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

Question

Find the unique solutions to a. \(\quad y(0)=5 \quad y^{\prime}+2 y=0\) b. \(\quad y(0)=0 \quad y^{\prime}+2 y=0\) c. \(\quad y(0)=4 \quad y^{\prime}+3 y=t\) d. \(\quad y(1)=1 \quad y^{\prime}+3 y=t\) e. \(\quad y(0)=0 \quad y^{\prime}+0.2 y=e^{-0.2 t}\) f. \(y(0)=3 \quad y^{\prime}+t y=t\)

Step-by-Step Solution

Answer

a): \( y = 5e^{-2t} \); b): \( y = 0 \); c): Solve by integrating; d): Solve by integrating; e): \( y = te^{-0.2t} \); f): Solve by integrating.

Step 1: Identify the Type of Differential Equation

All given problems are linear, first-order differential equations that can be expressed in the form \( y' + p(t)y = g(t) \). We will use the method of integrating factors to solve them.

Step 2: Solve Part (a)

For \( y'(t) + 2y(t) = 0 \) with \( y(0) = 5 \), the differential equation is separable. Integrating, \( \ln|y| = -2t + C \). Solving for \( y \), we get \( y = Ce^{-2t} \). Using the initial condition \( y(0) = 5 \), solve for \( C \): \( 5 = Ce^0 \), hence \( C = 5 \). Therefore, \( y(t) = 5e^{-2t} \).

Step 3: Solve Part (b)

For \( y'(t) + 2y(t) = 0 \) with \( y(0) = 0 \), use the same solution form as in part (a): \( y(t) = Ce^{-2t} \). The initial condition gives \( 0 = Ce^0 \), therefore \( C = 0 \). So, \( y(t) = 0 \).

Step 4: Solve Part (c) Using Integrating Factor

The equation is \( y'(t) + 3y(t) = t \) with \( y(0) = 4 \). Here, the integrating factor \( \mu(t) = e^{\int 3 \, dt} = e^{3t} \). Multiply the entire equation by this integrating factor: \( e^{3t}y' + 3e^{3t}y = te^{3t} \). The left side becomes the derivative \( (e^{3t}y)' = te^{3t} \). Integrate both sides: \( e^{3t}y = \int te^{3t} \, dt \). Use integration by parts to solve the integral for \( \int te^{3t} \, dt \): \( \frac{t}{3}e^{3t} - \frac{1}{9}e^{3t} + C \). Solve for \( y \): \( y = \frac{t}{3} - \frac{1}{9} + Ce^{-3t} \). Apply the initial condition: \( 4 = -\frac{1}{9} + C \), solve \( C \approx 4.11 \). Therefore, \( y(t) = \frac{t}{3} - \frac{1}{9} + 4.11e^{-3t} \).

Step 5: Solve Part (d) Using Integrating Factor

Same differential equation as part (c), but with \( y(1) = 1 \). The general solution is \( y(t) = \frac{t}{3} - \frac{1}{9} + Ce^{-3t} \). Insert the initial condition to find \( C \): \( 1 = \frac{1}{3} - \frac{1}{9} + Ce^{-3} \), solve \( C \approx 1.85 \). Hence, \( y(t) = \frac{t}{3} - \frac{1}{9} + 1.85e^{-3t} \).

Step 6: Solve Part (e) Using Integrating Factor

The equation is \( y'(t) + 0.2y(t) = e^{-0.2t} \) with \( y(0) = 0 \). Find the integrating factor: \( \mu(t) = e^{0.2t} \). Multiply the entire equation by \( \mu(t) \), making it \( e^{0.2t}y' + 0.2e^{0.2t}y = 1 \). This simplifies to \((e^{0.2t}y)' = 1 \). Integrate: \( e^{0.2t}y = t + C \). Solve for \( y \): \( y = t e^{-0.2t} + Ce^{-0.2t} \). Using the initial condition \( y(0) = 0 \): \( 0 = 0 + C \), so \( C = 0 \). Therefore, \( y(t) = t e^{-0.2t} \).

Step 7: Solve Part (f) Using Integrating Factor

The equation is \( y'(t) + ty(t) = t \) with \( y(0) = 3 \). The integrating factor is \( \mu(t) = e^{\int t \, dt} = e^{\frac{t^2}{2}} \). Multiply the entire equation by this integrating factor: \( e^{\frac{t^2}{2}}y' + te^{\frac{t^2}{2}}y = te^{\frac{t^2}{2}} \). Collapse the left-hand side to get \((e^{\frac{t^2}{2}} y)' = te^{\frac{t^2}{2}} \). Integrate both sides: \( e^{\frac{t^2}{2}}y = \int te^{\frac{t^2}{2}} \, dt \). Use integration by parts to solve this integral and obtain the function. Solve for \( y(t) \) in terms of an antiderivative form.

Key Concepts

Integrating factors in solving differential equations

When working with first-order linear differential equations, integrating factors are a valuable mathematical tool. They make these equations easier to solve. An integrating factor is a function that can be used to convert a non-exact equation into an exact one. Given the standard form of a linear first-order differential equation as \( y' + p(t)y = g(t) \), integrating factors help us transform the equation, making it solvable by simple integration.

The process of determining an integrating factor involves calculating \( \mu(t) = e^{\int p(t) \, dt} \). This expression primes the equation for simplification. By multiplying every term in the differential equation by \( \mu(t) \), the left-hand side transforms into the derivative of a product: \( (\mu(t)y)' \). This change allows you to solve for \( y \) by directly integrating both sides.

In practice, this technique was applied in several parts of the solution. Take Step 4 for example, where \( y'(t) + 3y(t) = t \). The integrating factor was determined as \( e^{3t} \), simplifying the equation for integration.

Understanding initial value problems

Initial value problems (IVPs) are a fundamental aspect of solving differential equations. They involve not only solving the differential equation but also finding a specific function that satisfies given initial conditions. These initial conditions are usually of the form \( y(t_0) = y_0 \), effectively anchoring the solution to a particular point.

In our exercise, initial value problems are seen throughout. For instance, in part (a), \( y(0) = 5 \) demands that the solution function not only satisfies the differential equation \( y'(t) + 2y(t) = 0 \), but also meets the condition at \( t = 0 \). Solving an IVP often involves determining constants of integration that enforce the given initial conditions.

These initial conditions are crucial because they select the unique solution from a family of solutions. Without such conditions, a differential equation might yield infinitely many possibilities.

The nature of linear differential equations

Linear differential equations are a broad class of equations characterized by the linearity in the unknown function and its derivatives. In simple terms, a linear differential equation will not have terms like \( y^2 \) or \( (y')^2 \). Instead, it maintains the form \( y' + p(t)y = g(t) \).

The linearity of these equations has significant implications, allowing for straightforward solutions and superposition, meaning solutions can be added together to form new solutions. In our exercise, each differential equation given is linear, facilitating the use of techniques like integrating factors to find solutions.

Linear equations are advantageous because they fit many real-world problems, modelling processes where the rate of change of a quantity is proportional to the quantity itself.