To determine for which values of m the function f(x) = x^m is a solution to the given differential equations, we will need to plug f(x) into each equation and analyze the results.
a) 3x² (d²y/dx²) + 11x (dy/dx) + 3y = 0
Let’s first calculate the derivatives of f(x) = x^m:
- First derivative: dy/dx = mx^(m-1)
- Second derivative: d²y/dx² = m(m-1)x^(m-2)
Now, substituting these derivatives into the differential equation:
3x²(m(m-1)x^(m-2)) + 11x(mx^(m-1)) + 3(x^m) = 0
Which simplifies to:
3m(m-1)x^m + 11mx^m + 3x^m = 0
Factoring out x^m gives:
x^m(3m(m-1) + 11m + 3) = 0
For this expression to hold true for all x, the coefficient must be zero:
3m(m-1) + 11m + 3 = 0
Expanding and simplifying:
3m^2 - 3m + 11m + 3 = 0 3m^2 + 8m + 3 = 0
Now, we can apply the quadratic formula:
m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-8 \pm 2\sqrt{7}}{6} = \frac{-4 \pm \sqrt{7}}{3}
This gives us two solutions for m:
m = \frac{-4 + \sqrt{7}}{3} \quad ext{and} \quad m = \frac{-4 - \sqrt{7}}{3}
b) x² (d²y/dx²) – x (dy/dx) + 5y = 0
We will repeat a similar process for this second equation. Using the same derivatives:
x²(m(m-1)x^(m-2)) - x(mx^(m-1)) + 5(x^m) = 0
Simplifying again leads to:
m(m-1)x^m - mx^m + 5x^m = 0
Factoring out x^m yields:
x^m(m(m-1) - m + 5) = 0
Setting the coefficient to zero gives us:
m(m-1) - m + 5 = 0
Solving this:
m^2 - m - m + 5 = 0
Which simplifies to:
m^2 - 2m + 5 = 0
This has no real roots (discriminant is less than zero), indicating that there are no values of m for which x^m solves this particular equation.
In summary:
- For the first equation (a), m values are \( m = \frac{-4 + \sqrt{7}}{3} \) and \( m = \frac{-4 – \sqrt{7}}{3} \).
- For the second equation (b), there are no real values for m.