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Studentka2010 [4]
3 years ago
6

Find the solution of the general equation of the differential equation:

Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Notice that the condition <em>x</em> ≠ 2<em>πk</em> for (presumably) integer <em>k</em> means cos(<em>x</em>) ≠ ±1, and in particular cos(<em>x</em>) ≠ 1 so that we could divide both sides by (1 - cos(<em>x</em>)) safely. Doing so lets us separate the variables:

(1 - cos(<em>x</em>)) <em>y'</em> - <em>y</em> sin(<em>x</em>) = 0

==>   (1 - cos(<em>x</em>)) <em>y'</em> = <em>y</em> sin(<em>x</em>)

==>   <em>y'</em>/<em>y</em> = sin(<em>x</em>)/(1 - cos(<em>x</em>))

==>   d<em>y</em>/<em>y</em> = sin(<em>x</em>)/(1 - cos(<em>x</em>)) d<em>x</em>

Integrate both sides and solve for <em>y</em>. On the right, substitute <em>u</em> = 1 - cos(<em>x</em>) and d<em>u</em> = sin(<em>x</em>) d<em>x</em>.

∫ d<em>y</em>/<em>y</em> = ∫ sin(<em>x</em>)/(1 - cos(<em>x</em>)) d<em>x</em>

∫ d<em>y</em>/<em>y</em> = ∫ d<em>u</em>/<em>u</em>

ln|<em>y</em>| = ln|<em>u</em>| + <em>C</em>

exp(ln|<em>y</em>|) = exp(ln|<em>u</em>| + <em>C </em>)

exp(ln|<em>y</em>|) = exp(ln|<em>u</em>|) exp(<em>C </em>)

<em>y</em> = <em>Cu</em>

<em>y</em> = <em>C</em> (1 - cos(<em>x</em>))

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