Problem 24 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d y}{d t}=y^{2}(1+t), \quad y=2 \text { when } t=1$$

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{t + \frac{t^2}{2} - 2} \).

Step 1: Separate the Variables

To use the separation of variables method, we start by rewriting the given differential equation \( \frac{d y}{d t} = y^{2}(1+t) \) in the form that allows each variable to be on separate sides of the equation. Divide both sides by \( y^2 \) and multiply by \( dt \) to get: \[ \frac{1}{y^2} \, dy = (1+t) \, dt. \]

Step 2: Integrate Both Sides

Integrate both sides of the equation obtained in Step 1. The left side becomes \[ \int \frac{1}{y^2} \, dy = \int (1+t) \, dt. \] The solution to these integrals is: \[ -\frac{1}{y} = t + \frac{t^2}{2} + C, \] where \( C \) is the constant of integration.

Step 3: Solve for the Constant of Integration

Apply the initial condition \( y = 2 \) when \( t = 1 \) to solve for \( C \). Substitute these values into the integrated equation: \[ -\frac{1}{2} = 1 + \frac{1^2}{2} + C. \] Simplify to find: \[ -\frac{1}{2} = 1 + \frac{1}{2} + C \Rightarrow C = -2. \]

Step 4: Complete the General Solution

Substitute the constant \( C = -2 \) back into the equation from Step 2, \[ -\frac{1}{y} = t + \frac{t^2}{2} - 2. \] Rearrange and solve for \( y \) to express the solution explicitly: \[ y = -\frac{1}{t + \frac{t^2}{2} - 2}. \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve a function and its derivatives. They play a crucial role in modeling real-world systems where the rate of change is significant, such as in physics, engineering, and biology. In differential equations, we often aim to find the function that satisfies the equation.
In the exercise, the differential equation is given as \( \frac{dy}{dt} = y^2(1+t) \). This equation describes how a quantity \( y \) changes with respect to another variable \( t \).
There are various methods to solve differential equations, and one such method is separation of variables. This technique is particularly useful when both sides of the equation can be expressed as a product of functions, each involving only one of the variables.
The goal is to isolate each variable on different sides of the equation, which simplifies the process of solving the equation through integration.

Integration

Integration is a fundamental concept in calculus, used to find a function's antiderivative or area under the curve. When solving differential equations, integration helps to find the function from its derivative.
In our exercise, once the variables were separated as \( \frac{1}{y^2} \, dy = (1+t) \, dt \), the next step was to integrate both sides.

• The left side: \( \int \frac{1}{y^2} \, dy = -\frac{1}{y} + C \)
• The right side: \( \int (1+t) \, dt = t + \frac{t^2}{2} + C \)

The integration process results in an equation \( -\frac{1}{y} = t + \frac{t^2}{2} + C \) where \( C \) is the constant of integration. Identifying \( C \) is critical, as it affects the particular solution of the problem.

Initial Conditions

Initial conditions are specific values given for the functions or their derivatives at a particular point. They are essential in determining the unique solution to a differential equation because integration can introduce an arbitrary constant, as seen with \( C \) in the integrated solution.
In this exercise, the initial condition \( y = 2 \) when \( t = 1 \) is used to find the value of \( C \). By substituting these initial values into the equation \( -\frac{1}{y} = t + \frac{t^2}{2} + C \), we get:

• \( -\frac{1}{2} = 1 + \frac{1}{2} + C \)
• Simplifying this gives \( C = -2 \)

Once \( C \) is determined, it can be substituted back into the integrated equation, leading to the particular solution \( y = -\frac{1}{t + \frac{t^2}{2} - 2} \). Hence, initial conditions provide the crucial link to tailor the general solution to a specific scenario.