Question 33 of 96intermediate🔧 ApplyLong Answer4 marks

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

dydx=sin⁡−1x\frac{dy}{dx} = \sin^{-1} x
Correct Answer
y=xsin⁡−1x+1−x2+Cy = x \sin^{-1}x + \sqrt{1 - x^2} + C

where CC is an arbitrary constant.

Exercise: EXERCISE 9.3 | Q: 9 | (Page 12)
For More Understanding

Explanation

This is a direct integration problem. The differential equation dydx=sin⁡−1x\frac{dy}{dx} = \sin^{-1}x can be solved by integrating both sides with respect to xx. The integration of sin⁡−1x\sin^{-1}x requires integration by parts, which is a standard technique taught in calculus.

Solution Steps

  1. Step 1: Write the given differential equation.

    dydx=sin⁡−1x\frac{dy}{dx} = \sin^{-1}x

  2. Step 2: Integrate both sides with respect to xx.

    ∫dy=∫sin⁡−1x dx\int dy = \int \sin^{-1}x \, dx y=∫sin⁡−1x dxy = \int \sin^{-1}x \, dx

  3. Step 3: Apply integration by parts to integrate sin⁡−1x\sin^{-1}x.

    Let u=sin⁡−1xu = \sin^{-1}x and dv=dxdv = dx

    Then du=11−x2 dxdu = \frac{1}{\sqrt{1-x^2}} \, dx and v=xv = x

  4. Step 4: Using ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du:

    ∫sin⁡−1x dx=x⋅sin⁡−1x−∫x1−x2 dx\int \sin^{-1}x \, dx = x \cdot \sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}} \, dx

  5. Step 5: For ∫x1−x2 dx\int \frac{x}{\sqrt{1-x^2}} \, dx, substitute t=1−x2t = 1 - x^2, dt=−2x dxdt = -2x \, dx:

    ∫x1−x2 dx=−1−x2\int \frac{x}{\sqrt{1-x^2}} \, dx = -\sqrt{1-x^2}

  6. Step 6: Combine results.

    y=x⋅sin⁡−1x+1−x2+Cy = x \cdot \sin^{-1}x + \sqrt{1-x^2} + C
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