Second-order homogeneous practice problems

You solve equations of the form ay'' + by' + cy = 0. You turn the equation into a quadratic, called the characteristic equation. You solve the quadratic. The roots tell you the answer: exponentials, sines and cosines, or an extra factor x. Recognition cue: the equation contains y'' with constant number coefficients, and the right side equals 0. You will practice: Distinct real roots, Repeated root, Complex roots, Initial values and the Wronskian.

11 problems, each with a worked answer. Work them on paper, in order — the difficulty ramps gently.

Write the characteristic equation of y'' - 3y' + 2y = 0. Use λ as the variable.

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Answer: $λ^{2} - 3 λ + 2 = 0$

Replace y'' with λ², y' with λ, and y with 1.

λ^{2} - 3 λ + 2 = 0

Find both roots of the characteristic equation of y'' - y = 0.

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Answer: $λ = 1, λ = - 1$

Write the characteristic equation and factor it.

λ^{2} - 1 = 0, (λ - 1) (λ + 1) = 0, λ = 1, λ = - 1

Find the general solution of y'' - 4y = 0.

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Answer: $y = C_{1} e^{2 x} + C_{2} e^{- 2 x}$

The characteristic equation has two different real roots.

λ^{2} - 4 = 0, λ = 2, λ = - 2

y = C_{1} e^{2 x} + C_{2} e^{- 2 x}

Find the general solution of y'' - 5y' + 6y = 0.

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Answer: $y = C_{1} e^{2 x} + C_{2} e^{3 x}$

Factor the characteristic equation.

λ^{2} - 5 λ + 6 = 0, (λ - 2) (λ - 3) = 0

y = C_{1} e^{2 x} + C_{2} e^{3 x}

Find the general solution of y'' + y' = 0.

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Answer: $y = C_{1} + C_{2} e^{- x}$

One root is zero, so one solution is the constant e^(0x) = 1.

λ^{2} + λ = 0, λ = 0, λ = - 1

y = C_{1} + C_{2} e^{- x}

Find the general solution of y'' + 9y = 0.

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Answer: $y = C_{1} cos 3 x + C_{2} sin 3 x$

The roots are pure imaginary, so α = 0 and β = 3.

λ^{2} + 9 = 0, λ = 3 i, λ = - 3 i

y = C_{1} cos 3 x + C_{2} sin 3 x

Find the general solution of y'' - 6y' + 9y = 0.

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Answer: $y = C_{1} e^{3 x} + C_{2} x e^{3 x}$

The quadratic is a perfect square, so the root repeats.

λ^{2} - 6 λ + 9 = 0, (λ - 3)^{2} = 0

y = C_{1} e^{3 x} + C_{2} x e^{3 x}

Compute the Wronskian W(x) of y₁ = e^(2x) and y₂ = e^(3x).

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Answer: $W = e^{5 x}$

Apply the Wronskian formula.

W = e^{2 x} \cdot 3 e^{3 x} - 2 e^{2 x} \cdot e^{3 x} = 3 e^{5 x} - 2 e^{5 x} = e^{5 x}

Find the general solution of y'' + 4y' + 5y = 0.

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Answer: $y = e^{- 2 x} (C_{1} cos x + C_{2} sin x)$

Use the quadratic formula. The discriminant is 16 - 20 = -4.

λ^{2} + 4 λ + 5 = 0, λ = - 2 + i, λ = - 2 - i

y = e^{- 2 x} (C_{1} cos x + C_{2} sin x)

10.

Solve the initial value problem y'' - y = 0 with y(0) = 3 and y'(0) = 3.

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Answer: $y = 3 e^{x}$

Find the general solution, then use both conditions.

y = C_{1} e^{x} + C_{2} e^{- x}, C_{1} + C_{2} = 3, C_{1} - C_{2} = 3

C_{1} = 3, C_{2} = 0, y = 3 e^{x}

11.

A damped spring-mass system satisfies u″ + 6u′ + 25u = 0. Is the motion overdamped, critically damped, or underdamped?

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Answer: $underdamped$

Check the sign of the discriminant of the characteristic equation r² + 6r + 25 = 0.

6^{2} - 4 \cdot 25 = 36 - 100 = - 64

The discriminant is negative, so the roots are complex: decaying oscillation, underdamped.

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Second-order homogeneous practice problems

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