Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
BITSAT2019MathematicsDifferential EquationsActual

If f ' x = f x + ∫ 0 1 f x d x , f 0 = 1 , then f x =

Options

  1. A2 e x 3 - e + 1 - e 3 - e
  2. Be x 3 - e + 1 + e 1 - e
  3. C3 e x 2 - e + 1 + e 1 - e
  4. D3 e x 2 - e + 1 - e 3 + e

Correct answer

A. 2 e x 3 - e + 1 - e 3 - e

Step-by-step solution

Given, f ' x = f x + ∫ 0 1 f x d x , f 0 = 1 . . . ( i ) ⇒   f ' ' x = f ' x + 0 ⇒   f ' ' x f ' x = 1 ⇒   ∫ f ' ' x f ' x d x = ∫ d x ⇒   log f ' x = x + C ⇒   f ' x = A e x ⇒   f x = A e x + K f 0 = A + K = 1 . . . ( ii ) ∴   A e x = A e x + K + ∫ 0 1 A e x + k d x ⇒   k + A e x + K x 0 1 = 0 ⇒   k + A e - A + K = 0 ⇒   A ( e - 1 ) + 2 k = 0 . . . ( iii ) From Eqs. ( ii ) and ( iii

Practice Differential Equations on Quantrex Academy →

More from Differential Equations

The solution of the differential equation (x+1) d y d x -y=e^ 3 x (x+1)^2 is 2024If d y d x -y _e 2=2^ x ( x-1) _e 2 , then y 2024The integrating factor of x d y d x -y=x^4-3 x is 2023The number of solutions of d y d x = y+1 x-1 , when y(1)=2 is 2023( d y d x ) x=y ^2 x+ x , find general solution 2022If the slope of the tangent to the curve at any point P (x, y) is y x - ² y x , then the equation of a curve passing through (1, 4 ) is 2021The general solution of the differential equation ( ⁻¹ y-x ) d y= (1+y² ) d x is 2021The solution of the differential equation (x+1) d y d x -y=e^ 3 x (x+1)² is 2020 Full Differential Equations list All BITSAT PYQs