Math, asked by tvlu9728, 11 months ago

Sin2x dy/dx=y+ tanx solve the linear differential equation.

Answers

Answered by radhu66
16

Answer:

dy/dx=y+tanx/sin2x

dy/dx= sin2x. dy/dx(y+tanx)-(y+tanx).dy/dx Sin2x/(sin2x) ²

=Sin2x(0+sec²X) -(y+tanx)cos2x.dy/dx2x/sin²2x

= sin2x. sec²x-2ycos2x-tanxcos2x/sin²x

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