Determine for which values of m the function Φ (x) = x^m is a solution to the given equation.
(A) 3x² d²y/dx² + 11x dy/dx - 3y = 0
(B) x² d²y/dx² - x dy/dx - 5y = 0

Solution:

Given, Φ(x) = x^m

We have to find the value of m.

y(x) = x^m

y’(x) = mx^{m - 1} = (m/x)x^m

y’’(x) = m(m - 1)x^{m - 2} = (m(m - 1)/x²)x^m

A) 3x² d²y/dx² + 11x dy/dx - 3y = 0

3x²(m(m - 1)/x²)x^m + 11x(m/x)x^m - 3x^m = 0

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

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

3(m² - m) + 11m - 3 = 0

3m² - 3m + 11m - 3 = 0

3m² + 8m - 3 = 0

3m² + 9m - m - 3 = 0

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

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

3m - 1 = 0 ⇒ 3m = 1 ⇒ m = 1/3

m + 3 = 0 ⇒ m = -3

The general solution is of the form y = Φ(x) = \(C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x}\)

Here, m₁ = 1/3 and m₂ = -3

So, \(C_{1}e^{\frac{1}{3}x}+C_{2}e^{-3x}\)

Therefore, the solution is \(C_{1}e^{\frac{1}{3}x}+C_{2}e^{-3x}\)

B) x² d²y/dx² - x dy/dx - 5y = 0

x²(m(m - 1)/x²)x^m - x (m/x)x^m - 5x^m = 0

x^m[m(m - 1) - m - 5] = 0

m² - m - m - 5 = 0

m² - 2m - 5 = 0

Using quadratic formula,

\(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here, a = 1, b = -2 and c = -5

\(x=\frac{-(-2)\pm \sqrt{(-2)^{2}-4(1)(-5)}}{2(1)}\)

\(x=\frac{2\pm \sqrt{4+20}}{2}\)

\(x=\frac{2\pm \sqrt{24}}{2}\)

\(x=\frac{2\pm 2\sqrt{6}}{2}\)

\(x=1\pm \sqrt{6}\)

The general solution is of the form y = Φ(x) = \(C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x}\)

Here, m₁ = 1 + √6 and m₂ = 1 - √6

So, \(C_{1}e^{(1-\sqrt{6})x}+C_{2}e^{(1-\sqrt{6})x}\)

Therefore, the solution is \(C_{1}e^{(1-\sqrt{6})x}+C_{2}e^{(1-\sqrt{6})x}\).

Therefore, the solution to

(A) 3x² d²y/dx² + 11x dy/dx - 3y = 0 is \(C_{1}e^{\frac{1}{3}x}+C_{2}e^{-3x}\)

(B) x² d²y/dx² - x dy/dx - 5y = 0 is \(C_{1}e^{(1-\sqrt{6})x}+C_{2}e^{(1-\sqrt{6})x}\).