If y=y(x) solves e^ y ydydx+e^ y x= x, y(0)=0, then 1+y(6)+√32y(3)+1√2y(4) is ______.

Question

If $y=y(x)$ solves $e^{\sin y}\cos y\dfrac{dy}{dx}+e^{\sin y}\cos x=\cos x$, $y(0)=0$, then $1+y\left(\dfrac{\pi}{6}\right)+\dfrac{\sqrt3}{2}y\left(\dfrac{\pi}{3}\right)+\dfrac{1}{\sqrt2}y\left(\dfrac{\pi}{4}\right)$ is \_\_\_\_\_\_.

Answer: 1

Accepted Answer

With $t=e^{\sin y}$: $\dfrac{dt}{dx}+t\cos x=\cos x$, linear with I.F.$=e^{\sin x}$; solving gives $te^{\sin x}=e^{\sin x}+C$, and $y(0)=0$ fixes $C=0$, so $e^{\sin y+\sin x}=e^{\sin x} \implies \sin y=0 \implies y\equiv0$.

So all the listed $y$-values are 0, and the expression equals $1$.

Solutions · 1