Question:
The complementary function of $y'' - 3y' + 2y = e^{3x}$ is}
The complementary function of $y'' - 3y' + 2y = e^{3x}$ is}
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Complementary functions depend ONLY on the left-hand side of the equation.
Updated On: Jul 3, 2026
- $y = c_{1}e^{x} + c_{2}e^{3x}$
- $y = c_{1}e^{-x} + c_{2}e^{-3x}$
- $y = c_{1}e^{-x} + c_{2}e^{-2x}$
- $y = c_{1}e^{2x} + c_{2}e^{x}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The complementary function (C.F.) is the general solution of the homogeneous part $y'' - 3y' + 2y = 0$.
Step 2: Meaning
The auxiliary equation is $m^{2} - 3m + 2 = 0$.
Step 3: Analysis
Factoring the equation: $(m - 2)(m - 1) = 0$, which gives roots $m_{1} = 2$ and $m_{2} = 1$.
Step 4: Conclusion
For distinct real roots, $C.F. = c_{1}e^{m_{1}x} + c_{2}e^{m_{2}x} = c_{1}e^{2x} + c_{2}e^{x}$. Final Answer: (D)
The complementary function (C.F.) is the general solution of the homogeneous part $y'' - 3y' + 2y = 0$.
Step 2: Meaning
The auxiliary equation is $m^{2} - 3m + 2 = 0$.
Step 3: Analysis
Factoring the equation: $(m - 2)(m - 1) = 0$, which gives roots $m_{1} = 2$ and $m_{2} = 1$.
Step 4: Conclusion
For distinct real roots, $C.F. = c_{1}e^{m_{1}x} + c_{2}e^{m_{2}x} = c_{1}e^{2x} + c_{2}e^{x}$. Final Answer: (D)
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