Question

Solve: \[ \frac{dy}{dx} = y\tan x \]

• A. \(y = C\sec x\)
• B. \(y = C\cos x\)
• C. \(y = C\sin x\)
• D. \(y = C\tan x\)

Hint:

Useful standard integrals: \[ \int \tan x\,dx = \ln |\sec x| + C \] For separable differential equations:
• first separate variables
• integrate both sides carefully
• absorb constants after exponentiation

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The given mathematical expression is a first-order ordinary differential equation (ODE).
In calculus, a differential equation relates a function with its derivatives.
This specific equation is categorized as a "separable" differential equation because the variables \( x \) and \( y \) can be completely isolated on opposite sides of the equality.
The goal is to find a function \( y(x) \) that satisfies this rate of change relationship.
Separation of variables is the most straightforward method for solving such first-order equations.
Step 2: Key Formula or Approach:
A separable equation follows the general structure:
\[ \frac{dy}{dx} = f(x)g(y) \]
We transform this into:
\[ \frac{1}{g(y)} dy = f(x) dx \]
Once separated, we integrate both sides independently to find the relationship between the variables.
Step 3: Detailed Explanation:
We begin with the initial equation:
\[ \frac{dy}{dx} = y \tan x \]
To separate the variables, we divide both sides by \( y \) and multiply both sides by \( dx \).
This yields the separated form:
\[ \frac{1}{y} dy = \tan x dx \]
Now, we apply the integration operator to both sides of the equation to eliminate the differentials:
\[ \int \frac{1}{y} dy = \int \tan x dx \]
Integrating the left side: The integral of \( 1/y \) with respect to \( y \) is the natural logarithm of the absolute value of \( y \).
\[ \ln |y| \]
Integrating the right side: The integral of \( \tan x \) is a standard trigonometric integral.
Recall that \( \tan x = \frac{\sin x}{\cos x} \). Using substitution where \( u = \cos x \), we find:
\[ \int \tan x dx = \ln |\sec x| + C_1 \]
Equating both sides, we have:
\[ \ln |y| = \ln |\sec x| + C_1 \]
To solve for \( y \), we exponentiate both sides (use the base \( e \)):
\[ e^{\ln |y|} = e^{\ln |\sec x| + C_1} \]
Applying the properties of exponents \( e^{a+b} = e^a \cdot e^b \) and the fact that \( e^{\ln z} = z \):
\[ |y| = e^{C_1} \cdot |\sec x| \]
Since \( e^{C_1} \) is a constant raised to a constant power, it is simply another arbitrary constant.
Let \( C = \pm e^{C_1} \). This allows us to remove the absolute value signs:
\[ y = C \sec x \]
Step 4: Final Answer:
The general solution to the differential equation is \( y = C \sec x \).
This matches the expression provided in Option (A).