Second Order Linear

Question

Solve the repeated-root equation \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0 and state the general solution that accounts for the double root of its auxiliary equation.

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For constant-coefficient linear equations the auxiliary equation determines the solution structure, and a repeated root requires a special modified form. Substituting y = e^{mx} converts \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 0 into m^2 - 6m + 9 = 0, which factors as (m-3)^2 = 0, giving a double root m = 3. The principle here is that when the auxiliary equation has a repeated root m, the two independent solutions are e^{mx} and x e^{mx}, because a single exponential cannot supply two independent solutions for a second-order equation. Therefore the general solution is y = (C_1 + C_2 x)e^{3x}. Option y = C_1 e^{3x} + C_2 e^{-3x} incorrectly assumes distinct roots 3 and -3. Option y = C_1 e^{3x} + C_2 e^{2x} invents a second root not present. Option (C_1 + C_2 x)e^{-3x} uses the wrong sign for the repeated root. This is the standard JEE Advanced repeated-root case. As a final consistency check, the discriminant 36 - 36 = 0 confirms a repeated root, validating the inclusion of the x e^{3x} term needed for a complete two-parameter solution family.

Second Order Linear

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