Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is . Then the differential equation, whose general solution is , where and are arbitrary constants, is [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is $e^{- a} + 4 a^{2} + a - 1$ . Then the differential equation, whose general solution is $y = c_{1} f (x) + c_{2}$ , where $c_{1}$ and $c_{2}$ are arbitrary constants, is [2024]

1 $(8 e^{x} - 1) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0$

2 $(8 e^{x} + 1) \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} = 0$

3 $(8 e^{x} + 1) \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = 0$

4 $(8 e^{x} - 1) \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} = 0$

Ans.
(3)

Given $\int_{0}^{a} f (x) d x = e^{- a} + 4 a^{2} + a - 1$

On differentiating both sides, we get $f (a) = - e^{- a} + 8 a + 1$

i.e., $f (x) = - e^{- x} + 8 x + 1$

Now, $y = c_{1} f (x) + c_{2}$

$\Rightarrow y' = c_{1} f' (x) = c_{1} (8 + e^{- x})$ ... (i)

Again, on differentiating, we get $c_{1} = - e^{x} y''$ ... (ii)

From (i) and (ii), we get

$y' = y'' e^{x} (- 8 - e^{- x}) \Rightarrow (8 e^{x} + 1) y'' + y' = 0$

Want Us to Email you About Special Offers & Updates?