# Shortcuts: Solution by Inspection for Two Special Cases A Lecture in ENGIANA AY 2014 – 2015.

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Shortcuts: Solution by Inspection for Two Special Cases A Lecture in ENGIANA AY 2014 – 2015

Solution by Inspection It is frequently easy to obtain a particular solution of a non-homogeneous equation by inspection.

Special Case 1: R(x) = R 0, b n  0 For example, if R(x) is a constant R 0 and if b n  0, is a solution of because all derivatives of y p are zero.

Special Case 2: R(x) = R 0, b n = 0 Suppose that b n = 0. Let D k y be the lowest-ordered derivative that actually appears in the differential equation. Then the equation may be written

Special Case 2: R(x) = R 0, b n = 0 Now D k x k = k! is a constant, so that all higher derivatives of x k are zero. Thus, it becomes evident that is a solution of

Examples: Find a particular solution by inspection. 1)(D 2 + 4)y = 12 2)(D 2 + 4D + 4)y = 8 3)(D 3 – 3D + 2)y = -7 4)(D 2 + 4D)y = 12 5)(D 3 + 5D)y = 15 6)(D 4 – 4D 2 )y = 24 7)(D 5 – D 3 )y = 24

Solution to finding a particular solution by inspection. 1)(D 2 + 4)y = 12y p = 12/4 = 3 2)(D 2 + 4D + 4)y = 8y p = 8/4 = 2 3)(D 3 – 3D + 2)y = -7y p = -7/2 4)(D 2 + 4D)y = 12

Examples: Find a particular solution by inspection. 5)(D 3 + 5D)y = 15 6)(D 4 – 4D 2 )y = 24 7)(D 5 – D 3 )y = 24

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