Differential equation Watch

karnten07
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#1
Report Thread starter 11 years ago
#1
dy/dx +ytanx = 0

In the solutions it says dy/y = -sinx/cosx dx

then dy/y = dcosx/cosx

then log y = log cosx + c

Im a bit confused as to what has been done. I see that -sinx is integrated to become cosx, but i dont understand how the dcosx/cosx is then integrated to become log cosx. Ive not seen the notation of dcosx/cosx before either so any explanations very welcome thnx
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generalebriety
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#2
Report 11 years ago
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It means d/dx (cos x) = -sin x, that is, the numerator is the derivative of the denominator. When you're integrating something in that form, f'(x) / f(x), you get simply ln f(x) (+ constant).
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SsEe
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Basically it's a separable differential equation. I'd write it like this:

 \frac{dy}{dx} = -y \tan(x)

\displaystyle \int \frac{1}{y} dy = \int \frac{-\sin(x)}{\cos(x)} dx

And then you recognise both parts as being in the form general just pointed out. So, integrating both sides gives the answer.
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karnten07
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Report Thread starter 11 years ago
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(Original post by generalebriety)
It means d/dx (cos x) = sin x, that is, the numerator is the derivative of the denominator. When you're integrating something in that form, f'(x) / f(x), you get simply ln f(x) (+ constant).
oh yes, i forgot about that identity, thanx
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generalebriety
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(Original post by karnten07)
oh yes, i forgot about that identity, thanx
Ahem. I was typing quickly. I do of course mean d/dx(cos x) = -sin x.
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