# Thread: How do I solve this linear equation?

1. ## How do I solve this linear equation?

How do I solve this linear equation?

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

The answer can be found here: y'+2ytan(x)=sin(x) - Wolfram|Alpha

But I would be really greatful if someone could explain the steps involved. 2. Originally Posted by jobloggs How do I solve this linear equation?

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

The answer can be found here: y'+2ytan(x)=sin(x) - Wolfram|Alpha

But I would be really greatful if someone could explain the steps involved. first thing you need to do is find you integrating factor which will be

$\displaystyle 2\tan{x}$

Next

$\displaystyle e^{\int{2\tan{x}}}dx = e^{-2\ln{\cos{x}}} = e^{\ln{\cos^{-2}{x}}} = \cos^{-2}{x} = \sec^2{x}$

so now

$\displaystyle y\sec^2{x} = \int \sin{x}\sec^2{x}dx = \int\sec{x}\tan{x}$

integrate that and you get

$\displaystyle y\sec^2{x} = \sec{x} + C$

divide by $\displaystyle \sec^2{x}$ and get

$\displaystyle y = \cos{x} +C\cos^2{x}$

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