Question 37 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

dydx=ytan⁡x;y=1 when x=0\frac{dy}{dx} = y \tan x; \quad y = 1 \text{ when } x = 0
Correct Answer

y = sec x

Exercise: EXERCISE 9.3 | Q: 14 | (Page 12)
For More Understanding

Explanation

This is a separable differential equation. The student needs to separate variables, integrate both sides, and then apply the initial condition y=1y = 1 when x=0x = 0 to find the particular solution.

Solution Steps

  1. Step 1: Separate the variables.

    Given:

    dydx=ytan⁡x\frac{dy}{dx} = y \tan x

    Separating:

    dyy=tan⁡x dx\frac{dy}{y} = \tan x \, dx

  2. Step 2: Integrate both sides.

    ∫1y dy=∫tan⁡x dx\int \frac{1}{y} \, dy = \int \tan x \, dx ln⁡∣y∣=ln⁡∣sec⁡x∣+C1\ln|y| = \ln|\sec x| + C_1

  3. Step 3: Express in exponential form.

    y=C⋅sec⁡x(where C=±eC1)y = C \cdot \sec x \quad (\text{where } C = \pm e^{C_1})

  4. Step 4: Apply the initial condition y=1y = 1 when x=0x = 0.

    1=C⋅sec⁡(0)1 = C \cdot \sec(0) 1=C⋅11 = C \cdot 1

    Therefore, C=1C = 1

  5. Step 5: Write the particular solution.

    ∴ y=sec⁡x\mathbf{y = \sec x}

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