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प्रश्न
\[\frac{dy}{dx} - y \tan x = e^x \sec x\]
उत्तर
We have,
\[\frac{dy}{dx} - y \tan x = e^x \sec x\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = - \tan x \]
\[Q = e^x \sec x\]
Now,
\[I . F . = e^{\int - \tan x\ dx} \]
\[ = e^{- \log\left| \left( \sec x \right) \right|} \]
\[ = e^{\log\left| \left( \cos x \right) \right|} \]
\[ = \cos x\]
So, the solution is given by
\[y \cos\ x = \int\left( \cos x\ e^x \sec x \right) dx + C\]
\[ \Rightarrow y \cos\ x = \int e^x dx + C\]
\[ \therefore y \cos\ x = e^x + C\]
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