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

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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 $xe^{y}=\int e^{-y} \sec^2 y e^y dy +c$
$xe^{y}=\tan y +c$
answered Sep 6, 2013 by meena.p
 
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