Determine for which values of m the function variant phi left parenthesis x right parenthesis equals x Superscript m
 
is a solution to the given equation.
​(a) 4 x squared StartFraction d squared y Over dx squared EndFraction minus 5 x StartFraction dy Over dx EndFraction plus 2 y equals 0
​(b) x squared StartFraction d squared y Over dx squared EndFraction plus 5 x StartFraction dy Over dx EndFraction minus 8 y equals 0

Let's solve each part separately.

### (a) \( 4x^2 \frac{d^2 y}{dx^2} - 5x \frac{dy}{dx} + 2y = 0 \)

Given that \( \phi(x) = x^m \), we need to substitute it into the equation and find the values of \( m \).

First, compute the necessary derivatives for \( \phi(x) = x^m \):
1. \( \frac{d\phi}{dx} = m x^{m-1} \)
2. \( \frac{d^2\phi}{dx^2} = m(m-1) x^{m-2} \)

Now, substitute these into the differential equation:
\[
4x^2 \frac{d^2\phi}{dx^2} - 5x \frac{d\phi}{dx} + 2\phi = 0
\]

Substituting \( \frac{d^2\phi}{dx^2} = m(m-1) x^{m-2} \), \( \frac{d\phi}{dx} = m x^{m-1} \), and \( \phi(x) = x^m \):
\[
4x^2 \cdot m(m-1) x^{m-2} - 5x \cdot m x^{m-1} + 2x^m = 0
\]
Simplify each term:
\[
4m(m-1) x^m - 5m x^m + 2x^m = 0
\]
Factor out \( x^m \):
\[
x^m \left( 4m(m-1) - 5m + 2 \right) = 0
\]
Since \( x^m \neq 0 \) for all \( x \), the expression inside the parentheses must be zero:
\[
4m(m-1) - 5m + 2 = 0
\]

Expand the terms:
\[
4m^2 - 4m - 5m + 2 = 0
\]
\[
4m^2 - 9m + 2 = 0
\]
Now solve this quadratic equation using the quadratic formula:
\[
m = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(4)(2)}}{2(4)}
\]
\[
m = \frac{9 \pm \sqrt{81 - 32}}{8}
\]
\[
m = \frac{9 \pm \sqrt{49}}{8}
\]
\[
m = \frac{9 \pm 7}{8}
\]
Thus, \( m = \frac{9 + 7}{8} = 2 \) or \( m = \frac{9 - 7}{8} = \frac{1}{4} \).

So, the values of \( m \) for part (a) are \( m = 2 \) and \( m = \frac{1}{4} \).

---

### (b) \( x^2 \frac{d^2 y}{dx^2} + 5x \frac{dy}{dx} - 8y = 0 \)

We proceed similarly for this equation. Using \( \phi(x) = x^m \), substitute the derivatives:
1. \( \frac{d\phi}{dx} = m x^{m-1} \)
2. \( \frac{d^2\phi}{dx^2} = m(m-1) x^{m-2} \)

Substituting these into the differential equation:
\[
x^2 \cdot m(m-1) x^{m-2} + 5x \cdot m x^{m-1} - 8x^m = 0
\]
Simplifying:
\[
m(m-1) x^m + 5m x^m - 8x^m = 0
\]
Factor out \( x^m \):
\[
x^m \left( m(m-1) + 5m - 8 \right) = 0
\]
Since \( x^m \neq 0 \), the term inside the parentheses must be zero:
\[
m(m-1) + 5m - 8 = 0
\]
Expand the terms:
\[
m^2 - m + 5m - 8 = 0
\]
\[
m^2 + 4m - 8 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
m = \frac{-4 \pm \sqrt{4^2 - 4(1)(-8)}}{2(1)}
\]
\[
m = \frac{-4 \pm \sqrt{16 + 32}}{2}
\]
\[
m = \frac{-4 \pm \sqrt{48}}{2}
\]
\[
m = \frac{-4 \pm 4\sqrt{3}}{2}
\]
\[
m = -2 \pm 2\sqrt{3}
\]

So, the values of \( m \) for part (b) are \( m = -2 + 2\sqrt{3} \) and \( m = -2 - 2\sqrt{3} \).

---

### Summary:
- **(a)** \( m = 2 \) or \( m = \frac{1}{4} \)
- **(b)** \( m = -2 + 2\sqrt{3} \) or \( m = -2 - 2\sqrt{3} \)

Do you want any further details or clarifications?

---

Here are five related questions:
1. How does the solution change if the differential equation has different coefficients?
2. Can you explain the relationship between the solution's exponent and the behavior of the function at different points?
3. What is the significance of solving differential equations using power functions like \( x^m \)?
4. How would the solution change if we considered complex values for \( m \)?
5. What methods can be used to solve second-order linear homogeneous differential equations with variable coefficients?

**Tip:** When solving quadratic equations, always check the discriminant first to understand the nature of the solutions (real vs complex).

Determine for which values of m the function φ(x) = x^m is a solution to the given equation: (a) 4x^2 d²y/dx² - 5x dy/dx + 2y = 0 (b) x^2 d²y/dx² + 5x dy/dx - 8y = 0

Explore how to find the values of m that make the function φ(x) = x^m a solution to second-order differential equations. This step-by-step solution includes using derivatives, quadratic equations, and factorization. Ideal for students studying calculus and differential equations.

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