Question 45 of 96intermediate🔧 ApplyLong Answer4 marks

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

dydx=4−y2(−2<y<2)\frac{dy}{dx} = \sqrt{4 - y^2} \quad (-2 < y < 2)

Correct Answer

y = 2 sin(x + C)

Exercise: EXERCISE 9.3 | Q: 2 | (Chapter: 11)
For More Understanding

Explanation

This is a separable differential equation. The student needs to separate the variables yy and xx on opposite sides, then integrate both sides. The integral on the left side uses the standard formula for the inverse sine function.

Solution Steps

  1. Step 1: Separate the variables — Rewrite the equation as:

    dy4−y2=dx\frac{dy}{\sqrt{4 - y^2}} = dx

  2. Step 2: Integrate both sides —

    ∫dy4−y2=∫dx\int \frac{dy}{\sqrt{4 - y^2}} = \int dx

  3. Step 3: Apply standard integral formula — Using ∫dxa2−x2=sin⁡−1 ⁣(xa)+C\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\!\left(\frac{x}{a}\right) + C, we get:

    sin⁡−1 ⁣(y2)=x+C\sin^{-1}\!\left(\frac{y}{2}\right) = x + C

  4. Step 4: Solve for yy —

    y2=sin⁡(x+C)\frac{y}{2} = \sin(x + C)

    ∴ y=2sin⁡(x+C)y = 2 \sin(x + C) is the general solution.

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