Differential Equations · Question 14

Let a differentiable function f satisfy the equation 036 f ( tx36 )dt = 4 f(x). If y = f(x) is a standard parabola passing through the points (2, 1) and (-4, ), then is equal to .

Let y = y(x) be a differentiable function in the interval (0, ) such that y(1) = 2, and t x ( t2y(x) - x2y(t)x - t ) = 3 for each x > 0. Then 2y(2) is equal to :

1. 27
2. 18
3. 23
4. 12

Let y = y(x) be the solution of the differential equation x4 dy + (4x3y + 2 x) dx = 0, x > 0, y ( 2 ) = 0. Then 4 y ( 3 ) is equal to :

1. 92
2. 72
3. 64
4. 81

Let f be a twice differentiable non-negative function such that (f(x))2 = 25 + 0x ((f(t))2 + (f'(t))2 )dt. Then the mean of f( e(1)), f( e(2)), , f( e(625)) is equal to .

If the solution curve y = f(x) of the differential equation (x2 - 4)y' - 2xy + 2x(4 - x2)2 = 0, x > 2, passes through the point (3, 15), then the local maximum value of f is

Let the solution curve of the differential equation x dy - y dx = x2 + y2 dx, x > 0, y(1) = 0, be y = y(x). Then y(3) is equal to

1. 6
2. 4
3. 1
4. 2

Let f : [1, ) R be a differentiable function. If 6 1x f(t) dt = 3xf(x) + x3 - 4 for all x 1, then the value of f(2) - f(3) is :

1. 4
2. 3
3. -4
4. -3

If y = y(x) satisfies the differential equation 16 ( 9x + x x ) (4 + 9 + x ) y dy = (1 + 2 y) dx, x > 0 and y(256) = 2, y(49) = , then 2 is equal to :

1. 2 2 - 1
2. 2 - 1
3. 2( 2 - 1)
4. 3( 2 - 1)

Let y = y(x) be the solution curve of the differential equation (1 + x2)dy + (y - -1 x)dx = 0, y(0) = 1. Then the value of y(1) is :

1. 4e /4 - 2 - 1
2. 2e /4 + 4 - 1
3. 4e /4 + 2 - 1
4. 2e /4 - 4 - 1

Let y = y(x) be the solution of the differential equation x dydx - 2y = 2 + 3 x, x (- 2, 2 ), y(0) = - 74. Then y ( 6 ) is equal to :

1. - 52
2. -3 2 - 7
3. - 54
4. -3 3 - 7