JEE Main20248 Apr 2024Morning ShiftMathematicsDifferential EquationsActual
Let y=y(x) be the solution of the differential equation (1+y^2 ) e^ x d x+ ^2 x (1+e^ 2 x ) d y=0, y(0)=1 . Then y ( 4 ) is equal to
Options
- A2 e
- B2 e^2
- C1 e
- D1 e^2
Correct answer
C. 1 e
Step-by-step solution
aligned & (1+y^2 ) e^ x d x+ ^2 x (1+e^ 2 x ) d y=0 & ^2 x e^ x 1+e^ 2 x d x+ d y 1+y^2 =C & ⁻¹ (e^ x )+ ⁻¹ y=C & for x=0, y=1, ⁻¹(1)+ ⁻¹ 1=C aligned C = 2 ⁻¹ (e^ x )+ ⁻¹ y= 2 Put x= , ⁻¹ e+ ⁻¹ y= 2 ⁻¹ y= ⁻¹ ey= 1 e