Question 25 of 96intermediate🔧 ApplyLong Answer4 marks

For each of the differential equations in Exercises 1 to 10, find the general solution:

dydx=1−cos⁡x1+cos⁡x\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}
Correct Answer

The general solution is:

y=2tan⁡(x2)−x+Cy = 2 \tan\left(\frac{x}{2}\right) - x + C

where CC is an arbitrary constant.

Exercise: EXERCISE 9.3 | Q: 1 | (Page 11)
For More Understanding

Explanation

The given differential equation is separable. By writing

dydx=1−cos⁡x1+cos⁡x\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}

we separate variables as

dy=1−cos⁡x1+cos⁡x dxdy = \frac{1 - \cos x}{1 + \cos x}\, dx

The integrand simplifies using the trigonometric identity

1−cos⁡x1+cos⁡x=tan⁡2(x2)\frac{1 - \cos x}{1 + \cos x} = \tan^2\left(\frac{x}{2}\right)

Then ∫tan⁡2(x2)dx\displaystyle\int \tan^2\left(\frac{x}{2}\right) dx is evaluated by expressing tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1, leading to the solution. The constant of integration CC appears after integration.

Solution Steps

  1. Step 1: Separate variables:

    dy=1−cos⁡x1+cos⁡x dxdy = \frac{1 - \cos x}{1 + \cos x}\, dx

  2. Step 2: Simplify the right side using the identity:

    1−cos⁡x1+cos⁡x=tan⁡2(x2)\frac{1 - \cos x}{1 + \cos x} = \tan^2\left(\frac{x}{2}\right)

  3. Step 3: Integrate both sides:

    ∫dy=∫tan⁡2(x2)dx\int dy = \int \tan^2\left(\frac{x}{2}\right) dx

  4. Step 4: Rewrite:

    tan⁡2(x2)=sec⁡2(x2)−1\tan^2\left(\frac{x}{2}\right) = \sec^2\left(\frac{x}{2}\right) - 1

  5. Step 5: Integrate term by term:

    ∫sec⁡2(x2)dx=2tan⁡(x2)\int \sec^2\left(\frac{x}{2}\right) dx = 2 \tan\left(\frac{x}{2}\right)

    since ddx[tan⁡(x2)]=12sec⁡2(x2)\frac{d}{dx}\left[\tan\left(\frac{x}{2}\right)\right] = \frac{1}{2}\sec^2\left(\frac{x}{2}\right), and

    ∫1 dx=x\int 1\, dx = x

  6. Step 6: Combine results:

    y=2tan⁡(x2)−x+Cy = 2 \tan\left(\frac{x}{2}\right) - x + C

    where CC is the constant of integration.

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