Solving y' - ytan(x) = 2sin(x): What Went Wrong with My Integrating Factor?

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SUMMARY

The discussion revolves around solving the differential equation y' - ytan(x) = 2sin(x) using an integrating factor. The user initially attempted to use the integrating factor e^{-\frac{1}{cos^{2}x}}, which was identified as incorrect. The correct approach involves finding an integrating factor where f' / f = -tan(x), leading to the conclusion that integrating tan(x) is necessary for the solution.

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okhjonas
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Hello, I've got a problem with a differential equation...

Homework Statement



y' - ytan(x) = 2sin(x)


The Attempt at a Solution



I've tried to multiply the equation with the integrating factor [tex]e^{-\frac{1}{cos^{2}x}}[/tex] and then express the equation as [tex]ye^{-\frac{1}{cos^{2}x}}=\int2sin(x)e^{-\frac{1}{cos^{2}x}}dx[/tex] but then I've got lost..

(excuse my bad English)..
 
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okhjonas said:
Hello, I've got a problem with a differential equation...

Homework Statement



y' - ytan(x) = 2sin(x)


The Attempt at a Solution



I've tried to multiply the equation with the integrating factor [tex]e^{-\frac{1}{cos^{2}x}}[/tex] and then express the equation as [tex]ye^{-\frac{1}{cos^{2}x}}=\int2sin(x)e^{-\frac{1}{cos^{2}x}}dx[/tex] but then I've got lost..

(excuse my bad English)..

Hi okhjonas! :smile:

hmm … I thought you'd got some sleep?

No, your integrating factor is wrong …

you need an f with f'/f = - tanx …

oh I see … you differentiated tanx … try integrating it! :wink:
 

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