99 Percentile Qs Bank for JEE MainMathematicsDifferential Equations
The solution of the initial value problem (2ln x ) d y d x + y x = 1 y cos x ,   y > 0 ,   x > 1 and y 3 π 2 = 0 is given by which of the following options?
Options
- Ay = a 1 - sin x ln x
- By = a 1 + sin x ln x
- Cy = a 1 - cos x ln x
- Dy = a 1 + cos x ln x
Correct answer
B. y = a 1 + sin x ln x
Step-by-step solution
Given differential equation is 2 y d y d x + y 2 1 x ln x = cos x ln x ⇒  d z d x + z x ln x = cos x ln x On substituting y 2 = z and 2 y d y d x = d z d x , we get ⇒   d z d x + z x ln x = cos x ln x Which is a LDE , ∴  y 2 ln x = z ln x = sin x + C Now, y 3 π 2 = 0   ⇒ C = 1 ∴  y = ± 1 + sin x ln x y = 1 + sin x ln x ∵   y > 0 Required solution y = a 1 + sin x ln x   where a is any constant.