Question

\(y=Ae^x+Be^{-2x}\) satisfies which of the following differential equations?

• A. \(\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}+2y=0\)
• B. \(\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}-y=0\)
• C. \(\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=0\)
• D. \(\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-2y=0\)

Hint:

If the general solution is \(y=Ae^{m_1x}+Be^{m_2x}\), then \(m_1\) and \(m_2\) are roots of the auxiliary equation.

Answer 1

The Correct Option is D

Step 1: Read off the characteristic roots.
The solution $y=Ae^{x}+Be^{-2x}$ comes from exponents $e^{1\cdot x}$ and $e^{-2x}$, so the auxiliary roots are $m=1$ and $m=-2$.
Step 2: Form the auxiliary equation.
$(m-1)(m+2)=0$, which expands to $m^2+m-2=0$.
Step 3: Convert to a differential equation.
Replacing $m^2$ by $\dfrac{d^2y}{dx^2}$, $m$ by $\dfrac{dy}{dx}$ and the constant by a multiple of $y$ gives $\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-2y=0$.
Step 4: Differentiate to verify.
$\dfrac{dy}{dx}=Ae^{x}-2Be^{-2x}$ and $\dfrac{d^2y}{dx^2}=Ae^{x}+4Be^{-2x}$.
Step 5: Substitute into the equation.
$(Ae^{x}+4Be^{-2x})+(Ae^{x}-2Be^{-2x})-2(Ae^{x}+Be^{-2x})=0$, which simplifies to $0$.
Step 6: Conclude.
The function satisfies $\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-2y=0$.
\[ \boxed{\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}-2y=0} \]