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प्रश्न
Solve the following differential equation:
(ey + 1)cos x dx + ey sin x dy = 0
उत्तर
(ey + 1) cos x dx + ey sin x dy = 0
ey sin x dy = – (ey + 1) cos x dx
`int ("e"^y "d"y)/("e"^y + 1) =-int (cosx "d"x)/sin x`
log (ey + 1) = – log sin x + log c
log [(ey + 1) + log sin x = log c
log (ey +1) sin x] = log c
(ey+ 1) sin x = c
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