Browse Questions

# Solve the following. $dx+x\;dy=e^{-y}\;\sec^{2}y\;dy$

Toolbox:
• Linear Differential equation.
• This is of the form $\large\frac{dy}{dx}$$+Py=Q where P and Q are functions of x only. • The integrating factor I= e^{\int \large pfd} and the G.S is ye^{\large pdx}=\int Q e^{\large pdx} dx+c Step 1: dx+ x\;dy=e^{-y} \sec^2 y \;dy divided by dy Step 2: \large\frac{dy}{dx}$$+x=e^{-y} \sec ^2 y$ this is linear in $\large\frac{dx}{dy}$
$P(y)=1,\quad Q(y)=e^{-y} \sec^2 y$
$e^{\large \int pdy}=e^{\large \int dy}=e^{\large y}$
Step 3:
The solution is $xe^{y}=\int e^{-y} \sec^2 y e^y dy +c$
$xe^{y}=\tan y +c$