Question

If $y=y(x)$ is the solution of the equaiton $e ^{\text{sin y}} \cos y \frac{ dy }{ dx }+ e ^{\text{sin y}} \cos x =\cos x , y (0)=0$; 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 1

Put $e ^{\text {siny }}= t$
$\Rightarrow e ^{\sin y} \cos y \frac{ dy }{ dx }=\frac{ dt }{ dx }$
$\Rightarrow D.E$ 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 $t.e ^{\sin x}=\int \cos x e^{\sin x}$
$\Rightarrow e^{\sin y} e^{\sin x}=e^{\sin x}+c$
$\because x=0, y=0 \Rightarrow c=0$
$\Rightarrow e^{\sin y}=1$
$\Rightarrow y=0$
$\Rightarrow 1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2} y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}} y\left(\frac{\pi}{4}\right)=1$