Question

cos^2ydx+(1+e^-x)sinydy=0

Answer 1

Diffrential equation is

$$\cos^2(y)dx+(1+e^{-x})\sin(y)dy=0$$

Now, divide the above equation by $\cos^2(y)(1+e^{-x})$ ,thus

$$\frac{dx}{1+e^{-x}}+\frac{\sin(y)}{\cos^2(y)}dy=0\\ \implies \frac{e^x}{1+e^x}dx+\frac{\sin(y)}{\cos^2(y)}dy=0\\ \implies \int\frac{e^x}{1+e^x}dx+\int\frac{\sin(y)}{\cos^2(y)}dy=\text{constant}\\ \implies \ln(1+e^x)-\int\frac{1}{u^2}du=\text{constant, where $u=\cos(y)$}\\ \implies \textcolor{blue}{\ln(1+e^x)+\frac{1}{\cos(y)}}=\text{\textcolor{blue}{constant}}$$