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JEE Main20246 Apr 2024Morning ShiftMathematicsDifferential EquationsActual

Let y=y(x) be the solution of the differential equation (1+x^2 ) d y d x +y=e^ ⁻¹ x , y(1)=0 . Then y(0) is

Options

  1. A1 2 (e^ / 2 -1 )
  2. B1 2 (1-e^ / 2 )
  3. C1 4 (1-e^ / 2 )
  4. D1 4 (e^ / 2 -1 )

Correct answer

B. 1 2 (1-e^ / 2 )

Step-by-step solution

aligned & d y d x + y 1+x^2 = e^ ⁻¹ x 1+x^2 & I.F. =e^ 1 1+x^2 d x =e^ ⁻¹ x & y e^ ⁻¹ x = ( e^ ⁻¹ x 1+x^2 ) e^ ⁻¹ x d x & Let ⁻¹ x=z d x 1+x^2 =d z & y e^z= e^ 2 z d z= e^ 2 z 2 +C & y e^ ⁻¹ x = e^ 2 ⁻¹ x 2 +C & y= e^ ⁻¹ x 2 + C e^ ⁻¹ x aligned aligned & y (1)=0 0= e ^ / 4 2 + C e ^ / 4 C = - e ^ / 2 2 & y = e ^ ⁻¹ x 2 - e ^ / 2 2 e ^ ⁻¹ x & y (0)= 1- e ^ / 2 2 aligned

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