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

Given, 1 - x 2 d y d x = x y + x 3 + 2 1 - x 2 ⇒ d y d x + - x 1 - x 2 y = x 3 + 2 1 - x 2 IF = e ∫ - x 1 - x 2 d x = 1 - x 2 y x · 1 - x 2 = x 4 4 + 2 x + c y 0 = 0 ⇒ c = 0 1 - x 2 y x = x 4 4 + 2 x So,required value = ∫ - 1 2 1 2 x 4 4 + 2 x d x - 1 4 · 2 ∫ 0 1 2 x 4 d x = 1 10 x 5 0 1 2 = 1 320 k - 1 = 320

Mathematics - Differential Equation Question with Solution | TestHub

MathematicsDifferential EquationLinear DE / Red. LDEHard2 minPYQ_2022

MathematicsHardnumerical range

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

Answer:

320.00

Solution:

Given,

$(1 - x^{2}) \frac{d y}{d x} = x y + (x^{3} + 2) \sqrt{1 - x^{2}}$

$\Rightarrow \frac{d y}{d x} + (\frac{- x}{1 - x^{2}}) y = \frac{x^{3} + 2}{\sqrt{1 - x^{2}}}$

$IF = e^{\int \frac{- x}{1 - x^{2}} d x} = \sqrt{1 - x^{2}}$

$y (x) \cdot \sqrt{1 - x^{2}} = \frac{x^{4}}{4} + 2 x + c$

$y (0) = 0 \Rightarrow c = 0$

$\sqrt{1 - x^{2}} y (x) = \frac{x^{4}}{4} + 2 x$

So,required value $= \int_{- \frac{1}{2}}^{\frac{1}{2}} (\frac{x^{4}}{4} + 2 x) d x - \frac{1}{4} \cdot 2 \int_{0}^{\frac{1}{2}} x^{4} d x$

$= \frac{1}{10} {(x^{5})}_{0}^{\frac{1}{2}} = \frac{1}{320}$

$k^{- 1} = 320$

Stream:JEESubject:MathematicsTopic:Differential EquationSubtopic:Linear DE / Red. LDE

⏱ 2mℹ️ Source: PYQ_2022

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