Differential Equation problem

[kcoe05]

How do I solve the following differential equation?

e^xcos^2y dx - (1-e^x)dy = 0

Not exactly sure how to even get started.

Thanks

 

[arthur ohlsten]

e^x cos^2 y dx - [1-e^e] dy =0
e^x cos^2y dx = [1-e^x] dy
e^x /[1-e^x] = dy/cos^2 y

right side of =
dy/cos^2 y = tan y

left side
let U=[1-e^x]
then dU=-e^x dx
-[-dU]/U = - ln[1-e^x]

Arthur

 

[BigGlenntheHeavy]

\(\displaystyle e^{x}cos^{2}(y)dx \ = \ (1-e^{x})dy\)

\(\displaystyle \frac{dy}{dx} \ = \ \frac{e^{x}cos^{2}(y)}{1-e^{x}}, \ remember \ this.\)

\(\displaystyle Now, \ this \ implies \ that \ \int\frac{dy}{cos^{2}(y)} \ = \ \int\frac{e^{x}}{1-e^{x}}dx\)

\(\displaystyle Hence, \ tan(y) \ = \ -ln|1-e^{x}|+C \ or \ y(x) \ = \ arctan[-ln|1-e^{x}|+C]\)

\(\displaystyle y(0) \ = \ arctan[-ln(0)+C] \ = \ \frac{\pi}{2}\)

\(\displaystyle Ergo, \ tan(\pi/2) \ = \ -ln(0)+C, \ \implies \ \infty \ = \ \infty \ +C, \ C \ = \ 0.\)

\(\displaystyle Therefore \ y(x) \ = \ arctan[-ln|1-e^{x}|]\)

\(\displaystyle Check: \ \frac{dy}{dx} \ = \ \frac{-e^{x}}{(e^{x}-1)((ln|1-e^{x}|)^{2}+1)} \ = \ \frac{-e^{x}}{(e^{x}-1)(tan^{2}(y)+1)}\)

\(\displaystyle \frac{dy}{dx} \ = \ \frac{-e^{x}}{(e^{x}-1)sec^{2}(y)} \ = \ \frac{e^{x}cos^{2}(y)}{1-e^{x}} \ QED\)

 

[kcoe05]

Thank you both for your help. I greatly appreciate it.