JEE Challenger
More from Differential Equations
MathMedAdv 2008 • Paper 2

Solution and Statements for First Order Differential Equation

Let a solution y=y(x)y = y(x) of the differential equation
xx2−1 dy−yy2−1 dx=0x \sqrt{x^2 - 1} \ dy - y \sqrt{y^2 - 1} \ dx = 0
satisfy y(2)=23y(2) = \frac{2}{\sqrt{3}}.

STATEMENT-1 : y(x)=sec⁡(sec⁡−1x−π6)y(x) = \sec \left( \sec^{-1} x - \frac{\pi}{6} \right)

and

STATEMENT-2 : y(x)y(x) is given by
1y=23x−1−1x2\frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}}

Options

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C

STATEMENT-1 is True, STATEMENT-2 is False

Correct
D

STATEMENT-1 is False, STATEMENT-2 is True

Topics & Concepts

Differential EquationsDifferential Equations

Step-by-Step Solution

To determine the correctness of STATEMENT-1 and STATEMENT-2, we begin by solving the given first-order differential equation:

xx2−1 dy−yy2−1 dx=0x \sqrt{x^2 - 1} \ dy - y \sqrt{y^2 - 1} \ dx = 0

Separating the variables, we get:
dyyy2−1=dxxx2−1\frac{dy}{y \sqrt{y^2 - 1}} = \frac{dx}{x \sqrt{x^2 - 1}}

Integrating both sides with respect to their corresponding variables:
∫dyyy2−1=∫dxxx2−1\int \frac{dy}{y \sqrt{y^2 - 1}} = \int \frac{dx}{x \sqrt{x^2 - 1}}

Using the standard integral formula ∫dzzz2−1=sec⁡−1z+C\int \frac{dz}{z \sqrt{z^2 - 1}} = \sec^{-1} z + C, we obtain:
sec⁡−1y=sec⁡−1x+C\sec^{-1} y = \sec^{-1} x + C

We are given the initial condition y(2)=23y(2) = \frac{2}{\sqrt{3}}. Substituting x=2x = 2 and y=23y = \frac{2}{\sqrt{3}} into the equation:
sec⁡−1(23)=sec⁡−1(2)+C\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \sec^{-1}(2) + C

Since sec⁡−1(23)=π6\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} and sec⁡−1(2)=π3\sec^{-1}(2) = \frac{\pi}{3}:
π6=π3+C  ⟹  C=−π6\frac{\pi}{6} = \frac{\pi}{3} + C \implies C = -\frac{\pi}{6}

Substituting C=−π6C = -\frac{\pi}{6} back into the equation yields:
sec⁡−1y=sec⁡−1x−π6\sec^{-1} y = \sec^{-1} x - \frac{\pi}{6}

Taking the secant of both sides:
y(x)=sec⁡(sec⁡−1x−π6)y(x) = \sec \left( \sec^{-1} x - \frac{\pi}{6} \right)

Thus, STATEMENT-1 is True.

Now, to evaluate STATEMENT-2, we take the reciprocal of y(x)y(x):
1y=cos⁡(sec⁡−1y)=cos⁡(sec⁡−1x−π6)\frac{1}{y} = \cos \left( \sec^{-1} y \right) = \cos \left( \sec^{-1} x - \frac{\pi}{6} \right)

Using the trigonometric identity cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B:
1y=cos⁡(sec⁡−1x)cos⁡(π6)+sin⁡(sec⁡−1x)sin⁡(π6)\frac{1}{y} = \cos(\sec^{-1} x) \cos\left(\frac{\pi}{6}\right) + \sin(\sec^{-1} x) \sin\left(\frac{\pi}{6}\right)

For x≥1x \ge 1:
cos⁡(sec⁡−1x)=1x\cos(\sec^{-1} x) = \frac{1}{x}
sin⁡(sec⁡−1x)=1−1x2\sin(\sec^{-1} x) = \sqrt{1 - \frac{1}{x^2}}

Substituting these values along with cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}:
1y=32x+121−1x2\frac{1}{y} = \frac{\sqrt{3}}{2x} + \frac{1}{2} \sqrt{1 - \frac{1}{x^2}}

Comparing this correct expression with the equation given in STATEMENT-2:
1y=23x−1−1x2\frac{1}{y} = \frac{2\sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^2}}

We see that STATEMENT-2 is incorrect. Thus, STATEMENT-2 is False.

Conclusion:
STATEMENT-1 is True, STATEMENT-2 is False.

Correct Option: C

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