Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Find an implicit or explicit expression for \(y(t)\) for each equation. Then use the given data point to evaluate the constant \(C\) of integration. The following derivative formulas will be helpful. \([\ln (y-1)]^{\prime}=\frac{1}{y-1} \quad\left[\ln \left(t^{2}+1\right)\right]^{\prime}=\frac{2 t}{t^{2}+1} \quad[\ln (1-y)]^{\prime}=-\frac{1}{y-1}\) a. \(\quad y^{\prime}=\frac{t}{y} \quad y(0)=2\) b. \(\quad y^{\prime}=\frac{y}{t} \quad y(1)=1\) c. \(\quad y^{\prime}=\frac{t-1}{y} \quad y(2)=1\) d. \(\quad y^{\prime}=\frac{t}{y\left(t^{2}+1\right)} \quad y(0)=1\) e. \(\quad y^{\prime}=y-1 \quad y(0)=1.5\) f. \(\quad y^{\prime}=1-y \quad y(0)=0.5\) g. \(\quad y^{\prime}=y^{2}-1 \quad y(0)=2\) h. \(\quad y^{\prime}=e^{t-y} \quad y(0)=1.5\)

Step-by-Step Solution

Answer

a. \( y(t) = \sqrt{t^2 + 4} \) b. \( y(t) = t \) c. \( y(t) = \sqrt{t^2 - 2t + 1} \) d. \( y^2 = \ln(t^2 + 1) + 1 \) e. \( y(t) = 1 + 0.5e^t \) f. \( y(t) = 1-0.5e^t \) g. \( y(t) = \frac{1+3e^{2t}}{1-3e^{2t}} \) h. \( y(t) = \ln(e^t + e^{1.5} - 1) \)

Step 1: Understand the given differential equation and data point

For each part, we begin by identifying the type of the differential equation and the initial condition provided. This will help determine the method of integration and solve for the constant of integration.

Step 2a: Solve Part (a); Separate Variables

For equation (a), separate the variables so that all terms involving \(y\) are on one side and \(t\) terms on the other:\[y' = \frac{t}{y} \implies y \cdot dy = t \cdot dt.\]

Step 3a: Integrate Both Sides (Part a)

Integrate both sides:\[\int y \cdot dy = \int t \cdot dt \\frac{y^2}{2} = \frac{t^2}{2} + C.\]

Step 4a: Solve For Integration Constant C (Part a)

Use the initial condition \( y(0) = 2 \) to determine \(C\):\[\frac{2^2}{2} = \frac{0^2}{2} + C \2 = C.\]

Step 5a: Explicit Solution (Part a)

The solution to the differential equation with the initial condition is:\[y^2 = t^2 + 4.\]So, \( y(t) = \sqrt{t^2 + 4} \).

Step 2b: Solve Part (b); Separate Variables

For equation (b), we have \(y' = \frac{y}{t}\). Separate the variables:\[\frac{dy}{y} = \frac{dt}{t}.\]

Step 3b: Integrate Both Sides (Part b)

Integrate each side:\[\int \frac{dy}{y} = \int \frac{dt}{t} \\ln |y| = \ln |t| + C.\]

Step 4b: Solve For Integration Constant C (Part b)

Using the initial condition \( y(1) = 1 \):\[\ln(1) = \ln (1) + C \0 = 0 + C.\]Thus, \( C = 0 \).

Step 5b: Explicit Solution (Part b)

The solution is:\[y = t.\]

Step 2c: Solve Part (c); Separate Variables

For equation (c):\[y' = \frac{t-1}{y} \implies y \cdot dy = (t-1) \cdot dt.\]

Step 3c: Integrate Both Sides (Part c)

Integrate both sides:\[\int y \cdot dy = \int (t-1) \cdot dt \\frac{y^2}{2} = \frac{t^2}{2} - t + C.\]

Step 4c: Solve For Integration Constant C (Part c)

Use the initial condition \( y(2) = 1 \):\[\frac{1^2}{2} = \frac{2^2}{2} - 2 + C \\frac{1}{2} = 2 - 2 + C \\frac{1}{2} = C.\]

Step 5c: Explicit Solution (Part c)

The solution is:\[y(t) = \sqrt{t^2 - 2t + 1}.\]

Step 2d: Solve Part (d); Separate Variables

For equation (d):\[y' = \frac{t}{y(t^2+1)} \implies y \, dy = \frac{t}{t^2+1} \, dt.\]

Step 3d: Integrate Using Known Derivative Formula (Part d)

Recognize the derivative \( \frac{2t}{t^2+1} \), adjust the integrand:\[\int y \, dy = \frac{1}{2} \int \frac{2t}{t^2+1} \, dt \\frac{y^2}{2} = \frac{1}{2} \ln(t^2+1) + C.\]

Step 4d: Solve For Integration Constant C (Part d)

Use the initial condition \( y(0) = 1 \):\[\frac{1^2}{2} = \frac{1}{2} \ln(0^2 + 1) + C \\frac{1}{2} = C.\]

Step 5d: Explicit Solution (Part d)

The solution is:\[y^2 = \ln(t^2 + 1) + 1.\]

Step 2e: Recognize Linear Growth Equation (Part e)

Equation (e) is in the form \(y' = y - 1\). Rearrange:\[dy = (y - 1) \, dt.\]

Step 3e: Integrate for Unit Step Growth (Part e)

