Let y=yx be the solution of the differential equation sin⁡xdy/dx+ycosx⁡=4x, x∈0, π. If y(π)/2=0, then y(π)/6 is equal to

Correct answer: (D) -8/9π^2. dy/dx+ycot x=4x/sinx, x∈0, π I.F. =e^(∫cotx)=e^(ln⁡sinx)=sinx ∴y sinx=∫4x/sinx.sinx dx ysinx=2x^2+c ∵y(π)/2=0 ∴0×1=2×π2^2+c ⇒c=-π^22 y(π)/6=(2×π6^2-π^22)/12=π^218-π^22×2 =(π^2)/9-π^2 =-(8π^2)/9

Q57 / 90 · JEE Main 2018 (08 Apr) · Mathematics

Let $y = y (x)$ be the solution of the differential equation $\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π) .$ If $y (\frac{π}{2}) = 0,$ then $y (\frac{π}{6})$ is equal to

A
$- \frac{4}{9} π^{2}$
B
$\frac{4}{9 \sqrt{3}} π^{2}$
C
$\frac{- 8}{9 \sqrt{3}} π^{2}$
Correct answer
D
$- \frac{8}{9} π^{2}$

Solution

\frac{d y}{d x} + y \cot x = \frac{4 x}{\sin x}, x \in (0, π)

I.F.

= e^{\int \cot x} = e^{\ln \sin x} = \sin x

∴ y \sin x = \int \frac{4 x}{\sin x} . \sin x d x

y \sin x = 2 x^{2} + c

∵ y (\frac{π}{2}) = 0

∴ 0 \times 1 = 2 \times {(\frac{π}{2})}^{2} + c

\Rightarrow c = - \frac{π^{2}}{2}

y (\frac{π}{6}) = \frac{2 \times {(\frac{π}{6})}^{2} - (\frac{π^{2}}{2})}{\frac{1}{2}} = [(\frac{π^{2}}{18}) - (\frac{π^{2}}{2})] \times 2

= \frac{π^{2}}{9} - π^{2}

= - \frac{8 π^{2}}{9}

Hence, the correct answer is option: 'D'.

Let y=yx be the solution of the differential equation sin⁡xdydx+ycosx⁡=4x, x∈0, π. If yπ2=0, then yπ6 is equal to

Solution

Relevant Questions

Let $y = y (x)$ be the solution of the differential equation $\sin x \frac{d y}{d x} + y \cos x = 4 x, x \in (0, π) .$ If $y (\frac{π}{2}) = 0,$ then $y (\frac{π}{6})$ is equal to