AP EAMCET202023 Sep 2020Morning ShiftMathematicsDifferential EquationsActual
Find the solution of differential equation given below: d y d x + y · cosec 2 ( x ) = cosec 2 ( x ) · cot ( x )
Options
- Ay e cot x = ( 1 + cot x ) e - cot x + c
- By e - cot x = ( 1 - cot x ) e - cot x + c
- Cy e cot x = ( 1 + cot x ) e cot x + c
- Dy e - cot x = ( 1 + cot x ) e - cot x + c
Correct answer
D. y e - cot x = ( 1 + cot x ) e - cot x + c
Step-by-step solution
Given the differential equation is, d y d x + cos e c 2 ( x ) · y = cos e c 2 ( x ) · c o t x . . . . . . . . . . ( 1 ) Comparing with (1), we get d y d x + p y = Q Which is a linear differential equation, P = cos e c 2 x Q = cos e c 2 x · c o t x I . F . = e ∫ P d x = e ∫ cos e c 2 x   d x = e - c o t x The solution of (1) is given by, y · I . F . = ∫ Q · I F + C y · e - c o t x = ∫ cos e c 2 x · c o t x · e - c o t x   d x + C . Now, put c