Let y=y(x) be a solution of the differential equation, \sqrt{1-x^{2}}\frac{dy}{dx}+\sqrt{1-y^{2}}=0,\left | x \right |<1. If y\left ( \frac{1}{2} \right )=\frac{\sqrt{3}}{2}, then y\left ( \frac{-1}{\sqrt{2}} \right ) is equal to :
Option: 1 -\frac{1}{\sqrt{2}}
Option: 2 -\frac{\sqrt{3}}{2}
Option: 3 \frac{1}{\sqrt{2}}
Option: 4 \frac{\sqrt{3}}{2}
 

Answers (1)

 

 

Formation of Differential Equation and Solutions of a Differential Equation -

This is the general solution of the differential equation (2), which represents the family of the parabola (when a = 1) and one member of the family of parabola is given in Eq (1).

Also, Eq (1) is a particular solution of the differential equation (2).

 

The solution of the differential equation is a relation between the variables of the equation not containing the derivatives, but satisfying the given differential equation.

 

A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation. 

 

Particular solution of the differential equation obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

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\sqrt{1-x^{2}} \frac{d y}{d x}=-\sqrt{1-y^{2}}

{\frac{d y}{\sqrt{1-y^{2}}}=-\frac{d x}{\sqrt{1-x^{2}}} \Rightarrow \sin ^{-1} y=-\sin ^{-1} x+c} \\ {x=\frac{1}{2}, y=\frac{\sqrt{3}}{2} \Rightarrow \frac{\pi}{2}=-\frac{\pi}{6}-c}

\begin{aligned} \sin ^{-1} y &=\frac{\pi}{2}-\sin ^{-1} x \\ &=\cos ^{-1} x \Rightarrow y=\sin (\cos^{-1}x) \end{aligned}

y\left ( \frac{1}{\sqrt{2}} \right ) =\frac{1}{\sqrt2}

Correct Option (3)

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