Question

A complete solution of \( y'' + a_1 y' + a_2 y = 0 \) is \( y = b_1 e^{-x} + b_2 e^{-3x} \), where \( a_1, a_2, b_1 \) and \( b_2 \) are constants, then the respective values of \( a_1 \) and \( a_2 \) are:

• A. 3, 3
• B. 3, 4
• C. 4, 3
• D. 4, 4

Hint:

Remember Vieta's formulas for a quadratic equation \( ax^2+bx+c=0 \). The sum of roots is \( -b/a \) and the product of roots is \( c/a \). For the characteristic equation \( m^2+a_1m+a_2=0 \), the sum of roots is \( -a_1 \) and the product is \( a_2 \). This provides a direct way to find the coefficients from the roots.

Answer 1

The Correct Option is C

Step 1: Problem Overview:
This problem links the solution of a second-order linear homogeneous differential equation with constant coefficients to the equation's coefficients. The solution's form reveals the characteristic equation's roots.

Step 2: Core Concepts:
The differential equation is \( y'' + a_1 y' + a_2 y = 0 \). Its corresponding characteristic equation is:\[ m^2 + a_1 m + a_2 = 0 \]The general solution is \( y = b_1 e^{-x} + b_2 e^{-3x} \). This solution form arises when the characteristic equation has two distinct real roots, which are the coefficients of \(x\) in the exponents: \( m_1 = -1 \) and \( m_2 = -3 \).

Step 3: Solution Breakdown:
Given the roots \( -1 \) and \( -3 \) of the characteristic equation \( m^2 + a_1 m + a_2 = 0 \), we can derive the equation. A quadratic equation with roots \( r_1 \) and \( r_2 \) can be expressed as \( (m - r_1)(m - r_2) = 0 \), which expands to \( m^2 - (r_1+r_2)m + r_1r_2 = 0 \).Here, \( r_1 = -1 \) and \( r_2 = -3 \).- Sum of roots: \( r_1 + r_2 = -1 + (-3) = -4 \)- Product of roots: \( r_1 r_2 = (-1)(-3) = 3 \)The characteristic equation is:\[ m^2 - (-4)m + (3) = 0 \]\[ m^2 + 4m + 3 = 0 \]Comparing this with \( m^2 + a_1 m + a_2 = 0 \), we find:- \( a_1 = 4 \)- \( a_2 = 3 \)
Step 4: Answer:
The values of \( a_1 \) and \( a_2 \) are 4 and 3.