Q2. (x+tan y)dy= Sin2y .dx
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Step-by-step explanation:
(x+tan y)dy= Sin2y .dx
⇒ dx / dy = (x+tan y) / sin2y
⇒ dx / dy = ( x / sin2y ) + ( tan y / sin2y )
⇒ ( dx / dy ) + ( - cosec2y )x = 1/2 sec²y
we all know that dx/dy + Rx = S
⇒ i.f = e^-∫cosec2y dy
= e^ -log | cosec2y - cot2y|
= e^ -log[ (1 - cos2y - cot2y ) / sin2y ]
= e^ -log[ 2sin² y / siny cosy )
= e^ -log coty
= coty
x. ( cot y ) = ∫ 1/2 sec²y * coty dy + c
⇒ x coty = ∫ ( 1 / 2siny cosy ) dy) + c
⇒ x coty = ∫cosec 2y dy + c
⇒ x cot y = 1/2 log ( cosec 2y - cot 2y ) + c
c means constant .
This is the solution of the given differential equation.
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