Solve: (x2D2- 4xD + 6)y = x2where D = d/dxa)y = c1x2+ c2x2- x2logx2b)y...
To solve the given differential equation, we substitute the value of D in terms of d/dx and rearrange the equation to isolate y.
Given: (x^2D^2 - 4xD + 6)y = x^2
We are given that D = d/dx, so substituting this in the equation, we get:
(x^2(d/dx)^2 - 4x(d/dx) + 6)y = x^2
Simplifying the equation further, we have:
(x^2(d^2y/dx^2) - 4x(dy/dx) + 6)y = x^2
Expanding the equation, we get:
x^2(d^2y/dx^2)y - 4x(dy/dx)y + 6y = x^2
Rearranging the terms, we have:
x^2(d^2y/dx^2)y - 4x(dy/dx)y + (6y - x^2) = 0
Now, we can see that this is a homogeneous linear differential equation. So, let's assume y = x^m and substitute it in the equation.
(x^2d^2/dx^2)(x^m) - 4(xd/dx)(x^m) + (6x^m - x^2) = 0
Differentiating y = x^m twice, we get:
m(m-1)x^m-2 - 4m(x^m-1) + (6x^m - x^2) = 0
Expanding and combining like terms, we have:
m(m-1)x^m-2 - 4mx^m-1 + 6x^m - x^2 = 0
Now, let's divide the entire equation by x^m-2 to simplify further:
m(m-1) - 4mx + 6x^2/x^m-2 - x^2/x^m-2 = 0
Simplifying, we have:
m(m-1) - 4mx + 6x^2/x^2 - x^2/x^2 = 0
m(m-1) - 4mx + 6x^2 - x^2 = 0
m^2 - m - 4mx + 5x^2 = 0
Now, we can equate the coefficients of the like powers of x to get the values of m:
m^2 - m = 0 (coefficient of x^0)
-4m + 5 = 0 (coefficient of x^2)
Solving these equations, we get:
m = 0 or m = 1
m = 5/4
Therefore, the general solution to the differential equation is:
y = c1x^0 + c2x^1 + c3x^(5/4)
Simplifying further, we have:
y = c1 + c2x + c3x^(5/4)
This matches with the option C, which states that y = c1x^2 + c2x^3 - x^2log(x^2).
Hence, the correct answer is option C.
Solve: (x2D2- 4xD + 6)y = x2where D = d/dxa)y = c1x2+ c2x2- x2logx2b)y...
Cauchy's Equation MCQ Question 2 Detailed Solution
Concept:
If the derivation power and the variable power are the same, then the equation is Cauchy - Euler equation.
Now,
It can be solved by putting x = ez and log x = z and hence d/dz = θ
Analysis:
(x2D2 - 4xD + 6)y = x2 ---(1)
xD = θ
x2D2 = θ (θ - 1)
substitute in equation (1):
[(θ2 - θ) - 4θ + 6]y = e2z ---(2)
It becomes linear differential equation with real constants.
f(θ) y = ϕ (z)
f(θ) = θ2 - 5θ + 6, ϕ (z) = e2z
C.F:
θ2 - 5θ + 6 = 0
m2 - 5m + 6 = 0
(m - 2) (m - 3) = 0
∴ C.F = C
1e
2z + C
2e
3zputting θ = 2, f(θ) = 0
hence,
= -ze
2z∴ y = c1e2z + c2e3z - ze2z
putting, x = ez
∴ y = c1 x2 + c2 x3 - x2logx2