Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Show that if \(y_{h}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=0\) and \(y_{p}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f(t)\) then for any number \(C, y_{g}(t)=y_{p}(t)+C y_{h}(t)\) solves \(y^{\prime \prime}(t)+p y^{\prime}(t)+q y(t)=f(t)\)

Step-by-Step Solution

Answer

The function \( y_g(t) = y_p(t) + C y_h(t) \) solves the given non-homogeneous differential equation.

Step 1: Identify Given Functions

We are given two functions, \( y_h(t) \) and \( y_p(t) \). The function \( y_h(t) \) solves the homogeneous equation \( y''(t) + p y'(t) + q y(t) = 0 \). The function \( y_p(t) \) solves the non-homogeneous equation \( y''(t) + p y'(t) + q y(t) = f(t) \). We need to prove that \( y_g(t) = y_p(t) + C y_h(t) \) solves the non-homogeneous equation as well.

Step 2: Calculate Derivatives of \( y_g(t) \)

First, calculate the first and second derivatives of \( y_g(t) = y_p(t) + C y_h(t) \). The first derivative is:\[ y_g'(t) = y_p'(t) + C y_h'(t) \]The second derivative is:\[ y_g''(t) = y_p''(t) + C y_h''(t) \]

Step 3: Substitute Derivatives into the Equation

Substitute \( y_g(t), y_g'(t), \) and \( y_g''(t) \) into the differential equation \( y''(t) + p y'(t) + q y(t) \):\[y_g''(t) + p y_g'(t) + q y_g(t) = (y_p''(t) + C y_h''(t)) + p (y_p'(t) + C y_h'(t)) + q (y_p(t) + C y_h(t))\]

Step 4: Simplify the Equation

Simplify the expression:\[ (y_p''(t) + p y_p'(t) + q y_p(t)) + C (y_h''(t) + p y_h'(t) + q y_h(t)) \]Using the fact that \( y_p(t) \) solves the non-homogeneous equation and \( y_h(t) \) solves the homogeneous equation, we get:\[ y_p''(t) + p y_p'(t) + q y_p(t) = f(t) \]\[ y_h''(t) + p y_h'(t) + q y_h(t) = 0 \]

Step 5: Final Expression to Verify the Solution

Substitute back:\[ f(t) + C \cdot 0 = f(t) \]This confirms that \( y_g(t) = y_p(t) + C y_h(t) \) solves the differential equation \( y''(t) + p y'(t) + q y(t) = f(t) \).

Key Concepts

Homogeneous Equations

A homogeneous differential equation is a vital concept in understanding differential equations. In these equations, all the terms depend on the function or its derivatives, and importantly, there is no independent term, which means it equals zero. So every solution curves pass through the origin. For instance, a second-order homogeneous differential equation can be written as:\[ y''(t) + p y'(t) + q y(t) = 0 \]Here, any solution of this equation is called the complementary function, often denoted as \( y_h(t) \). It mainly characterizes systems where forces act on one another without any external influence being present.

• The general solution includes terms of possible sin, cos, and exponential functions, depending on the characteristic equation's roots.
• The principle of superposition applies, meaning any linear combination of solutions is also a solution.

Understanding and solving homogeneous equations is crucial because they often form the foundation upon which we build solutions for more complex expressions in differential equations.

Non-Homogeneous Equations

Non-homogeneous differential equations are slightly more complex because they involve external forces or inputs. Simply put, they equal something other than zero. These equations model systems influenced by external forces, resulting in solutions that behave differently from homogeneous equations.The form of a non-homogeneous second-order differential equation can be expressed as:\[ y''(t) + p y'(t) + q y(t) = f(t) \]The function \( f(t) \) denotes the non-zero part of the equation, marking it as non-homogeneous.

• Non-homogeneous equations require a particular solution, \( y_p(t) \), which accounts for the effect of the external term \( f(t) \).
• The general solution combines both the particular solution \( y_p(t) \) and the complementary solution \( y_h(t) \): \( y(t) = y_p(t) + C y_h(t) \), where \( C \) is an arbitrary constant.
• This formulation allows the general solution to incorporate the natural response of the system, as well as the forced response.

Understanding how to solve these equations helps in addressing real-world problems where systems are subjected to outside influences.

Solution Verification

Verification of solutions is a crucial step in dealing with differential equations. It ensures that the functions identified as solutions genuinely satisfy the equation in question.In the exercise above, we verify whether \( y_g(t) = y_p(t) + C y_h(t) \) qualifies as a solution for the differential equation \( y''(t) + p y'(t) + q y(t) = f(t) \).

• First, identify derivatives of \( y_g(t) \) and substitute these into the original differential equation.
• Upon substitution, each solution component corresponds to distinct terms in the equation: \( y_p(t) \) handles the non-zero part, \( f(t) \), while \( y_h(t) \) handles the zero homogeneous part.
• The simplification demonstrates that \( f(t) + C \cdot 0 = f(t) \), confirming that the expression satisfies the equation.

Verification is essential as it ensures that theoretical solutions are actually applicable in modeling real systems, linking mathematical solutions with practical realities.