If \( y=y(x) \) is the solution of the equation\( e^{\sin y} \cos y \f

1.	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\( 1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right) \) is equal to
Answer» e^siny cos y \(\frac{dy}{dx}\) + e^siny cos x = cos x e^siny = t e^siny. cos y \(\frac{dy}{dx}\) = \(\frac{dt}{dx}\) \(\therefore\) \(\frac{dt}{dx}\) + cos x t = cos x \(\therefore\) I.F. = \(e^{\int cosxdx}\) = e^sinx \(\therefore\) t x I.F. = \(\int\)(I.F.) \(\times\) α dx ⇒ t.e^sinx = \(\int e^{sinx}.cos x dx\) = e^sinx + c ⇒ t = 1 + c e^-sinx ⇒ e^siny = 1 + c e^-sinx y(0) = 0 ⇒ c + 1 = e^o = 1 ⇒ c = 0 \(\therefore\) e^siny = 1 is solution of given differential equation. ⇒ sin y = ln 1 = 0 ⇒ y = sin^-10 = 0 \(\therefore\) y = 0 is solution of given differential equation. \(\therefore\) y(π/3) = y(π/4) = y(π/6) = 0 \(\therefore\) 1 + y(π/6) + \(\frac{\sqrt3}2y(\pi/3)\) + \(\frac1{\sqrt2}y(π/4)\) = 1 + 0 = 1

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\( 1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right) \) is equal to

Answer»

e^siny cos y \(\frac{dy}{dx}\) + e^siny cos x = cos x

e^siny = t

e^siny. cos y \(\frac{dy}{dx}\) = \(\frac{dt}{dx}\)

\(\therefore\) \(\frac{dt}{dx}\) + cos x t = cos x

\(\therefore\) I.F. = \(e^{\int cosxdx}\) = e^sinx

\(\therefore\) t x I.F. = \(\int\)(I.F.) \(\times\) α dx

⇒ t.e^sinx = \(\int e^{sinx}.cos x dx\)

= e^sinx + c

⇒ t = 1 + c e^-sinx

⇒ e^siny = 1 + c e^-sinx

y(0) = 0

⇒ c + 1 = e^o = 1

⇒ c = 0

\(\therefore\) e^siny = 1 is solution of given differential equation.

⇒ sin y = ln 1 = 0

⇒ y = sin^-10 = 0

\(\therefore\) y = 0 is solution of given differential equation.

\(\therefore\) y(π/3) = y(π/4) = y(π/6) = 0

\(\therefore\) 1 + y(π/6) + \(\frac{\sqrt3}2y(\pi/3)\) + \(\frac1{\sqrt2}y(π/4)\) = 1 + 0

= 1