Second-order homogeneous equations
Objective. Solve auxiliary equations with distinct and repeated roots.
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- 2. Worked example
- 3. Practice
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Learn: the key idea
A second-order homogeneous linear differential equation with constant coefficients, a(d²y/dx²) + b(dy/dx) + cy = 0, is solved by forming the auxiliary equation am² + bm + c = 0 and finding its roots. Two distinct real roots m₁, m₂ give the general solution y = Ae^(m₁x) + Be^(m₂x); a repeated root m gives y = (A+Bx)e^(mx); and complex roots p±qi give y = e^(px)(A cos(qx) + B sin(qx)). For example, y''−5y'+6y=0 has auxiliary equation m²−5m+6=0 with roots 2 and 3, giving y = Ae^(2x)+Be^(3x).
Worked example
Solve the second-order homogeneous equation y'' − 6y' + 9y = 0.
- 1Form the auxiliary equation: m² − 6m + 9 = 0
- 2Factorise: (m−3)² = 0, giving a repeated root m = 3
- 3For a repeated root, the general solution is y = (A+Bx)e^(mx)
- 4Substitute m=3: y = (A+Bx)e^(3x)
Answer: y = (A + Bx)e^(3x)
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Practice
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Start practisingCommon mistake. Students use the distinct-roots form Ae^(m₁x)+Be^(m₂x) even when the auxiliary equation has a repeated root, forgetting the extra factor of x needed: y=(A+Bx)e^(mx). Always check the discriminant of the auxiliary equation first.
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