MHT CET2019Evening ShiftMathematicsDifferential EquationsActual
The particular solution of the differential equation log d y d x = x , when x =0, y = 1 is …..
Options
- Ay = e x + 2
- By = - e x
- Cy = - e x + 2
- Dy = e x
Correct answer
D. y = e x
Step-by-step solution
We have, differential equations, log d y d x = x ⇒ d y d x = e x ⇒ d y = e x d x Integrating on both sides, we get ∫ d y = ∫ e x d x ⇒ y = e x + C … (i) On putting x = 0, y = 1 is Eq. (i), we get 1 = e 0 + C ⇒ C = 0 Now, particular solution of the given differential is y = e x