Question

The solution of the differential equation $\frac{d^3 y}{dx^3} + \frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 4y = 0$ is given by}

• A. $y = C_1 e^x + C_2 \cos 2x + C_3 \sin 2x$
• B. $y = C_1 e^{-x} + C_2 \cos 2x + C_3 \sin 2x$
• C. $y = C_1 e^{2x} + C_2 \cos 2x + C_3 \sin 2x$
• D. $y = C_1 e^{-2x} + C_2 \cos 2x + C_3 \sin 2x$

Hint:

Group terms immediately in polynomial auxiliary equations. Recognizing that $m = -1$ is a root allows factoring the cubic equation into a linear and a quadratic term without synthetic division.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To solve a linear homogeneous ordinary differential equation with constant coefficients, we find the roots of its characteristic (auxiliary) equation.
Key Formula or Approach:
If the auxiliary equation has a real root $m_1$ and a pair of complex conjugate roots $\alpha \pm i\beta$, the general solution is: \[ y = C_1 e^{m_1 x} + e^{\alpha x} (C_2 \cos \beta x + C_3 \sin \beta x) \]

Step 2: Detailed Explanation:
The given differential equation is: \[ y''' + y'' + 4y' + 4y = 0 \]
Write down the auxiliary equation: \[ m^3 + m^2 + 4m + 4 = 0 \]
Factor the equation by grouping terms: \[ m^2(m + 1) + 4(m + 1) = 0 \] \[ (m^2 + 4)(m + 1) = 0 \]
Setting each factor to zero gives the roots: \[ m + 1 = 0 \implies m_1 = -1 \] \[ m^2 + 4 = 0 \implies m^2 = -4 \implies m_{2,3} = \pm 2i \]
Here, we have one real root $m_1 = -1$ and a pair of purely imaginary roots with $\alpha = 0$ and $\beta = 2$.
Substitute these roots into the general solution template: \[ y = C_1 e^{-x} + e^{0x} (C_2 \cos 2x + C_3 \sin 2x) \] \[ y = C_1 e^{-x} + C_2 \cos 2x + C_3 \sin 2x \]

Step 3: Final Answer:
The general solution of the differential equation is $y = C_1 e^{-x} + C_2 \cos 2x + C_3 \sin 2x$.