What is the equation of a curve passing through (0,1) and whose differential equation is given by dy = y tan x dx ?

Options

A. y = cos x
B. y = sin x
C. y = sec x
D. y = cosec x

Answer

To solve the differential equation \( dy = y \tan x \, dx \) and find the equation of the curve passing through the point \( (0, 1) \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given differential equation:
\[ dy = y \tan x \, dx \]
We can rearrange this to separate the variables:
\[ \frac{dy}{y} = \tan x \, dx \]

### Step 2: Integrate Both Sides
Next, we integrate both sides. The left side integrates to:
\[ \int \frac{dy}{y} = \ln |y| + C_1 \]
The right side integrates to:
\[ \int \tan x \, dx = -\ln |\cos x| + C_2 \]
Thus, we have:
\[ \ln |y| = -\ln |\cos x| + C \]
where \( C = C_2 - C_1 \) is a constant.

### Step 3: Simplify the Equation
We can exponentiate both sides to eliminate the logarithm:
\[ |y| = e^{-\ln |\cos x| + C} = e^C \cdot \frac{1}{|\cos x|} = \frac{C_1}{|\cos x|} \]
Letting \( C_1 = e^C \) (a positive constant), we can write:
\[ y = \frac{C_1}{\cos x} \]

### Step 4: Determine the Constant Using Initial Condition
We know the curve passes through the point \( (0, 1) \). Thus, we substitute \( x = 0 \) and \( y = 1 \):
\[ 1 = \frac{C_1}{\cos(0)} = \frac{C_1}{1} \]
This gives us \( C_1 = 1 \).

### Step 5: Write the Final Equation
Substituting \( C_1 \) back into the equation gives us:
\[ y = \frac{1}{\cos x} \]
or equivalently:
\[ y = \sec x \]

### Final Answer
The equation of the curve is:
\[ y = \sec x \]

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