For $\dfrac{dy}{dx}=\dfrac{e^y\sin^2x}{y\sec x}$, we separate the variables to have $$\int \frac{y}{e^y}\,dy=\int \frac{\sin^2x}{\sec x}\,dx$$

evaluate each integral we have$$\int \frac{y}{e^y} \,dy=-e^{-y}(y+1)+C_1$$ and $$\int \frac{\sin^2x}{\sec x}\, dx=\frac{1}{3}\sin^3x+C_2$$

equate them to get$-e^{-y}(y+1)+C_1=\frac{1}{3} \sin^3x+C_2$.

The thing is, how to solve this equation for $y$ in terms of $x$?

$$ e^{-y}(y+1)=\frac{1}{3} \sin^3x+C_3$$ where $C_3=C_1 -C_2$

  • $\begingroup$ Chances are that this equation is transcendental. You have solved the differential equation, tough. It's called an implicit solution. $\endgroup$ – InertialObserver Sep 15 '17 at 2:54

You properly arrived to $$e^{-y}(y+1)=\frac{1}{3} \sin^3(x)+C\tag 1$$

Sooner or later, you will learn that any equation which can write or rewrite as $$A+Bt+C\log(D+Et)=0$$ as solution(s) which can be expressed in terms of Lambert function.

So, for your case, rewriting $(1)$ as $$e^{-(y+1)}(y+1)=\frac{1}{3e} \sin^3(x)+C\tag 2$$ we get $$y=-W\left(\frac{\sin ^3(x)}{3 e}+C\right)-1\tag 3$$

Use Wolfram Alpha and just type (for $C=0$)

plot -1-LambertW(sin[x]^3/(3e)) from x=0 to x=2pi

to see the plot of the function.

  • $\begingroup$ what is W in (3)? $\endgroup$ – bbw Sep 15 '17 at 14:59
  • $\begingroup$ Lambert function. Just google for it. It is one of the most interesting functions after the logarithm and the exponential. On this site, search for Lambert : you will find a lot. Cheers. $\endgroup$ – Claude Leibovici Sep 15 '17 at 16:34

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