Let y = y x be a solution of the differential equation, 1 - x 2 d y d x + 1 - y 2 = 0 , x < 1 . If y 1 2 = 3 2 , then y - 1 2 is equal to

d y 1 - y 2 + d x 1 - x 2 = 0 ⇒ sin - 1 y + sin - 1 x = c At, x = 1 2 , y = 3 2 ⇒ c = π 2 ⇒ s i n - 1 y = c o s - 1 x Hence, y - 1 2 = s i n c o s - 1 - 1 2 = 1 2

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationVariable separableMedium2 minPYQ_2020

MathematicsMediumnumerical range

Let $y = y (x)$ be a solution of the differential equation, $\sqrt{1 - x^{2}} \frac{d y}{d x} + \sqrt{1 - y^{2}} = 0, |x| < 1 .$ If $y (\frac{1}{2}) = \frac{\sqrt{3}}{2},$ then $y (\frac{- 1}{\sqrt{2}})$ is equal to

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Answer:

Solution:

$\frac{d y}{\sqrt{1 - y^{2}}} + \frac{d x}{\sqrt{1 - x^{2}}} = 0$
$\Rightarrow \sin^{- 1} y + \sin^{- 1} x = c$
At, $x = \frac{1}{2}, y = \frac{\sqrt{3}}{2} \Rightarrow c = \frac{π}{2} \Rightarrow {s i n}^{- 1} y = {c o s}^{- 1} x$
Hence, $y (\frac{- 1}{\sqrt{2}}) = s i n ({c o s}^{- 1} \frac{- 1}{\sqrt{2}}) = \frac{1}{\sqrt{2}}$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Variable separable

⏱ 2mℹ️ Source: PYQ_2020

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