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

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 e^{-\frac{1}{cos^{2}x}} and then express the equation as ye^{-\frac{1}{cos^{2}x}}=\int2sin(x)e^{-\frac{1}{cos^{2}x}}dx 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 e^{-\frac{1}{cos^{2}x}} and then express the equation as ye^{-\frac{1}{cos^{2}x}}=\int2sin(x)e^{-\frac{1}{cos^{2}x}}dx 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:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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