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Variation of Parameters and the Wronskianmd a0d6313
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Variation of Parameters and the Wronskian

date2026-07-16document_iddoc_7db9f0ee18d79fdd6073e08a3574c010description定数変化法と Wronskian を、連続係数をもつ二階線型非同次方程式の一般的な特殊解構成法として説明する。prerequisites非同次方程式と未定係数法 / ベクトル空間と基底type講義content_typelecturestatusactiverelateddata/lecture/math/differential-equations/nonhomogeneous-equations-and-undetermined-coefficients.lecture.n.md / data/lecture/math/differential-equations/euler-cauchy-and-higher-order-equations.lecture.n.md / data/lecture/math/linear-algebra/vector-spaces-and-bases.lecture.n.md / data/exercise/math/differential-equations/nonhomogeneous-equations-and-particular-solutions.exercise.n.md
mathdifferential-equationsvariation-of-parameterslecture

1Introduction

This lecture develops variation of parameters, which constructs a particular solution of a nonhomogeneous equation from a fundamental pair of homogeneous solutions. It formulates nonvanishing of the Wronskian as an interval-wide condition and explains the role of constants of integration.

2Hypotheses and a Fundamental Pair

On an interval I, consider

y'+P(x)y+Q(x)y=R(x),

where P,Q,R are continuous on I. Let y1,y2 be a fundamental pair for the corresponding homogeneous equation; that is, they are linearly independent solutions on I.

Define the {Wronskian} by

W(x)=|y1(x)y2(x)y1(x)y2(x)|=y1y2-y1y2.

Abel's identity states that

W(x)=W(x0)exp(-x0xP(s)ds),x,x0I.

Thus, if W(x0)0 at one point of I, then W(x)0 throughout I. Conversely, vanishing at one point implies vanishing throughout the interval. For a homogeneous equation with continuous coefficients, nonvanishing need not be assumed separately at every point.

3Derivation of Variation of Parameters

Set

yp=u1(x)y1(x)+u2(x)y2(x)

and impose

u1y1+u2y2=0.

Then yp=u1y1+u2y2. Differentiating once more and substituting into the original equation leaves

u1y1+u2y2=R.

Therefore,

(y1y2y1y2)(u1u2)=(0R).

Since W0 on I, Cramer's rule gives

u1=-y2RW,u2=y1RW.

4Definite-Integral Formula

Fix a base point x0I. Then

u1(x)=-x0xy2(s)R(s)W(s)ds,u2(x)=x0xy1(s)R(s)W(s)ds,

so a particular solution is

yp(x)=-y1(x)x0xy2(s)R(s)W(s)ds+y2(x)x0xy1(s)R(s)W(s)ds.

This choice satisfies u1(x0)=u2(x0)=0. If indefinite integrals are used instead, their arbitrary constants contribute C1y1+C2y2 and are absorbed into the homogeneous solution. Hence the general solution is

y=C1y1+C2y2+yp.

5Example: y'+y=tanx

Fix any interval on which tanx is continuous,

Ik=(-π2+kπ,π2+kπ),kZ.

The functions y1=cosx and y2=sinx form a fundamental pair on Ik, with W=1. Thus,

u1=-sinxtanx,u2=sinx.

One choice of antiderivatives is

u1=-log|secx+tanx|+sinx,u2=-cosx.

The cross terms cancel in u1y1+u2y2, yielding

yp=-cosxlog|secx+tanx|.

This expression is defined on the selected interval Ik, and direct substitution verifies yp'+yp=tanx. Different integration constants merely add a homogeneous solution.

6Method Selection and Limitations

Undetermined coefficients has lower computational cost when its hypotheses hold. Variation of parameters is less dependent on the form of R, but a fundamental pair must already be known, and the resulting integrals need not be elementary. The method must be applied separately on each interval where P,Q,R are simultaneously continuous.

An analogous formula using a fundamental matrix exists for first-order linear systems. Its details are deferred to the subsequent lectures on systems.

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8Next Lecture

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9Exercises

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