If y=yx is the solution of the equation e^sinycosydy/dx+e^sinycosx=cosx,y0=0; then 1+y(π)/6+√32y(π)/3+1√2y(π)/4 is equal to _______.

Correct answer: 1. Put e^siny=t ⇒e^sinycosydy/dx=dt/dx ⇒D.E is dt/dx+tcosx=cosx I.F.=e^(∫cosxdx)=e^sinx ⇒ solution is t.e^sinx=∫cosxe^sinx ⇒e^sinye^sinx=e^sinx+c ∵ x=0,y=0⇒c=0 ⇒e^siny=1 ⇒y=0 ⇒1+y(π)/6+√32y(π)/3+1√2y(π)/4=1

Q54 / 90 · JEE Main 2021 (26 Feb Shift 1) · Mathematics

If $y = y (x)$ is the solution of the equation $e^{\sin y} \cos y \frac{d y}{d x} + e^{\sin y} \cos x = \cos x, y (0) = 0$ ; then $1 + y (\frac{π}{6}) + \frac{\sqrt{3}}{2} y (\frac{π}{3}) + \frac{1}{\sqrt{2}} y (\frac{π}{4})$ is equal to _______.

Correct answer: 1

Solution

Put

e^{\sin y} = t

$\Rightarrow e^{\sin y} \cos y \frac{d y}{d x} = \frac{d t}{d x}$

$\Rightarrow D . E$ is $\frac{d t}{d x} + t \cos x = \cos x$

$I . F . = e^{\int \cos x d x} = e^{\sin x}$

$\Rightarrow$ solution is $t . e^{\sin x} = \int \cos x e^{\sin x}$

$\Rightarrow e^{\sin y} e^{\sin x} = e^{\sin x} + c$

$∵ x = 0, y = 0 \Rightarrow c = 0$

$\Rightarrow e^{\sin y} = 1$

$\Rightarrow y = 0$

$\Rightarrow 1 + y (\frac{π}{6}) + \frac{\sqrt{3}}{2} y (\frac{π}{3}) + \frac{1}{\sqrt{2}} y (\frac{π}{4}) = 1$

Hence, the correct answer is 1.

If y=yx is the solution of the equation esinycosydydx+esinycosx=cosx,y0=0; then 1+yπ6+32yπ3+12yπ4 is equal to _______.

Solution

Relevant Questions

If $y = y (x)$ is the solution of the equation $e^{\sin y} \cos y \frac{d y}{d x} + e^{\sin y} \cos x = \cos x, y (0) = 0$ ; then $1 + y (\frac{π}{6}) + \frac{\sqrt{3}}{2} y (\frac{π}{3}) + \frac{1}{\sqrt{2}} y (\frac{π}{4})$ is equal to _______.