JEE Main20254 Apr 2025Evening ShiftMathematicsDifferential EquationsActual
If a curve y=y(x) passes through the point (1, 2 ) and satisfies the differential equation (7 x^4 y-e^x cosec y ) d x d y =x^5, x 1 , then at x=2 , the value of cosy is:
Options
- A2 e ^2- e 64
- B2 e ^2+ e 64
- C2 e ^2- e 128
- D2 e ^2+ e 128
Correct answer
C. 2 e ^2- e 128
Step-by-step solution
aligned & d y d x = 7 y x - e^x cosec y x^5 & d y d x = 7 y y x - e^x y x^5 & y d y d x - y 7 x = -e^x x^5 & let - y=t & y d y d x = d t d x aligned aligned & d t d x + 7 t x = -e^x x^5 & I.F. =x^7 & t. x^7=- x^2 e^x d x & cosy x^7=x^2 e^x-2 x e^x d x & cosy x^7=x^2 e^x-2 x e^x+2 e^x+c & x=1, y= 2 , c=-e & cosy = 2 e^2-e 128 & option (3) aligned