3. For x²y'' + 4xy' + 2y = 0, the characteristic equation is:
Correct Answer: b) r² + 3r + 2 = 0
Explanation:
1. The standard form of an **Euler-Cauchy equation** is:
\( x^2y'' + bxy' + cy = 0 \)
2. To solve it, we use the substitution:
\( y = x^r \)
3. Then the derivatives become:
\( y' = rx^{r-1}, \quad y'' = r(r-1)x^{r-2} \)
4. Substituting these into the given equation:
\( x^2 \cdot r(r-1)x^{r-2} + 4x \cdot rx^{r-1} + 2x^r = 0 \)
5. Simplify:
\( r(r-1)x^r + 4rx^r + 2x^r = 0 \)
6. Factor out \( x^r \) (assuming \( x \neq 0 \)):
\( x^r \left( r^2 + 4r + 2 \right) = 0 \)
7. The characteristic equation is:
\( r^2 + 4r + 2 = 0 \)
Hence, the correct answer is: Option (a).