Problem 35 - Calculus Early Transcendentals

Question

Determine whether the following equations are separable. If so, solve the given initial value problem. $$\frac{d y}{d t}=t y+2, y(1)=2$$

Step-by-Step Solution

Answer

Question: Determine whether the given equation $$\frac{dy}{dt} = ty + 2$$ is separable. If it is, solve the equation and obtain the solution satisfying the initial condition y(1)=2. Answer: The given equation is separable, and the solution satisfying the initial condition y(1)=2 is $$y(t) = e^{\frac{1}{2}t^2} - 2$$.

Step 1: Check if the equation is separable

To check if the equation is separable, we'll try to rewrite the given equation in the form $$\frac{dy}{y+2} = t dt$$ both sides should be the function of y and t only. The given equation is $$\frac{dy}{dt} = ty + 2$$ Now, we try to separate the variables. $$\frac{dy}{ty + 2} = dt$$ We now have the equation in the form we desired, and the ODE is separable.

Step 2: Integrate both sides

Integrate both sides of the equation with respect to their respective variables: $$\int_{ }^{ }\frac{dy}{y + 2} = \int_{ }^{ }t dt$$ On the left side, the integral is straightforward, and on the right side, we need to find the antiderivative of t with respect to t. $$\ln|y + 2| = \frac{1}{2}t^2 + C$$ where C is the constant of integration.

Step 3: Solve for y(t)

Now, we want to isolate y and write its expression as a function of t: $$y + 2 = e^{\frac{1}{2}t^2 + C}$$ Move 2 to the other side: $$y(t) = e^{\frac{1}{2}t^2 + C} - 2$$

Step 4: Apply the initial condition

To find the constant C, we'll use the initial condition y(1) = 2: $$2 = e^{\frac{1}{2}(1)^2 + C} - 2$$ Solving for C, we get: $$e^{C} = 1 \implies C = 0$$

Step 5: Write the final solution

Now that we have the constant C, let's substitute it back into the equation for y(t): $$y(t) = e^{\frac{1}{2}t^2} - 2$$ The final solution to the initial value problem is: $$y(t) = e^{\frac{1}{2}t^2} - 2$$

Key Concepts

Initial Value Problem

In mathematics, an initial value problem involves an ordinary differential equation (ODE) alongside a specified initial condition. The purpose is to find a function that satisfies both the differential equation and the initial value. In our problem, the equation \( \frac{dy}{dt} = ty + 2 \) with the initial condition \( y(1) = 2 \) defines the initial value problem.
This setup means that when \( t = 1 \), the value of the function \( y(t) \) must be 2. We use this condition to determine any constants in the solution once the ODE is solved.

To solve such problems, these are necessary steps:

• Identify if the equation can be rewritten for easier solving, such as checking if it’s separable.
• Use integration techniques to solve the ODE.
• Apply the initial conditions to find the specific solution of the differential equation.

This approach ensures that the function aligns perfectly with the condition provided initially, giving us a precise solution suitable for real-world applications.

Integration

Integration is a core process in solving differential equations, especially when trying to find a function that describes a system's behavior over time. In the case of separable differential equations, integrating each side of the equation separately is essential.

In our example, after separating variables, we arrived at the equation \( \frac{dy}{y+2} = t \, dt \). This separation allows us to integrate both sides individually:

• Left side: \( \int \frac{dy}{y+2} \), which uses the natural logarithm \( \ln|y + 2| \).
• Right side: \( \int t \, dt \), which results in \( \frac{1}{2}t^2 \).

Integration generally involves finding the antiderivative, which is the reverse process of differentiation, and adding a constant of integration \( C \) because integration reflects an infinite number of potential functions. This constant is later determined with the initial condition.
It is helpful to be familiar with common integrals and techniques such as substitution and integration by parts to tackle more complex differential equations.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) describe relations between a function and its derivatives. They are called "ordinary" because they involve functions of a single variable, as opposed to partial differential equations which involve multiple variables.

In our example, \( \frac{dy}{dt} = ty + 2 \) is a first-order ODE involving the function \( y(t) \) and its derivative \( \frac{dy}{dt} \). Solving an ODE typically means finding the function that satisfies the given relationship.

There are several methods to approach solving ODEs:

• Separable Equations: These can be rewritten so all terms involving one variable are on one side of the equation and all terms involving the other variable are on the opposite side.
• Linear Equations: Involves constant coefficients and can often be tackled with techniques like integration factors.
• Exact Equations: Requires finding a potential function from which the differential form of the ODE derives.

Practicing these techniques aids in addressing various ordinary differential equations, making it crucial for students to understand and apply them correctly, as they are foundational in fields like physics, engineering, and beyond.