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: Identify the given general solution.
The given function is \[ y=Ae^x+Be^{-2x} \] This is the general solution of a second-order linear homogeneous differential equation with constant coefficients.
The terms \(e^x\) and \(e^{-2x}\) show that the roots of the auxiliary equation are \[ m=1 \] and \[ m=-2 \]

Step 2: Form the auxiliary equation.
If the roots are \(1\) and \(-2\), then the auxiliary equation is \[ (m-1)(m+2)=0 \] Expanding, \[ m^2+m-2=0 \]

Step 3: Convert auxiliary equation into differential equation.
For a linear differential equation with constant coefficients, replace \(m^2\) by \[ \frac{d^2y}{dx^2} \] replace \(m\) by \[ \frac{dy}{dx} \] and the constant term multiplies \(y\).
Thus, from \[ m^2+m-2=0 \] we get \[ \frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=0 \]

Step 4: Verify by direct differentiation.
Given, \[ y=Ae^x+Be^{-2x} \] Differentiate once: \[ \frac{dy}{dx}=Ae^x-2Be^{-2x} \] Differentiate again: \[ \frac{d^2y}{dx^2}=Ae^x+4Be^{-2x} \] Now substitute in \[ \frac{d^2y}{dx^2}+\frac{dy}{dx}-2y \] \[ = (Ae^x+4Be^{-2x})+(Ae^x-2Be^{-2x})-2(Ae^x+Be^{-2x}) \] \[ = Ae^x+4Be^{-2x}+Ae^x-2Be^{-2x}-2Ae^x-2Be^{-2x} \] \[ = 0 \] Hence, the given function satisfies the differential equation.

Step 5: Final Answer.
Therefore, the required differential equation is \[ \boxed{\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=0} \]