Question

Determine for which values of m the function variant Φ(x) = x^m is a solution to the given equation. a. 3x^2 (d^2y/dx^2) + 11x(dy/dx) - 3y = 0 b. x^2 (d^2y/dx^2) - x(dy/dx) - 5y = 0

Answer 1

a) m = -3 or 1/3

b) m = 1 + root 6 or 1 - root 6

Explanation:

The step by step calculation is as shown in the attachment.

Answer 2

a) m = -9 or m = 1

b) m = 1 + √6 or m = 1 -√6

Explanation:

for

Φ(x) = x^m

then

dΦ/dx (x) = m*x^(m-1)

d²Φ/dx² (x) = m*(m-1)*x^(m-2)

then

for a

3x^2 (d^2y/dx^2) + 11x(dy/dx) - 3y = 0

3x^2*m*(m-1)*x^(m-2) + 11*x* m*x^(m-1) - 3*x^m = 0

3*m*(m-1)*x^m + 11*m*x^m- 3*x^m = 0

dividing by x^m

3*m*(m-1) + 11*m - 3 =0

3*m² + 8 m - 3 =0

m= [-8 ± √(64 + 4*3*3)]/2 = (-8±10)/2

m₁ = -9 , m₂= 1

then Φ(x) = x^m is a solution for the equation a , when m = -9 or m = 1

for b)

x^2 (d^2y/dx^2) - x(dy/dx) - 5y = 0

x^2*m*(m-1)*x^(m-2) - x* m*x^(m-1) - 5*x^m = 0

m*(m-1)*x^m -m *x^m- 5*x^m = 0

dividing by x^m

m*(m-1) -m - 5 =0

m² - 2 m - 5 =0

m= [2 ± √(4 + 4*1*5)]/2 = (2±√24)/2 = 1 ±√6

m₁ = 1 + √6 , m₂ = 1 - √6

then Φ(x) = x^m is a solution for the equation b , when m = 1 + √6 or m = 1 - √6