Question

Which of the following differential equation is satisfied by $y_{1}(x)=e^{x}$, $y_{2}(x)=x~e^{x}$ and $y_{3}=e^{2x}?$

• A. $\frac{d^{3}y}{dx^{3}}+\frac{4~d^{2}y}{dx^{2}}+\frac{5~dy}{dx}+2y=0$
• B. $\frac{d^{3}y}{dx^{3}}-\frac{4d^{2}y}{dx^{2}}+\frac{5dy}{dx}-2y=0$
• C. $\frac{d^{3}y}{dx^{3}}+\frac{4~d^{2}y}{dx^{2}}-\frac{5dy}{dx}-2y=0$
• D. $\frac{d^{3}y}{dx^{3}}-\frac{4~d^{2}y}{dx^{2}}+\frac{5~dy}{dx}+2y=0$

Hint:

If you see a term $x^k e^{ax}$ in a solution, it always means the root $a$ is repeated at least $k+1$ times. Reconstructing the polynomial from roots is the most reliable way to find the original differential equation.

Answer 1

The Correct Option is B

The given functions $y_1, y_2, y_3$ are the basis of the solution set. We can reconstruct the differential equation by finding the roots of the auxiliary (characteristic) equation.

Step 1: \color{redIdentify Roots from the Functions
The general solution of a linear homogeneous equation with constant coefficients is related to the roots $m$ of the auxiliary equation.
- $y_1 = e^x$ corresponds to a root $m = 1$.
- $y_2 = x e^x$ implies that the root $m = 1$ is a repeated root (multiplicity 2).
- $y_3 = e^{2x}$ corresponds to a root $m = 2$.
The roots are 1, 1, and 2.

Step 2: \color{redConstruct the Auxiliary Equation
The characteristic equation is formed by $(m - \text{root}_1)(m - \text{root}_2)(m - \text{root}_3) = 0$:
$(m - 1)^2 (m - 2) = 0$
$(m^2 - 2m + 1)(m - 2) = 0$

Step 3: \color{redExpand the Polynomial
$m(m^2 - 2m + 1) - 2(m^2 - 2m + 1) = 0$
$m^3 - 2m^2 + m - 2m^2 + 4m - 2 = 0$
$m^3 - 4m^2 + 5m - 2 = 0$.

Step 4: \color{redConvert back to Differential Equation
Replace $m^k$ with $\frac{d^k y}{dx^k}$:
$\frac{d^3 y}{dx^3} - 4\frac{d^2 y}{dx^2} + 5\frac{dy}{dx} - 2y = 0$.
Comparing this result with the options, it perfectly matches Option (2).