Question

The solution of the differential equation \[ \frac{d^3 y}{dx^3}-5.5\,\frac{d^2 y}{dx^2}+9.5\,\frac{dy}{dx}-5\,y=0 \] is expressed as $y=C_1 e^{2.5x}+C_2 e^{\alpha x}+C_3 e^{\beta x}$, where $C_1,C_2,C_3,\alpha,\beta$ are constants, with $\alpha$ and $\beta$ distinct and not equal to $2.5$. Which of the following options is correct for the values of $\alpha$ and $\beta$?

• A. 1 and 2
• B. $-1$ and $-2$
• C. 2 and 3
• D. $-2$ and $-3$

Hint:

When one exponential term is given in a linear ODE's solution, use it to factor the characteristic polynomial. Divide out the known root first, then solve the remaining quadratic for the other exponents.

Answer 1

The Correct Option is A

Step 1: Write the characteristic (auxiliary) equation.
Assume a trial solution $y=e^{rx}$; substituting gives
$r^3-5.5r^2+9.5r-5=0.$

Step 2: Use the given root $r=2.5$.
Since $e^{2.5x}$ is part of the solution, $(r-2.5)$ is a factor.
Divide $r^3-5.5r^2+9.5r-5$ by $(r-2.5)$ (synthetic division):
Coefficients $1,\ -5.5,\ 9.5,\ -5$ $\Rightarrow$ bring down $1$;
$1\times 2.5=2.5$, add to $-5.5$ gives $-3$;
$-3\times 2.5=-7.5$, add to $9.5$ gives $2$;
$2\times 2.5=5$, add to $-5$ gives $0$ (remainder).
Hence the quadratic factor is $r^2-3r+2=0$.

Step 3: Solve for the remaining roots.
$r^2-3r+2=(r-1)(r-2)=0 \Rightarrow r=1,\,2.$
Therefore $\alpha$ and $\beta$ are $1$ and $2$ (distinct and $\neq 2.5$).
\[ \boxed{\alpha=1,\ \beta=2} \]