((t, y) = (0, 0), (0, 1), (1, 0)\).1. For \((t, y) = (0, 0)\): \[ g(0) \times h(0) = 0 + 0 = 0 \] Therefore, \(g(0) \times h(0) = 0\).2. For \((t, y) = (0, 1)\): \[ g(0) \times h(1) = 0 + 1 = 1 \] Therefore, \(g(0) \times h(1) = 1\).3. For \((t, y) = (1, 0)\): \[ g(1) \times h(0) = 1 + 0 = 1 \] Therefore, \(g(1) \times h(0) = 1\).

Step 3: Identify Inconsistency

From Step 2, we have:- \(g(0) \times h(0) = 0\)- \(g(0) \times h(1) = 1\)- \(g(1) \times h(0) = 1\)These results are inconsistent because:- \(g(0) eq 0\) or \(h(0) eq 0\) cannot make both \(g(0) \times h(0) = 0\) and \(g(0) \times h(1) = 1\) hold simultaneously.- The requirement \(g(0) \times h(1) = 1\) implies \(g(0)\) and \(h(1)\) are non-zero, contradicting \(g(0) \times h(0) = 0\) which implies at least one of them is zero.

Step 4: Conclusion

Since the equations are inconsistent, the assumption of separability is invalid. Thus, there cannot exist functions \(g(t)\) and \(h(y)\) that separate the variables in the given differential equation.

Key Concepts

variable separability

In solving a differential equation like \( y^{\prime}(t)=t+y \), one might wonder if it is possible to express the equation in a form where one function depends solely on \( t \) and another on \( y \). This is known as variable separability. The goal is to express the equation as \( g(t) \times h(y) = t + y \), where \( g(t) \) is a function of \( t \) alone and \( h(y) \) is a function of \( y \) alone.

To check separability, we assume the existence of such functions. However, not all differential equations can be separated this way. It requires that the equation can be restructured into parts that, when multiplied, equal the original equation. This method is particularly useful when it can be applied, as it often simplifies finding a solution.

In this particular case, after testing certain values for \( t \) and \( y \), the assumed functions \( g(t) \) and \( h(y) \) show inconsistencies, meaning that such a separated form is not possible here.

inconsistent equations

When checking the separability of the differential equation \( y^{\prime}(t)=t+y \), we encounter the term "inconsistent equations." This means the equations derived from our assumptions conflict with each other, making it impossible to satisfy all conditions simultaneously.

Here's how that happens: Let's examine specific points, say \((t,y)=(0,0),\,(0,1),\,(1,0)\). For these values, if we assume \( g(t) \times h(y) = t + y \), we get three different equations:

• \( g(0) \times h(0) = 0 \)
• \( g(0) \times h(1) = 1 \)
• \( g(1) \times h(0) = 1 \)

These are inconsistent. You cannot have both \( g(0) \times h(0) = 0 \) and \( g(0) \times h(1) = 1 \) as these presume contradictory conditions about \( g(0) \) and \( h(0) \). Essentially, at least one term must be zero to satisfy the first equality, but both must be non-zero to satisfy the second.

Being inconsistent means there's no possible way to choose \( g(t) \) and \( h(y) \) such that they meet all these criteria. The consequence is that the original equation cannot be simplified into separable forms.

calculus

Calculus plays a significant role in determining if differential equations can be simplified, such as through the separability method. Differential equations, in simple terms, relate a function with its derivatives, encompassing many real-world phenomena, from physics to biology.

The original equation, \( y^{\prime}(t)=t+y \), is a basic form of a differential equation. When we differentiate or seek integration for solutions, understanding calculus becomes essential. Techniques in calculus allow us to solve or transform complex expressions, though not all provide straightforward paths.

In this context, while calculus provides powerful tools, we learn that it also tells us when some methods will or will not work. Here, the attempt to separate variables failed not because the rules of calculus weren't followed, but due to the inherent nature of the equation itself. The inconsistency observed means calculus guides us to conclude that, sometimes, what seems possible algebraically can't happen. Thus, calculus doesn't just help in solving equations; it also dictates what's theoretically feasible.