Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

The complete solution to Exercise 7.4 .7 is given in the Solutions Section on the web. Try your hand with showing that $$ y(t)=e^{-b t} \sin \left(\sqrt{c-b^{2}} t\right) $$ solves $$ y^{\prime \prime}(t)+2 b y^{\prime}(t)+c y(t)=0 \quad \text { for } \quad c-b^{2}>0 \quad \text { for } \quad b=1, \quad c=2 $$ Get a big piece of paper.

Step-by-Step Solution

Answer

Yes, substituting and simplifying confirm that it solves the equation.

Step 1: Differentiate y(t)

The function provided is \( y(t) = e^{-b t} \sin(\sqrt{c-b^2} \, t) \). Start by finding the first derivative with respect to \( t \), using the product rule. Let \( u(t) = e^{-b t} \) and \( v(t) = \sin(\sqrt{c-b^2} \, t) \). So, \( y'(t) = u'(t) v(t) + u(t) v'(t) \).Calculate \( u'(t) = -b e^{-b t} \) and \( v'(t) = \cos(\sqrt{c-b^2} \, t) \cdot \sqrt{c-b^2} \).Thus,\[ y'(t) = -b e^{-b t} \sin(\sqrt{c-b^2} \, t) + e^{-b t} \cos(\sqrt{c-b^2} \, t) \cdot \sqrt{c-b^2} \].

Step 2: Differentiate y'(t) to Find y''(t)

Now, differentiate \( y'(t) \) to find \( y''(t) \). Again, apply the product rule for both terms in \( y'(t) \).For the first term: \( -b e^{-b t} \sin(\sqrt{c-b^2} \, t) \): - Differentiate \( -b e^{-b t} \) yielding \( b^2 e^{-b t} \) and keep \( \sin(\sqrt{c-b^2} \, t) \) the same.For the second term: \( e^{-b t} \cos(\sqrt{c-b^2} \, t) \cdot \sqrt{c-b^2} \): - Differentiate \( e^{-b t} \) to get \( -b e^{-b t} \), and \( \cos(\sqrt{c-b^2} \, t) \) to get \( -\sin(\sqrt{c-b^2} \, t) \cdot (c-b^2) \).This provides the second derivative, \( y''(t) \).

Step 3: Substitute and Simplify the Left Side of the Differential Equation

Substitute \( y(t) \), \( y'(t) \), and \( y''(t) \) into the differential equation: \( y''(t) + 2b y'(t) + c y(t) = 0 \).After substituting, you will get an expression involving \( e^{-b t} \). Simplify each term considering \( b = 1 \) and \( c = 2 \).

Step 4: Simplify the Expression and Validate the Equation

After substituting \( b = 1 \) and \( c = 2 \), the exponential and trigonometric terms should cancel out such that\[ y''(t) + 2b y'(t) + c y(t) = 0 \] becomes 0 on simplification.Use trigonometric identities if necessary and simplify carefully to show equality to 0. By the problem statement, it should result in the expected form proving that \( y(t) \) satisfies the differential equation.

Key Concepts

Second-Order Differential Equations

Second-order differential equations are mathematical expressions involving derivatives of an unknown function up to the second order. In general, these equations help model diverse scientific phenomena like mechanical vibrations and electrical circuits. The standard form of a second-order differential equation is given as:

• \( y'' + p(t) y' + q(t) y = g(t) \)

Here, \( y'' \) is the second derivative, \( y' \) is the first derivative, and \( y \) is the function itself. \( p(t) \), \( q(t) \), and \( g(t) \) are coefficient functions of the independent variable \( t \).
This problem involves a homogenous second-order linear differential equation, meaning \( g(t) = 0 \). Such equations typically take the form:

• \( y'' + 2b y' + cy = 0 \)

Our task is to verify if a given function is a solution to this equation, which requires derivatives, understanding of coefficients and simplifications.

Product Rule

The product rule is an essential tool in calculus for differentiating functions expressed as a product of two separate functions. If you have a function in the form \( y(t) = u(t) \, v(t) \), the derivative \( y'(t) \) is calculated using the product rule as follows:

• \( y'(t) = u'(t) \, v(t) + u(t) \, v'(t) \)

This rule ensures that both components of the product are accounted for in the differentiation process.
For example, the differentiation of \( y(t) = e^{-b t} \sin(\sqrt{c-b^2} \, t) \) involves defining:

• \( u(t) = e^{-b t} \)
• \( v(t) = \sin(\sqrt{c-b^2} \, t) \)

To differentiate, calculate each derivative separately as \( u'(t) = -b e^{-b t} \) and \( v'(t) = \cos(\sqrt{c-b^2} \, t) \, \sqrt{c-b^2} \). These components are then inserted back into the product rule to find the first derivative and similarly handle higher derivatives.

Trigonometric Functions

Trigonometric functions, like sine and cosine, frequently appear in differential equations because they describe periodic behaviors. In this problem, \( \sin(\sqrt{c-b^2} \, t) \) represents a significant part of the solution.

• The sine function is periodic, meaning it repeats its values in cycles, which suits modelling waves or oscillatory motion.
• The argument \( \sqrt{c-b^2} \, t \) affects the frequency of oscillation. A higher value results in a quicker oscillation frequency and vice versa.

In the solution, its derivative is \( \cos(\sqrt{c-b^2} \, t) \, \sqrt{c-b^2} \). Expressing derivatives in terms of one another helps to continually derive and simplify solutions in differential equations.
Identities and transformations involving trigonometric functions, such as those seen here, provide flexibility in manipulating terms to satisfy required conditions in equations.

Exponential Functions

Exponential functions involve expressions of the form \( e^{x} \) where \( e \) is Euler's number. In differential equations, they frequently describe growth, decay, and damping behaviors.

• The function \( e^{-bt} \) indicates an exponential decay due to the negative exponent \(-b\).
• This function is key in modeling damping, which describes how oscillations decrease over time in a mechanical system or signal.

When differentiating \( e^{-b t} \), the derivative results in \(-b e^{-bt} \). This property of exponential derivatives—a constant multiplied by the original function—facilitates easy manipulation in solving differential equations.
Combining exponential functions with trigonometric components allows the modeling of damped oscillatory systems, where exponential terms govern the decay and trigonometric terms describe oscillation.