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Magazine Chapter 11: Problem 27
In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d w}{d \psi}=-w^{2} \tan \psi, \quad w(0)=2$$
Short Answer
Expert verified
The solution is \( w = \frac{1}{\frac{1}{2} - \ln |\cos \psi|} \).
Step by step solution
01
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: \[ \frac{1}{w^2} dw = - \tan \psi \, d\psi \]
02
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.
03
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, \( C = -\frac{1}{2} \).
04
Solve for \(w\)
With \(C\) found, substitute it back into the equation:\[ -\frac{1}{w} = \ln |\cos \psi| -\frac{1}{2} \]Now solve for \(w\):\[ \frac{1}{w} = - \ln |\cos \psi| + \frac{1}{2} \]\[ w = \frac{1}{\frac{1}{2} - \ln |\cos \psi|} \]
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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.
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.
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鈥檚 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.
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鈥檚 go through how it鈥檚 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.
