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Differential Equations

JEE Main 2024 (Online) 8th April Evening ShiftNumericalMedium📈 Topic in 7% papers+4
Ques. 1JEE 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

Answer

4

Solution

Divide by xy: x dy - y dxxy + (x dy + y dx) = 0

dyy - dxx + d(xy) = 0

|y| - |x| + xy = c

y(1) = 2 c = 2 + 2

|y|2|x| = 2 - xy 2|x| = |y|exy - 2

= 2, = 2

∴ + = 4

Answer: 4

Keep practising — Differential Equations

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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 .

Ques. 7JEE Main 2024, 5 April (Morning Shift)

If y = y(x) is the solution of the differential equation dydx + 2y = (2x), y(0) = 34, then y ( 8 ) is equal to :

  1. e- /8
  2. e /4
  3. e- /4
  4. e /8
Ques. 8JEE Main 2024, 5 April (Morning Shift)

Let f be a differentiable function in the interval (0, ) such that f(1) = 1 and t x t2 f(x) - x2 f(t)t - x = 1 for each x > 0. Then 2f(2) + 3f(3) is equal to .

Ques. 9JEE Main 2024, 5 April (Evening Shift)

The differential equation of the family of circles passing through the origin and having centre at the line y = x is :

  1. (x2 - y2 + 2xy )dx = (x2 - y2 + 2xy )dy
  2. (x2 + y2 - 2xy )dx = (x2 + y2 + 2xy )dy
  3. (x2 + y2 + 2xy )dx = (x2 + y2 - 2xy )dy
  4. (x2 - y2 + 2xy )dx = (x2 - y2 - 2xy )dy
Ques. 10JEE Main 2024, 5 April (Evening Shift)

Let y = y(x) be the solution of the differential equation dydx + 2x (1 + x2 )2y = xe 1 (1 + x2 ); y(0) = 0. Then the area enclosed by the curve f(x) = y(x)e- 1 (1 + x2 ) and the line y - x = 4 is .

Ques. 11JEE Main 2024, 4 April (Morning Shift)

If the solution y=y(x) of the differential equation (x4+2x3+3x2+2x+2 )dy- (2x2+2x+3 )dx=0 satisfies y(-1)=- 4, then y(0) is equal to :

  1. - 12
  2. 2
  3. 0
  4. 4

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