Second Order Linear
Find the general solution of the second-order constant-coefficient equation \frac{d^2y}{dx^2} - 5\frac{dy}{dx} + 6y = 0 using its auxiliary characteristic equation.
Select the correct option:
Solution
y=C1e2x+C2e3x
A linear homogeneous equation with constant coefficients is solved through its auxiliary equation, obtained by substituting the trial solution y = e^{mx}, which converts derivatives into powers of m. The justification is that exponential functions reproduce themselves under differentiation, so the differential equation becomes a polynomial equation in m. Replacing \frac{d^2y}{dx^2} with m^2, \frac{dy}{dx} with m, and y with 1 gives the characteristic equation m^2 - 5m + 6 = 0. Factoring yields (m-2)(m-3) = 0, so the roots are m = 2 and m = 3, two distinct real values. Distinct real roots give a general solution that is a linear combination of the corresponding exponentials: y = C_1 e^{2x} + C_2 e^{3x}. Option y = C_1 e^{-2x} + C_2 e^{-3x} would arise from roots -2 and -3, contradicting the middle coefficient sign. Option (C_1 + C_2 x)e^{2x} corresponds to a repeated root, which does not occur here. Option with cosine and sine requires complex roots, which the real discriminant rules out. This is the canonical JEE Advanced constant-coefficient method. As a final check, the discriminant 25 - 24 = 1 is positive, confirming two distinct real roots and validating the exponential form.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- second order linear
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
y=C1e2x+C2e3x
A linear homogeneous equation with constant coefficients is solved through its auxiliary equation, obtained by substituting the trial solution y = e^{mx}, which converts derivatives into powers of m. The justification is that exponential functions reproduce themselves under differentiation, so the differential equation becomes a polynomial equation in m. Replacing \frac{d^2y}{dx^2} with m^2, \frac{dy}{dx} with m, and y with 1 gives the characteristic equation m^2 - 5m + 6 = 0. Factoring yields (m-2)(m-3) = 0, so the roots are m = 2 and m = 3, two distinct real values. Distinct real roots give a general solution that is a linear combination of the corresponding exponentials: y = C_1 e^{2x} + C_2 e^{3x}. Option y = C_1 e^{-2x} + C_2 e^{-3x} would arise from roots -2 and -3, contradicting the middle coefficient sign. Option (C_1 + C_2 x)e^{2x} corresponds to a repeated root, which does not occur here. Option with cosine and sine requires complex roots, which the real discriminant rules out. This is the canonical JEE Advanced constant-coefficient method. As a final check, the discriminant 25 - 24 = 1 is positive, confirming two distinct real roots and validating the exponential form.
This medium difficulty mathematics question is from the chapter differential equations, covering the topic of second order linear. It appeared in the 2025 exam.
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