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Differential Equations

JEE Main 2024 (Online) 9th April Evening ShiftMCQMedium📈 Topic in 33% papers+4−1
Ques. 1JEE Main 2024, 9 April (Evening Shift)

Let 0x 1 - (y'(t))2 dt = 0x y(t) dt, 0 x 3, y 0, y(0) = 0. Then at x = 2, y'' + y + 1 is equal to

  1. 2
  2. 2
  3. 1/2
  4. 1

Answer

(4) 1

Solution

Differentiating both sides w.r.t. x (Leibnitz rule): 1 - (y')2 = y

(y')2 = 1 - y2, so dy 1 - y2 = dx

Integrating, -1y = x + c, and y(0) = 0 gives c = 0.

Since y 0 for 0 x 3, we must take -1y = x, i.e. y = x.

Hence y' = x and y'' = - x, so y'' + y + 1 = - x + x + 1 = 1 for every x, and in particular at x = 2.

Answer: (4)

Keep practising — Differential Equations

Ques. 2JEE Main 2024, 9 April (Evening Shift)

For a differentiable function f : R R, suppose f'(x) = 3f(x) + , where R, f(0) = 1 and x - f(x) = 7. Then 9f(- e 3) is equal to .

Ques. 3JEE Main 2024, 8 April (Morning Shift)

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 + 4a2 + a - 1. Then the differential equation, whose general solution is y = c1 f(x) + c2, where c1 and c2 are arbitrary constants, is

  1. (8ex + 1 ) d2ydx2 - dydx = 0
  2. (8ex + 1 ) d2ydx2 + dydx = 0
  3. (8ex - 1 ) d2ydx2 - dydx = 0
  4. (8ex - 1 ) d2ydx2 + dydx = 0
Ques. 4JEE Main 2024, 8 April (Morning Shift)

Let y = y(x) be the solution of the differential equation (1 + y2)e xdx + 2 x (1 + e2 x )dy = 0, y(0) = 1. Then y ( 4 ) is equal to

  1. 1e2
  2. 2e2
  3. 2e
  4. 1e
Ques. 5JEE Main 2024, 8 April (Evening Shift)

Let y = y(x) be the solution curve of the differential equation y dydx + 2x y = x3 y, y(1) = 0. Then y( 3) is equal to:

  1. 6
  2. 12
  3. 3
  4. 4
Ques. 6JEE Main 2024, 8 April (Evening Shift)

Let |x| = |y|exy - , , N be the solution of the differential equation x dy - y dx + xy(x dy + y dx) = 0, y(1) = 2. Then + is equal to

Ques. 7JEE Main 2024, 6 April (Morning Shift)

Let y = y(x) be the solution of the differential equation (2x e x) dydx + 2y = 3x e x, x > 0 and y (e-1 ) = 0. Then, y(e) is equal to

  1. - 3e
  2. - 32e
  3. - 23e
  4. - 2e
Ques. 8JEE Main 2024, 6 April (Morning Shift)

Let y = y(x) be the solution of the differential equation (1 + x2 ) dydx + y = e -1x, y(1) = 0. Then y(0) is

  1. 14 (e /2 - 1 )
  2. 12 (1 - e /2 )
  3. 14 (1 - e /2 )
  4. 12 (e /2 - 1 )
Ques. 9JEE Main 2024, 6 April (Morning Shift)

Let a conic C pass through the point (4, -2) and P(x, y), x 3, be any point on C. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3, -5). If the focal distance of the point (7, 1) on C is d, then 12d equals .

Ques. 10JEE Main 2024, 6 April (Evening Shift)

Suppose the solution of the differential equation dydx = (2 + )x - y + 2 x - 2 y - ( - 4 ) represents a circle passing through origin. Then the radius of this circle is :

  1. 17
  2. 2
  3. 172
  4. 12
Ques. 11JEE Main 2024, 6 April (Evening Shift)

If the solution y(x) of the given differential equation (ey + 1 ) x dx + ey x dy = 0 passes through the point ( 2, 0 ), then the value of ey ( 6 ) is equal to .

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