The solution of the differential equation $\frac{d y}{d x}-y \tan x=e^x \sec x$ is
The solution of the differential equation $\frac{d y}{d x}-y \tan x=e^x \sec x$ is
- $y=e^x \cos x+c$
- $y \cos x=e^x+c$
- $y=e^x \sin x+c$
- $y \sin x=e^x+c$
Solution
Given linear differential equation is
$
\begin{aligned}
\frac{d y}{d x}-y \tan x & =e^x \sec x \\
\therefore \quad \quad \quad \mathrm{IF}=e^{\int-\tan x d x} & =e^{-\log \sec x} \\
& =\frac{1}{\sec x}
\end{aligned}
$
$\therefore$ Complete solution is
$
\begin{array}{rlrl}
& & y \cdot \frac{1}{\sec x} & =\int e^x \sec x \cdot \frac{1}{\sec x} d x \\
\Rightarrow & & \frac{y}{\sec x} & =e^x+c \\
\Rightarrow & y \cos x & =e^x+c
\end{array}
$
Asked in: AP EAMCET 2008
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