Determine the parameter k such that y=x^k is a particular…

Determine the parameter $k$ such that $y=x^k$ is a particular solution of the Euler equation $6x^2y'' + 5xy' - 2y = 0$.

Step 1: Substituting the ansatz $y=x^k$

For $y=x^k$ ($x>0$), the first two derivatives are

\[ y' = kx^{k-1}, \qquad y'' = k(k-1)x^{k-2}. \]

Substituting into the equation: $6x^2\cdot k(k-1)x^{k-2} + 5x\cdot kx^{k-1} - 2x^k = \bigl[\,6k(k-1) + 5k - 2\,\bigr]x^k = 0$.

Step 2: The indicial equation

Since $x^k\neq0$ for $x>0$, the bracketed factor must vanish identically. This is the characteristic (indicial) equation attached to the Euler equation: \[6k(k-1) + 5k - 2 = 0 \;\Longrightarrow\; 6k^2 - 6k + 5k - 2 = 0 \;\Longrightarrow\; 6k^2 - k - 2 = 0.\]

Step 3: Solving the indicial equation

By the quadratic formula, $k = \frac{1 \pm \sqrt{1 + 48}}{12} = \frac{1\pm7}{12}$.

which produces two distinct real roots: \[k_1 = \frac{8}{12} = \frac23, \qquad k_2 = \frac{-6}{12} = -\frac12.\]

Step 4: General solution

Because $k_1\neq k_2$, the powers $x^{2/3}$ and $x^{-1/2}$ are linearly independent on $x>0$ and form a fundamental system for this (real, distinct-roots) Euler equation, so

\[ y = C_1x^{2/3} + C_2x^{-1/2}. \]

Step 5: Verification

For $y_1=x^{2/3}$: $y_1'=\tfrac23x^{-1/3}$, $y_1''=-\tfrac29x^{-4/3}$. Substituting into the left-hand side, \[6x^2\left(-\frac29x^{-4/3}\right) + 5x\left(\frac23x^{-1/3}\right) - 2x^{2/3} = \left(-\frac43+\frac{10}{3}-2\right)x^{2/3} = 0.\ \checkmark\]

For $y_2=x^{-1/2}$: $y_2'=-\tfrac12x^{-3/2}$, $y_2''=\tfrac34x^{-5/2}$. Substituting, \[6x^2\left(\frac34x^{-5/2}\right) + 5x\left(-\frac12x^{-3/2}\right) - 2x^{-1/2} = \left(\frac92-\frac52-2\right)x^{-1/2} = 0.\ \checkmark\]

Both power functions satisfy the equation exactly, confirming the two indicial roots.