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Example 2. Solve the differential equation (5.19) y'' - 6y' + 9y = 0. We can write the equation as (5.20) $(D^2 - 6D + 9)y = 0$ or $(D - 3)(D - 3)y = 0$. Since the roots of the auxiliary equation are equal, we know that the solution is of the form (5.15) and we simply write the result (5.21) y = (Ax + B)$e^{3x}$.

          Example 2. Solve the differential equation
(5.19)
y'' - 6y' + 9y = 0.
We can write the equation as
(5.20)
$(D^2 - 6D + 9)y = 0$ or $(D - 3)(D - 3)y = 0$.
Since the roots of the auxiliary equation are equal, we know that the solution is of
the form (5.15) and we simply write the result
(5.21)
y = (Ax + B)$e^{3x}$.

Example 2. Solve the differential equation
(5.19)
y” - 6y' + 9y = 0.
We can write the equation as
(5.20)
(D^2 - 6D + 9)y = 0 or (D - 3)(D - 3)y = 0.
Since the roots of the auxiliary equation are equal, we know that the solution is of
the form (5.15) and we simply write the result
(5.21)
y = (Ax + B)e^3x.

Added by Kevin F.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Arjun Singh

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Step 1: Start with the given differential equation: y - 6y + 9y = 0 Show more…

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I have this solved example, but I need more details. Step by step, please. Book: Mathematical Methods in the Physical Sciences, p. 411. Example 2. Solve the differential equation: (5.19) y - 6y + 9y = 0 We can write the equation as: (5.20) D^2 - 6D + 9y = 0 or D^2 - 3D - 3y = 0 Since the roots of the auxiliary equation are equal, we know that the solution is of the form 5.15, and we simply write the result: (5.21) y = Ax + Be^(3x)