99 Percentile Qs Bank for JEE MainMathematicsDifferential Equations
If 2 x - y + c log x - 2 y - 4 = k is the general solution of d y d x = 2 x - 4 y - 5 x - 2 y + 2 then c =
Options
- A4
- B2
- C3
- D- 4
Correct answer
C. 3
Step-by-step solution
d y d x = 2 x - 4 y - 5 x - 2 y + 2 = 2 x - 2 y + 2 - 9 x - 2 y + 2 Let x - 2 y + 2 = t ⇒ 1 - 2 d y d x = d t d x ⇒ d y d x = 1 2 - 1 2 d t d x i.e. 1 2 - 1 2 d t d x = 2 t - 9 t ⇒ d t d x = - 2 2 t - 9 t - 1 2 = 18 - 4 t t + 1 ⇒ d t d x = 18 - 3 t t ⇒ t 18 - 3 t d t = d x ⇒ - 6 - t - 6 6 - t d t = 3 d x ⇒ - d t + 6 6 - t d t = 3 d x Integrating, we get - t - 6 ln 6 - t = 3 x + C - x + 2 y - 2 - 6 ln 6 - x + 2 y - 2 = 3 x + C - 4 x + 2 y - 2 - 6 ln 4 - x + 2 y = C 2 x - y +