A dy y cos x 2 y0 1 dx 1 y c dy 1

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Unformatted text preview: s
are
on
the
other
side.

 € Separable
EquaMons
 Next,
integrate
both
sides
of
the
equa+on:
 ∫ h( y)dy = ∫ g( x)dx This
equaMon
defines
y
implicitly
as
a
funcMon
of
 x.

 € Finally,
solve
this
equa+on
for...
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This note was uploaded on 08/13/2013 for the course MATH 1LT3 taught by Professor Erinclements during the Fall '13 term at McMaster University.

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