Question

The solution of the differential equation \(\frac{d^3y}{dx^3} + 3\frac{d^2y}{dx^2} + 2\frac{dy}{dx} = 0\) is

• A. \(y = a + be^{-x} + ce^{-2x}\)
• B. \(y = a + be^x + ce^{2x}\)
• C. \(y = ae^{-x} + be^{-2x} + ce^x\)
• D. \(y = a + be^{-2x} + ce^{-3x}\)

Hint:

For homogeneous linear DEs with constant coefficients, the process is always the same: form the auxiliary equation, find its roots, and write the solution based on the type of roots. A root of \(m=0\) always contributes a simple constant term to the solution.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the general solution for a third-order homogeneous linear differential equation with constant coefficients.

Step 2: Key Formula or Approach:
We solve this by finding the roots of the auxiliary (or characteristic) equation.
1. Form the auxiliary equation by replacing \(\frac{d^ny}{dx^n}\) with \(m^n\).
2. Find the roots of the resulting polynomial equation.
3. The form of the general solution depends on the nature of these roots (real and distinct, real and repeated, or complex).

Step 3: Detailed Explanation:
The given differential equation is \(y''' + 3y'' + 2y' = 0\).
1. The auxiliary equation is:
\[ m^3 + 3m^2 + 2m = 0 \]
2. Factor the polynomial to find the roots:
\[ m(m^2 + 3m + 2) = 0 \]
Factor the quadratic part:
\[ m(m+1)(m+2) = 0 \]
The roots are \(m_1 = 0\), \(m_2 = -1\), and \(m_3 = -2\).
3. Since we have three distinct real roots, the general solution is of the form:
\[ y = c_1 e^{m_1 x} + c_2 e^{m_2 x} + c_3 e^{m_3 x} \]
Substituting the roots we found:
\[ y = c_1 e^{0x} + c_2 e^{-1x} + c_3 e^{-2x} \]
Since \(e^{0x} = 1\), the solution is:
\[ y = c_1(1) + c_2 e^{-x} + c_3 e^{-2x} \]
Using the arbitrary constants \(a, b, c\) from the options, this becomes:
\[ y = a + be^{-x} + ce^{-2x} \]

Step 4: Final Answer:
The general solution is \(y = a + be^{-x} + ce^{-2x}\).