Intuition
Undetermined coefficients needs luck: constant coefficients and a forcing of a few shapes. Variation of parameters needs none. Take the general solution of the homogeneous equation and let its two constants vary, as functions. One condition on them is free to choose, and choosing it well leaves two linear equations for their derivatives, whose determinant is the Wronskian.
Steering between two landmarks. The homogeneous solutions are fixed reference directions; the method asks how much of each is needed at every moment to follow the forced path, and the answer always exists because the Wronskian never vanishes.
Let the constants vary
For with a fundamental pair , and Wronskian , a particular solution is , where and solve and . By Cramer's rule they are the formulas below. The equation must be in standard form, with the coefficient of equal to one, before is read off.
Using it
- on : , and , so .
The two conditions give a particular solution
Differentiate the trial solution. The terms containing the derivatives of the u are removed by the free condition, so the first derivative looks as though the u were constants. Differentiate again: now the derivatives of the u appear once, in exactly the combination the second condition fixes. Substituting into L, the parts where the u behave as constants give u1 times L of y1 plus u2 times L of y2, which vanish, and what remains is the second condition, which is g.
Proof steps
Let the constants of the homogeneous solution vary.
Differentiate, and use the free condition to drop the terms in u prime.
Differentiate again: the derivatives of the u appear once.
Substitute and group the terms.
The homogeneous solutions give zero and the second condition gives the forcing.
Applications
Practice
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Which equation needs variation of parameters rather than undetermined coefficients?
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To use the formulas for , the forcing must first be divided by two.
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For with and , what is ?
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For with and , what is the Wronskian in the denominators?
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Why impose the condition ?
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Constants of integration in and can be dropped when finding a particular solution.
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For with , and , what is ?
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What particular solution does variation of parameters give for on ?
Final checkpoint
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Variation of parameters works for equations with variable coefficients, provided a fundamental pair of homogeneous solutions is known.
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What is the determinant of the system , ?
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For , both methods give particular solutions containing . What is ?
Completion
Lesson complete
Great work! You now know how to:
- set up variation of parameters from a fundamental pair
- derive the two conditions and solve them with the Wronskian
- use it where undetermined coefficients cannot go
- check that the two methods agree where both apply