Nonhomogeneous equations practice problems

In this topic you solve equations like y'' + 4y = 8 sin x, where the right side is not zero. You solve the homogeneous version first, then find one particular solution that produces the right side, then add the two parts. Recognition cue: a linear equation with a nonzero function on the right side, such as 6, e^(2x), 5x, or sin 3x. You will practice: undetermined coefficients, modification rule, variation of parameters, superposition.

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

Find a particular solution of y'' + y = 6.

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Answer: $y = 6$

The forcing is a constant, so try a constant y_p = A. Then y'' = 0.

0 + A = 6, y_{p} = 6

Find a particular solution of y'' - y = 6e^(2x).

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Answer: $y = 2 e^{2 x}$

Try y_p = A e^(2x) and substitute.

4 A e^{2 x} - A e^{2 x} = 3 A e^{2 x} = 6 e^{2 x}

A = 2, y_{p} = 2 e^{2 x}

Find a particular solution of y'' + 4y = 20x.

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Answer: $y = 5 x$

The forcing is degree 1, so try y_p = Ax + B.

0 + 4 (A x + B) = 20 x, A = 5, B = 0

Find a particular solution of y'' + 5y = 8 sin x.

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Answer: $y = 2 sin x$

Try y_p = A sin x + B cos x. Substitution gives 4A sin x + 4B cos x.

4 A = 8, 4 B = 0, y_{p} = 2 sin x

Find the general solution of y'' - 3y' + 2y = 4x.

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

The characteristic roots are 1 and 2. Then try y_p = Ax + B.

r^{2} - 3 r + 2 = (r - 1) (r - 2)

2 A x - 3 A + 2 B = 4 x, A = 2, B = 3

A cup of coffee cools according to T' + 2T = 40, with t in hours and T in degrees Celsius. T(0) = 90. Find T(t).

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Answer: $T = 20 + 70 e^{- 2 t}$

The constant particular solution is T_p = 20. The homogeneous solution is Ce^(-2t).

2 T_{p} = 40, T_{p} = 20

T (0) = C + 20 = 90, C = 70

In an RL circuit the current satisfies I' + 4I = 12 with I(0) = 0. Find I(t).

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Answer: $I = 3 - 3 e^{- 4 t}$

The constant particular solution is I_p = 3. Add the homogeneous part and use I(0) = 0.

4 I_{p} = 12, I_{p} = 3

I (0) = C + 3 = 0, C = - 3

Find a particular solution of y'' + y = 4e^x + 3.

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

Split the forcing into 4e^x and 3, solve each, then add.

A e^{x} + A e^{x} = 4 e^{x}, A = 2

0 + B = 3, y_{p} = 2 e^{x} + 3

Find a particular solution of y'' - 9y = 12e^(3x).

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

e^(3x) solves the homogeneous equation, so multiply the trial by x.

y_{p} = A x e^{3 x}, (A x e^{3 x})^{''} - 9 A x e^{3 x} = 6 A e^{3 x}

6 A = 12, A = 2

10.

A mass on a spring satisfies x'' + 4x = 8 with x(0) = 0 and x'(0) = 0. Find x(t).

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Answer: $x = 2 - 2 cos (2 t)$

The particular solution is the constant 2. The homogeneous part uses cos 2t and sin 2t.

x = C_{1} cos (2 t) + C_{2} sin (2 t) + 2

x (0) = C_{1} + 2 = 0, x^{'} (0) = 2 C_{2} = 0

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Nonhomogeneous equations practice problems

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