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How can we solve (if a closed form expression for f(x) can be found) the following first-order linear differential equation?

$$f'(x)=f(x)\cdot (\cos x+\tan x)$$

I have found that one function which validates this equation is: $$f(x)=\frac{e^{\sin x}}{\cos x}$$

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HINT Separation of variables yields $$ \int \frac{f'(x)dx}{f(x)} = \int (\cos x + \tan x) dx $$ and LHS integrates to $\ln f(x) + C$.

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  • $\begingroup$ So, RHS integrates to $ \sin x + \ln {(\sec x)}$, ignoring constants because we already have $C$. Finally, we have $$f(x)=C_1 \cdot \frac{e^{\sin x}}{\cos x}$$ $\endgroup$ – Jason Jul 31 '15 at 16:01
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    $\begingroup$ i think you get $\sin x + \ln |\sec x|$, and must carry the absolute value unless you restrict yourself to the interval where $\cos x \ge 0$... $\endgroup$ – gt6989b Jul 31 '15 at 16:03
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    $\begingroup$ Yes, you are right! So, RHS integrates to $ \sin x + \ln {\left| \sec x \right|}$, ignoring constants because we already have $C$. Finally, we have: $$f(x)=C_1 \cdot \frac{e^{\sin x}}{\left| \cos x \right|}$$ $\endgroup$ – Jason Jul 31 '15 at 16:07
  • $\begingroup$ @Jason yes, indeed $\endgroup$ – gt6989b Jul 31 '15 at 16:08

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