Rearranging using known derivative:\[\int \frac{dy}{y-1} = \int dt \\ln|y-1| = t + C.\]

Step 4e: Solve For Integration Constant C (Part e)

Use the initial condition \( y(0) = 1.5 \):\[\ln(1.5-1) = 0 + C \\ln(0.5) = C.\]

Step 5e: Explicit Solution (Part e)

The solution is:\[y - 1 = e^{t + \, ln(0.5)} = 0.5e^t \y(t) = 1 + 0.5e^t.\]

Step 2f: Recognize Decay Equation (Part f)

Equation (f) is \(y' = 1-y\). Rearrange:\[dy = (1-y) \, dt.\]

Step 3f: Integrate Exponential Decay (Part f)

Using the derivative form given:\[\int \frac{dy}{1-y} = \int dt \-\ln|1-y| = t + C.\]

Step 4f: Solve For Integration Constant C (Part f)

Use the initial condition \( y(0) = 0.5 \):\[-\ln(1-0.5) = 0 + C \-\ln(0.5) = C.\]

Step 5f: Explicit Solution (Part f)

The solution is:\[|1-y| = e^{-t - \, ln(0.5)} \y(t) = 1-0.5e^t.\]

Step 2g: Recognize Quadratic Equation (Part g)

Equation (g) can be expressed as \(y' = (y-1)(y+1)\). Separate variables:\[dy = (y^2-1) \, dt.\]

Step 3g: Partial Fraction Decomposition (Part g)

Decompose \( \frac{1}{y^2-1} \):\[\int \left( \frac{1}{2(y-1)} - \frac{1}{2(y+1)} \right) dy = \int 1 \, dt.\]

Step 4g: Integrate Both Sides (Part g)

Integrate using the decomposition:\[\frac{1}{2} [\ln|y-1| - \ln|y+1|] = t + C.\]

Step 5g: Solve For Integration Constant C (Part g)

Use the initial condition \( y(0) = 2 \):\[\frac{1}{2} [\ln(2-1) - \ln(2+1)] = 0 + C \C = \ln|1/3| = \ln(1/3).\]

Step 6g: Simplify to Explicit Solution (Part g)

The solution simplifies to:\[| \frac{y-1}{y+1}|^{1/2} = e^{t+c}, \taking exp: \y(t) = \frac{1+3e^{2t}}{1-3e^{2t}}.\]

Step 2h: Understand the Nonlinear Exponential Equation (Part h)

For equation (h): \(y' = e^{t-y}\). Separate the variables:\[e^y \, dy = e^t \, dt.\]

Step 3h: Integrate Both Sides (Part h)

Integrate the equation:\[\int e^y \, dy = \int e^t \, dt \e^y = e^t + C.\]

Step 4h: Solve For Integration Constant C (Part h)

Use the initial condition \( y(0) = 1.5 \):\[e^{1.5} = e^0 + C \C = e^{1.5} - 1.\]

Step 5h: Explicit Solution (Part h)

The solution becomes:\[y(t) = \ln(e^t + e^{1.5} - 1).\]

Key Concepts

Implicit Solution

In differential equations, an implicit solution refers to expressions where the dependent variable, say \( y \), cannot be directly isolated on one side of the equation. Instead, it remains part of a relationship or equation involving both \( y \) and the independent variable, often \( t \).

This might resemble something like \( y^2 + t^2 = C \), where \( C \) is a constant found through solving initial conditions. Implicit solutions are essential because they often arise in equations that aren't easily solvable for the dependent variable explicitly. They can capture complex relationships that might be simplified or broken down through further manipulation.

• In some problems, implicit solutions provide a complete, simpler representation of the relationship without the need for radical simplification.
• Inherent complexity in differentiating implicitly can often be advantageous in handling complex boundary conditions.

Explicit Solution

An explicit solution is when the dependent variable is isolated on one side of the equation. It takes the form \( y = f(t) \), making it straightforward to evaluate \( y \) for any given \( t \).

Finding explicit solutions is often the goal when solving differential equations because they provide clear, direct relationships between variables. This type of solution facilitates easy computations and predictions of future states. For example, if you find \( y(t) = \sqrt{t^2 + 4} \), you have a clear expression showing how \( y \) changes as \( t \) varies.

• They allow easy substitution back into constraints or other equations.
• Explicit solutions are often easier to graph and interpret in practical applications.

Separation of Variables

Separation of variables is a traditional and powerful technique used to solve ordinary differential equations where all the terms involving one variable are moved to one side and those involving another variable are moved to the opposite side.

It relies on the structure of the differential equation to allow integration on both sides separately. Consider an equation like \( y' = \frac{t}{y} \). You can rearrange this to \( y \, dy = t \, dt \), and by integrating both sides, solve for the function.

• This method works well for equations where variables can be distinctly separated.
• Each side's integration can then handle different functions or inputs, often leading to further analytical simplification.

Initial Conditions

Initial conditions in differential equations provide specific values for the solution at a particular point. This is crucial for determining the constants of integration that appear when you solve differential equations.

Without an initial condition, the solution could represent an entire family of curves. Initial conditions narrow these down to a single, specific solution. In our example, if the initial condition is \( y(0) = 2 \), we use this to find the constant in the integrated solution.

• These known values ensure that the solution meets specific practical or theoretical requirements.
• The use of initial conditions is a key aspect that differentiates ordinary differential equations from partial differential equations.