Problem 49 - Calculus Early Transcendentals

Question

Determine whether the following statements are true and give an explanation or counterexample. a. The general solution of \(y^{\prime}(t)=20 y\) is \(y=e^{20 t}\). b. The functions \(y=2 e^{-2 t}\) and \(y=10 e^{-2 t}\) do not both satisfy the differential equation \(y^{\prime}+2 y=0\). c. The equation \(y^{\prime}(t)=t y+2 y+2 t+4\) is not separable. d. A solution of \(y^{\prime}(t)=2 \sqrt{y}\) is \(y=(t+1)^{2}\).

Step-by-Step Solution

Answer

Answer: Yes, the given solution is a correct solution for the given differential equation.

Step 1: a. Verifying the general solution

Firstly, we need to find the general solution of the differential equation, which is: \(y^{\prime}(t)=20 y\). Upon inspection, we can see that this is a first-order linear homogeneous ODE. One form of the general solution for this type of ODE is \(y(t)=Ce^{rt}\), where \(r=20\) and \(C\) is a constant. Thus, the general solution is \(y(t)=Ce^{20t}\). The given function \(y=e^{20t}\) is just a specific solution when \(C=1\). So, the statement is false.

Step 2: b. Verifying the functions

We are given two functions, \(y_1(t)=2e^{-2t}\) and \(y_2(t)=10e^{-2t}\). Let's check if they both satisfy the given differential equation \(y^{\prime}+2y=0\). For \(y_1(t)\): $$y_{1}^{\prime}(t) = -4e^{-2t}$$ $$y_{1}^{\prime}(t) + 2y_1(t) = -4e^{-2t}+4e^{-2t} = 0$$ So, \(y_1(t)\) satisfies the equation. For \(y_2(t)\): $$y_{2}^{\prime}(t) = -20e^{-2t}$$ $$y_{2}^{\prime}(t) + 2y_2(t) = -20e^{-2t}+20e^{-2t} = 0$$ So, \(y_2(t)\) also satisfies the equation. Therefore, the statement is false, as both functions satisfy the given differential equation.

Step 3: c. Determining if the equation is separable

We need to determine if the given equation \(y^{\prime}(t) = ty+2y+2t+4\) is separable or not. A separable equation can be written in the form \(\frac{dy}{dt} = g(t)h(y)\), where we can separate the \(y\)-terms and the \(t\)-terms. In this case, the equation cannot be written as a product of two functions where one function depends only on \(t\) and the other depends only on \(y\). Therefore, the statement is true – the given equation is not separable.

Step 4: d. Checking the solution

We need to verify if the given solution \(y(t)=(t+1)^2\) is correct for the given differential equation \(y^{\prime}(t) = 2\sqrt{y}\). Let's find the derivative of the given solution: $$y^{\prime}(t) = 2(t+1)$$ To find if this solution is correct, we need to substitute this into the given differential equation and simplify: $$2(t+1) = 2\sqrt{(t+1)^2}$$ $$2(t+1) = 2(t+1)$$ As the resulting equation is true for all values of \(t\), the statement is true – the given solution is a correct solution for the given differential equation.

Key Concepts

First-Order Linear Homogeneous ODE

A first-order linear homogeneous ordinary differential equation (ODE) is a type of differential equation commonly seen in mathematical studies. It takes the form \( y'(t) = ay(t) \), where \( y' \) denotes the derivative of \( y \) with respect to \( t \), and \( a \) is a constant coefficient. These equations are called "homogeneous" because they are set to zero when all terms are transferred to one side.
The solution of a first-order linear homogeneous ODE can be expressed as \( y(t) = Ce^{at} \), where \( C \) is an arbitrary constant. This general form shows that the solutions are exponential functions, and they are often determined by initial conditions provided in specific problems. For instance, when \( a = 20 \), the general solution becomes \( y(t) = Ce^{20t} \). If an initial condition like \( y(0)=1 \) is given, \( C \) would be determined to be 1, making \( y(t) = e^{20t} \) the specific solution.
As demonstrated in the example, the statement that \( y = e^{20t} \) is the general solution is incorrect because it assumes \( C = 1 \), making it a specific case of the broader general solution.

Separable Equations

Separable equations are a special class of differential equations that can be written in the form \( \frac{dy}{dt} = g(t) h(y) \). The crucial feature that defines them is the ability to factor the equation into two separate parts: one involving only \( y \) and the other involving only \( t \). This separation allows us to solve both sides independently, usually by integrating separately each part.
For example, if you have \( y' = ty + 2y + 2t + 4 \), determining if it’s separable requires expressing it as the product of a function of \( t \) and a function of \( y \). In this case, rearranging doesn’t accomplish this division, indicating the equation isn’t separable.

• Non-Separable Form: Ensures terms can't be isolated easily on either side of the ODE.
• Verification: Attempts to write it in \( g(t) h(y) \) form will show it’s not possible.

This nature of non-separability in certain equations implies that alternative approaches are necessary to solve them, such as recognizing certain substitution methods or using numerical solutions in more advanced contexts.

Solution Verification

Solution verification is an essential part of working with differential equations, ensuring that proposed solutions satisfy the initial equation. To verify whether a function is indeed a solution, it involves substituting the function back into the original differential equation and simplifying to check if the two sides equate.
For example, let's consider the differential equation \( y'(t) = 2 \sqrt{y} \) and a proposed solution \( y(t) = (t+1)^2 \). To verify, you need to:

• Find the derivative \( y'(t) = 2(t+1) \).
• Substitute back into the differential equation: \( 2(t+1) = 2 \sqrt{(t+1)^2} \).
• Simplifying gives \( 2(t+1) = 2(t+1) \), which holds true.

Since the expression remains consistent for all values of \( t \), we confirm that \( y(t) = (t+1)^2 \) is indeed a valid solution. This process highlights the importance of careful verification to ensure understanding and correctness in differential equations.