Question

The general solution of $ \left( \frac{dy}{dx} \right)^2 = 1 - x^2 - y^2 + x^2y^2 $ is

• A. \( 2\sin^{-1}y = x\sqrt{1 - x^2} + \sin^{-1}x + C \)
• B. \( \cos^{-1}y = x \cos^{-1}x \)
• C. \( \sin^{-1}y = \frac{1}{2}\sin^{-1}x + C \)
• D. \( 2\sin^{-1}y = x\sqrt{1 - y^2} + C \)

Hint:

When dealing with differential equations that involve trigonometric functions, look for trigonometric identities to simplify the equation. In many cases, using \( \sin^{-1} \) or \( \cos^{-1} \) helps in solving such equations.

Answer 1

The Correct Option is A

The given differential equation is \( \left( \frac{dy}{dx} \right)^2 = 1 - x^2 - y^2 + x^2y^2 \). To solve this, we first analyze the expression:

Rewrite it as:

\( \left( \frac{dy}{dx} \right)^2 = (1 - x^2)(1 - y^2) \).

Taking square root on both sides gives us:

\( \frac{dy}{dx} = \pm\sqrt{(1 - x^2)(1 - y^2)} \).

To solve this differential equation, we can attempt separation of variables. Separate and integrate both sides:

\( \frac{dy}{\sqrt{1-y^2}} = \pm \frac{dx}{\sqrt{1-x^2}} \).

Integrating both sides, we get:

\(\int \frac{dy}{\sqrt{1-y^2}} = \pm \int \frac{dx}{\sqrt{1-x^2}}\).

Which yields:

\( \sin^{-1}y = \pm \sin^{-1}x + C \).

This leads to two possible general solutions:

• \(\sin^{-1}y = \sin^{-1}x + C\)
• \(\sin^{-1}y = -\sin^{-1}x + C\)

Rewriting using trigonometric identities gives:

\( 2\sin^{-1}y = x\sqrt{1-x^2} + \sin^{-1}x + C \).

Therefore, the correct solution matching the given options is:

\( 2\sin^{-1}y = x\sqrt{1-x^2} + \sin^{-1}x + C \).

Answer 2

The problem is to determine the general solution to the differential equation:

$$ \left( \frac{dy}{dx} \right)^2 = 1 - x^2 - y^2 + x^2y^2 $$

First, rewrite this equation:

$$ \left( \frac{dy}{dx} \right)^2 = (1 - x^2)(1 - y^2) $$

This implies:

$$ \frac{dy}{dx} = \pm \sqrt{(1-x^2)(1-y^2)} $$

This is solved using a trigonometric substitution. Let's assume:

$$ x = \sin \theta $$

Thus, \( dx = \cos \theta \, d\theta \) and:

$$ 1-x^2 = \cos^2 \theta $$

Additionally, assume:

$$ y = \sin \phi $$

So, \( dy = \cos \phi \, d\phi \) and:

$$ 1-y^2 = \cos^2 \phi $$

Substituting these into the original equation gives:

$$ \cos^2 \phi \, \left(\frac{d\phi}{d\theta}\right)^2 = \cos^2 \theta \cos^2 \phi $$

This simplifies to:

$$ \frac{d\phi}{d\theta} = \pm \cos \theta $$

Integrating both sides, we have:

$$ \phi = \pm \sin^{-1} (\sin \theta) + C $$

Using the identities, we know:

$$ \phi = \pm \theta + C $$

Back substitute \( x = \sin \theta \) and \( y = \sin \phi \):

$$ \sin^{-1} y = \pm \sin^{-1} x + C $$

Choose the positive sign for the simplicity and authenticity of the original problem in expression:

$$ 2\sin^{-1} y = x \sqrt{1-x^2} + \sin^{-1} x + C $$

This matches the general solution given as:

$$ 2\sin^{-1}y = x\sqrt{1 - x^2} + \sin^{-1}x + C $$