Answer:
a) m = -9 or m = 1
b) m = 1 + √6 or m = 1 -√6
Step-by-step 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
