3.8 Q-15

Question Statement

Solve the differential equation:

$$1 + \cos x \tan y \frac{d y}{d x} = 0$$

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Background and Explanation

This is a first-order separable differential equation. The goal is to separate the variables $x$ and $y$ on opposite sides of the equation so that we can integrate each side independently. The key concepts used here involve trigonometric identities and basic integration techniques.

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Solution

1. Start with the given equation:

   The equation is:

$$1 + \cos x \tan y \frac{d y}{d x} = 0$$

2. Rearrange the equation:

   Move all terms involving $\frac{dy}{dx}$ to one side:

$$\cos x \tan y \frac{d y}{d x} = -1$$

3. Separate the variables:

   Divide both sides to isolate $dy$ on one side and $dx$ on the other:

$$\tan y , dy = -\frac{1}{\cos x} , dx$$

This can be rewritten as:

$$\frac{-\sin y}{\cos y} , dy = \sec x , dx$$

4. Integrate both sides:

   Now, we integrate both sides of the equation.

   • For the left-hand side, use the identity $\frac{d}{dy} (\cos y) = -\sin y$ to simplify the integral:

$$\int \frac{-\sin y}{\cos y} , dy = \int \sec x , dx$$

• The integral of $\frac{-\sin y}{\cos y}$ is $\ln |\cos y|$, and the integral of $\sec x$ is $\ln |\sec x + \tan x|$. Thus, we have:

$$\ln |\cos y| = \ln |\sec x + \tan x| + \ln |C|$$

5. Simplify the equation:

   Combine the logarithms on the right-hand side:

$$\ln |\cos y| = \ln |\sec x + \tan x| + \ln |C|$$

Exponentiate both sides to remove the logarithms:

$$|\cos y| = C |\sec x + \tan x|$$

6. Final equation:

   The solution is:

$$\cos y = C (\sec x + \tan x)$$

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Key Formulas or Methods Used

• Separation of Variables: Rearranged the equation so that terms involving $y$ were on one side and terms involving $x$ were on the other.
• Trigonometric Identities: Used basic trigonometric identities such as $\sec x = \frac{1}{\cos x}$ and $\frac{d}{dy}(\cos y) = -\sin y$.
• Integration: Applied standard integrals for $\sec x$ and $\frac{-\sin y}{\cos y}$.

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Summary of Steps

1. Start with the equation: $1 + \cos x \tan y \frac{d y}{d x} = 0$.
2. Rearrange to: $\cos x \tan y \frac{d y}{d x} = -1$.
3. Separate the variables: $\tan y , dy = -\frac{1}{\cos x} , dx$.
4. Integrate both sides: $\int \frac{-\sin y}{\cos y} , dy = \int \sec x , dx$.
5. Simplify: $\ln |\cos y| = \ln |\sec x + \tan x| + \ln |C|$.
6. Final solution: $\cos y = C (\sec x + \tan x)$.