Problem 41 - Calculus

Question

Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d y}{d t}=k y^{2}\left(1+t^{2}\right)$$

Step-by-Step Solution

Answer

\( y = -\frac{1}{kt + \frac{kt^3}{3} + C} \)

Step 1: Separate Variables

The given differential equation is \( \frac{dy}{dt} = ky^2(1+t^2) \). To solve it, we begin by separating the variables. We will move all terms involving \( y \) to one side and all terms involving \( t \) to the other side. This gives us: \( \frac{1}{y^2} \, dy = k(1+t^2) \, dt \).

Step 2: Integrate Both Sides

Next, we will integrate both sides of the equation. The left side with respect to \( y \) and the right side with respect to \( t \). The integral of \( \frac{1}{y^2} \) is \( -\frac{1}{y} + C_1 \), and the integral of \( k(1+t^2) \) is \( kt + \frac{kt^3}{3} + C_2 \). Thus, we have: \( -\frac{1}{y} = kt + \frac{kt^3}{3} + C \), where \( C \) includes both integration constants \( C_1 \) and \( C_2 \).

Step 3: Solve for y

To find the expression for \( y \), we solve \( -\frac{1}{y} = kt + \frac{kt^3}{3} + C \) for \( y \). Taking the reciprocal, we have \( y = -\frac{1}{kt + \frac{kt^3}{3} + C} \).

Step 4: Simplify the Expression

Finally, we simplify the expression for \( y \), if possible. However, in its current form, \( y = -\frac{1}{kt + \frac{kt^3}{3} + C} \) is as simplified as it can be without specific values for \( k \) and \( C \).

Key Concepts

Separation of Variables

Separation of Variables is a method used to solve differential equations. It is a powerful technique because it allows us to break down complex equations into simpler parts. This is essential for solving equations like \(\frac{dy}{dt} = ky^2(1+t^2)\).
The aim is to separate the variables, typically \(y\) and \(t\) in our case, to opposite sides of the equation. This conversion simplifies the integration process later on.
Here's the step-by-step breakdown:

• Identify all terms involving \(y\) and \(t\).
• Organize them so that all \(y\)-related terms are on one side, and all \(t\)-related terms are on the other side.
• This allows us to separate the equation into the form \( \frac{1}{y^2} \, dy = k(1+t^2) \, dt \).

After achieving this form, each side can be integrated individually with respect to its own variable.

Integration

Integration is a fundamental concept in calculus. It's the process of finding the antiderivative or the "integral" of a function. This is crucial in solving the separated variable differential equation.
Once we have the separated form \( \frac{1}{y^2} \, dy = k(1+t^2) \, dt \), we integrate both sides:

• The left side, \( \int \frac{1}{y^2} \, dy \), results in \( -\frac{1}{y} + C_1 \). This is because \( \frac{1}{y^2} \) integrates to \( -\frac{1}{y} \).


• The right side, \( \int k(1+t^2) \, dt \), simplifies to \( kt + \frac{kt^3}{3} + C_2 \), as each term's integral is taken separately and combined.

After integration, the solution combines the constants from both sides into a single constant \( C \). These constants account for the indefinite nature of integration, as they can vary to satisfy different initial conditions.

Initial Conditions

Initial conditions are often provided in differential equations to find a specific solution to a problem. They help determine the integration constant \( C \) and refine the general solution obtained through integration.
In our example, initial conditions would allow us to solve for \( C \) by substituting known values of \( y \) and \( t \). This step is crucial when a particular solution is desired rather than a family of solutions.
Without specific initial conditions, our solution remains in its general form:

• \( y = -\frac{1}{kt + \frac{kt^3}{3} + C} \).

Thus, incorporating initial conditions gives more precision to our solution, making it applicable to real-world problems where initial data is usually known or measured. Initial conditions align the mathematical model with specific scenarios or experiments.