Step 1: Understanding the Concept:
This is a third-order linear homogeneous differential equation with constant coefficients. Such equations are solved by finding the roots of their corresponding auxiliary (or characteristic) equation.
Step 2: Key Formula or Approach:
1. Write down the auxiliary equation by replacing $\frac{d^ny}{dx^n}$ with $m^n$.
2. Solve the auxiliary equation for its roots ($m_1, m_2, m_3, \dots$).
3. The form of the general solution depends on the nature of these roots:
- If roots are real and distinct ($m_1, m_2, \dots$), the solution is $y = C_1e^{m_1x} + C_2e^{m_2x} + \dots$.
- If roots are real and repeated (e.g., $m_1$ is a root of multiplicity k), the corresponding part of the solution is $(C_1 + C_2x + \dots + C_kx^{k-1})e^{m_1x}$.
- If roots are complex conjugates ($\alpha \pm i\beta$), the corresponding part of the solution is $e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))$.
Step 3: Detailed Explanation:
The differential equation is $y''' + 3y'' + 2y' = 0$.
1. The auxiliary equation is:
\[ m^3 + 3m^2 + 2m = 0 \]
2. Solve for the roots. First, factor out m:
\[ m(m^2 + 3m + 2) = 0 \]
Factor the quadratic expression:
\[ m(m+1)(m+2) = 0 \]
The roots are $m_1 = 0$, $m_2 = -1$, and $m_3 = -2$.
3. The roots are real and distinct. Therefore, the general solution is of the form $y = C_1e^{m_1x} + C_2e^{m_2x} + C_3e^{m_3x}$.
Substituting the roots:
\[ y = C_1e^{0x} + C_2e^{-1x} + C_3e^{-2x} \]
Since $e^{0x} = 1$, the solution is:
\[ y = C_1(1) + C_2e^{-x} + C_3e^{-2x} \]
Using the constants a, b, and c as in the options:
\[ y = a + be^{-x} + ce^{-2x} \]
Step 4: Final Answer:
The solution of the differential equation is $y = a + be^{-x} + ce^{-2x}$. Therefore, option (A) is correct.