Problem 27 - Calculus

Question

In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d w}{d \psi}=-w^{2} \tan \psi, \quad w(0)=2$$

Step-by-Step Solution

Answer

The solution is \( w = \frac{1}{\frac{1}{2} - \ln |\cos \psi|} \).

Step 1: Rewrite the Differential Equation

The equation is given as \( \frac{d w}{d \psi} = -w^2 \tan \psi \). To solve it using separation of variables, we first rewrite it in a form that allows us to separate the variables. Multiply both sides by \(d\psi\) and divide by \(w^2\), resulting in: \[ \frac{1}{w^2} dw = - \tan \psi \, d\psi \]

Step 2: Integrate Both Sides

Now, integrate both sides of the equation. Integrate the left side with respect to \(w\) and the right side with respect to \(\psi\):\[ \int \frac{1}{w^2} \, dw = \int - \tan \psi \, d\psi \]This gives us:\[ -\frac{1}{w} = \ln |\cos \psi| + C \] where \(C\) is the constant of integration.

Step 3: Solve for the Constant of Integration

Use the initial condition \(w(0) = 2\) to find the constant \(C\). Substitute \(\psi = 0\) and \(w = 2\) into the integrated equation:\[ -\frac{1}{2} = \ln |\cos 0| + C \]Since \(\cos 0 = 1\), \(|\cos 0| = 1\), so:\[ -\frac{1}{2} = 0 + C \]Thus, \( C = -\frac{1}{2} \).

Step 4: Solve for \(w\)

With \(C\) found, substitute it back into the equation:\[ -\frac{1}{w} = \ln |\cos \psi| -\frac{1}{2} \]Now solve for \(w\):\[ \frac{1}{w} = - \ln |\cos \psi| + \frac{1}{2} \]\[ w = \frac{1}{\frac{1}{2} - \ln |\cos \psi|} \]

Step 5: Verify the Solution

Finally, verify that the solution satisfies the original differential equation and the initial condition. Substituting \(\psi = 0\) back into the equation, confirm that \(w(0) = 2\) is satisfied by the function\[ w(0) = \frac{1}{\frac{1}{2} - \ln |1|} = \frac{1}{\frac{1}{2}} = 2 \]This confirms that our solution is consistent with the initial condition.

Key Concepts

Separation of Variables

Separation of variables is a powerful technique when solving differential equations, especially when dealing with equations that can be rearranged into a form where each variable occurs on one side. In our given equation, \( \frac{d w}{d \psi} = -w^2 \tan \psi \), it is essential to separate the variables \(w\) and \(\psi\).
Here's how it works:

• First, manipulate the equation to move all terms involving \(w\) to one side and all terms involving \(\psi\) to the other.
• Perform algebraic operations such as multiplying or dividing both sides by necessary variables and differentials.
• In this case, multiply both sides by \(d\psi\) and divide by \(w^2\) to achieve: \(\frac{1}{w^2} dw = - \tan \psi \ d\psi.\)
• This allows each side to be integrated separately, a crucial step towards finding a solution.

Executing the separation correctly helps simplify the integration process and lays the groundwork for incorporating initial conditions effectively.

Initial Conditions

Initial conditions are critical in the process of solving differential equations as they allow us to find a specific solution from the broad set of possible solutions. In our problem, the initial condition provided is \(w(0) = 2\).
This means that when \(\psi = 0\), the value of \(w\) should be 2.

• After integrating both sides, we encounter a constant of integration \(C\).
• To determine the value of \(C\), substitute the initial condition values into the integrated equation.
• For this, you replace \(\psi\) with 0 and \(w\) with 2 in the equation \(-\frac{1}{w} = \ln |\cos \psi| + C \).
• This step ensures we find the right constant \(C = -\frac{1}{2}\), making the solution specific to the initial condition.

Without initial conditions, our solution would be one of many potential solutions and wouldn’t be useful for solving real-world problems that usually start from a known state.

Integration

Integration is the mathematical process that reverses differentiation, playing a crucial role in solving differential equations. After separating the variables, integration becomes the next step. Let’s briefly look into how this works:

• We integrate both sides of the separated equation: \(\int \frac{1}{w^2} \, dw = \int - \tan \psi \, d\psi\).
• The left integral, \(\int \frac{1}{w^2} \, dw\), simplifies through integration to \(-\frac{1}{w}\), since integrating \(w^{-2}\) results in \(-w^{-1}\).
• The right integral, \(\int - \tan \psi \, d\psi\), becomes \(\ln |\cos \psi|\), using the integral results of trigonometric identities.
• Combining these, we have \(-\frac{1}{w} = \ln |\cos \psi| + C\), a crucial equation for our solution.

Integration transforms the separated differential equation into an expression that we can solve algebraically. It bridges the gap between the differential equation and the solution.

Solution Verification

Solution verification is essential to confirm that the derived solution satisfies both the original differential equation and the given initial conditions. In this context, let’s go through how it’s done:

• Substitute the determined expression for \(w\) back into the original differential equation \(\frac{d w}{d \psi} = -w^2 \tan \psi\) to ensure it holds true. This checks if our steps were correct.
• Incorporate the initial condition back to verify: ensuring \(w(0) = 2\) is satisfied.
• Use the constant obtained from integrating and condition to substitute: \(w = \frac{1}{\frac{1}{2} - \ln |\cos \psi|}\).
• Check if substituting \(\psi = 0\) gives \(w(0)=2\). In this case, it should simplify neatly, reaffirming accuracy.

Verification guarantees that you have not only followed the right steps but also solved correctly. This final step gives confidence in the exactness of your mathematical solution.