Question

Q2. (x+tan y)dy= Sin2y .dx

Answer 1

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Answer 2

Answer:

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.