Question 85 of 96intermediate🔧 ApplyLong Answer4 marks

Find the general solution of the differential equation dydx+1−y21−x2=0\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0.

Correct Answer

The given differential equation is:

dydx+1−y21−x2=0\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0

To find the general solution, we first rearrange the terms to separate the variables xx and yy. Moving the square root term to the right side gives:

dydx=−1−y21−x2\frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}

Separating the variables xx and yy, we write the equation as:

dy1−y2=−dx1−x2\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}

Integrating both sides of the equation with respect to their respective variables, we get:

∫dy1−y2=−∫dx1−x2\int \frac{dy}{\sqrt{1-y^2}} = - \int \frac{dx}{\sqrt{1-x^2}}

Using the standard integration formula for inverse trigonometric functions, we obtain:

sin⁡−1y=−sin⁡−1x+C\sin^{-1} y = -\sin^{-1} x + C

where CC is the constant of integration. This can also be written as:

sin⁡−1y+sin⁡−1x=C\sin^{-1} y + \sin^{-1} x = C

This is the general solution of the given differential equation.

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 4 | (Page 34)
For More Understanding

Explanation

This question is solved using the method of separation of variables, which is explicitly demonstrated in Example 5 of the provided context. Example 5 shows the process of rearranging a differential equation to group yy terms with dydy and xx terms with dxdx, followed by integration.

In this specific problem, separating the variables yields integrals of the form ∫dz1−z2\int \frac{dz}{\sqrt{1-z^2}}, which corresponds to the inverse sine function. This mirrors the logic in Example 5 where the integral ∫dz1+z2\int \frac{dz}{1+z^2} corresponds to the inverse tangent function.

The final solution includes the constant of integration CC, consistent with the general solutions found in the textbook examples.

Solution Steps

  1. Step 1: Rearrange the given differential equation to isolate the derivative term and separate variables.

  2. Step 2: Write the equation in the form dy1−y2=−dx1−x2\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}.

  3. Step 3: Integrate both sides of the equation.

  4. Step 4: Apply the standard integral formula ∫dz1−z2=sin⁡−1z\int \frac{dz}{\sqrt{1-z^2}} = \sin^{-1} z and add the constant of integration CC.

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