Advertisements
Advertisements
प्रश्न
Form the differential equation having for its general solution y = ax2 + bx
उत्तर
y = ax2 + bx ......(1)
Since we have two arbitrary constants, differentiative twice.
`("d"y)/("d"x) = 2"a"x + "b"` ......(2)
`("d"^2y)/("d"x^2)` = 2a
a = `1/2 ("d"^2y)/("d"x^2)` .......(3)
From (2) and (3)
`("d"y)/("d"x) = ("d"^2y)/("d"x^2) (x) +"b"`
b = `("d"y)/("d"x) - x ("d"^2y)/("d"x^2)` .........(4)
Substitute the value of a and b in equation (1)
Equation (1)
⇒ y = `x^2/2 ("d"^2y)/("d"x^2) + (("d"y)/("d"x) - x ("d"^2y)/("d"x^2))x`
y = `x^2/2 ("d"^2y)/("d"x^2) + x ("d"y)/("d"x) - x^2 ("d"^2y)/("d"x^2)`
Multiply each term by 2
2y = `x^2 ("d"^2y)/("d"x^2) + 2x ("d"y)/("d"x) - 2x^2 ("d"^2y)/("d"x^2)`
2y = `- x^2 ("d"^2y)/("d"x^2) + 2x ("d"y)/("d"x)`
⇒ `x^2 ("d"^2y)/("d"x^2) - 2x ("d"y)/(d") + 2y`= 0
APPEARS IN
संबंधित प्रश्न
If F is the constant force generated by the motor of an automobile of mass M, its velocity V is given by `"M""dv"/"dt"` = F – kV, where k is a constant. Express V in terms of t given that V = 0 when t = 0
The velocity v, of a parachute falling vertically satisfies the equation `"v" (dv)/(dx) = "g"(1 - v^2/k^2)` where g and k are constants. If v and are both initially zero, find v in terms of x
Solve the following differential equation:
`(ydx - xdy) cot (x/y)` = ny2 dx
Solve: `y(1 - x) - x ("d"y)/("d"x)` = 0
Solve: `("d"y)/("d"x) + "e"^x + y"e"^x = 0`
Solve the following:
If `("d"y)/("d"x) + 2 y tan x = sin x` and if y = 0 when x = `pi/3` express y in term of x
Solve the following:
`("d"y)/("d"x) + y/x = x"e"^x`
Choose the correct alternative:
The solution of the differential equation `("d"y)/("d"x) + "P"y` = Q where P and Q are the function of x is
Choose the correct alternative:
A homogeneous differential equation of the form `("d"x)/("d"y) = f(x/y)` can be solved by making substitution
Choose the correct alternative:
Which of the following is the homogeneous differential equation?