Question

The solution of the differential equation \(\frac{d^3y}{dx^3} - 5\frac{d^2y}{dx^2} + 8\frac{dy}{dx} - 4y = 0\) is

• A. y \(=\) c\(_1\)e\(^x\) \(+\) (c\(_2\) \(+\) c\(_3\)x)e\(^{2x}\)
• B. y \(=\) c\(_1\)e\(^{2x}\) \(+\) (c\(_2\) \(+\) c\(_3\)x)e\(^x\)
• C. y \(=\) (c\(_1\) \(+\) c\(_2\)x \(+\) c\(_3\)x\(^2\))e\(^x\)
• D. y \(=\) (c\(_1\) \(+\) c\(_2\)x \(+\) c\(_3\)x\(^2\))e\(^{2x}\)

Hint:

Standard Auxiliary Root Rules:
- Real, distinct roots \( m_1, m_2 \rightarrow c_1 e^{m_1 x} + c_2 e^{m_2 x} \).
- Real, repeated roots \( m_1 = m_2 = m \rightarrow (c_1 + c_2 x)e^{mx} \).
- Complex conjugate roots \( \alpha \pm i\beta \rightarrow e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x) \).

Answer 1

The Correct Option is A

Step 1: Use Vieta's relations instead of testing roots one by one.
For \( m^3 - 5m^2 + 8m - 4 = 0 \), the sum of the three roots is 5, the sum of products taken two at a time is 8, and the product of all three roots is 4.

Step 2: Guess a repeated-root pattern and check it against these relations.
Suppose the roots are \( r, r, s \) (a repeated root \( r \) and a single root \( s \)). Trying \( r = 2, s = 1 \): the sum gives \( 2+2+1=5 \), the pairwise sum gives \( 4+2+2=8 \), and the product gives \( 4 \times 1 = 4 \), and all three match.

Step 3: Write the solution from these roots.
So the roots are \( m = 2 \) (twice) and \( m = 1 \) (once), giving the general solution:
\[ y = c_1 e^{x} + (c_2 + c_3 x)e^{2x} \]

Step 4: Final Answer.
\[ \boxed{y = c_1 e^{x} + (c_2 + c_3 x)e^{2x}} \]
Therefore, option (A) is correct.