on inhomogeneous second-order linear ODE with constant coefficients


Let’s consider solving the ordinary second-order linear differential equation

d2ydx2+adydx+by=R(x) (1)

which is inhomogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation), i.e. R(x)0.

For obtaining the general solution of (1) we have to add to the general solution of the corresponding homogeneous equation (http://planetmath.org/SecondOrderLinearODEWithConstantCoefficients)

d2ydx2+adydx+by= 0 (2)

some particular solution (http://planetmath.org/SolutionsOfOrdinaryDifferentialEquation) of the inhomogeneous equation (1).  A latter one can always be gotten by means of the variation of parametersMathworldPlanetmath, but in many cases there exist simpler ways to find a particular solution of (1).

1:  R(x) is a nonzero constant function xc.  In this case, apparently  y=cb is a solution of (1), supposing that  b0.  If  b=0  but a0,  a particular solution is y=cax.  If  a=b=0,  a solution is gotten via two consecutive integrations.

2:  R(x) is a polynomial function of degree n1.  Now (1) has as solution a polynomialPlanetmathPlanetmath which can be found by using indetermined coefficients.  If  b0,  the polynomial is of degree n and is uniquely determined.  If  b=0  and  a0,  the degree of the polynomial is n+1 and its constant term is arbitrary. If  a=b=0  the polynomial is of degree n+2 and is gotten via two integrations.

3:  Let R(x) in (1) be of the form αsinnx+βcosnx with α, β, n constants.  We try to find a solution of the same form and put into (1) the expression

y:=Asinnx+Bcosnx. (3)

Then the left hand side of (1) attains the form

[(b-n2)A-anB]sinnx+[anA+(b-n2)B]cosnx.

This must equal R(x), i.e. we have the conditions

(b-n2)A-anB=αandanA+(b-n2)B=β.

These determine uniquely the values of A and B provided that the determinantMathworldPlanetmath

|b-n2-ananb-n2|=a2n2+(b-n2)2

does not vanish.  Then we obtain the particular solution (3).  The determinant vanishes only if  a=0 and  b=n2, in which case the differential equation (1) reads

d2ydx2+n2y=αsinnx+βcosnx. (4)

Unless we have  α=β=0, the equation (4) has no solution of the form (3), since

d2dx2(Asinnx+Bcosnx)+n2(Asinnx+Bcosnx)= 0 (5)

identically.  But we find easily a solution of (4) when we differentiate the identity (5) with respect to n.  Changing the order of differentiations we get

d2dx2(Axcosnx-Bxsinnx)+n2(Axcosnx-Bxsinnx)=-2nAsinnx-2nBcosnx.

The right hand side coincides with the right hand side of (4) iff  -2nA=α  and  -2nB=β, and thus (4) has the solution

y:=-α2nxcosnx+β2nxsinnx.

4:  Let R(x) in (1) now be αekx where α and k are constants.  Denote the left hand side of (1) briefly d2ydx2+adydx+by=:F(y).  We seek again a solution of the same form Aekx as R(x).

First we have

F(Aekx)=A(k2+ak+b)f(k)ekx=Af(k)ekx.

Thus A can be determined from the condition  Af(k)=α.  If  f(k)0,  i.e. k is not a root of the characteristic equationMathworldPlanetmathPlanetmathPlanetmathf(r)=0  corresponding the homogeneous equation (2), then we obtain the particular solution

y:=αf(k)ekx=αk2+ak+bekx

of the inhomogeneous equation (1).

If  f(k)=0, then ekx and Aekx satisfy the homogeneous equation  F(y)=0.  Now we may start from the identity

F(Aerx)=Af(r)erx

and differentiate it with respect to r.  Changing again the order of differentiations we can write first

F(Axerx)=Aerx[f(r)+xf(r)], (6)

and differentiating anew,

F(Ax2erx)=Aerx[f′′(r)+2xf(r)+x2f(r)]. (7)

If k is a simple root of the equation  f(r)=0,  i.e. if  f(k)=0  but  f(k)0,  then  r:=k  makes the right hand side of (6) to Af(k)ekx, which equals to  R(x)=αekx  by choosing  A:=αf(k).  Then we have found the particular solution

y:=αf(k)xekx=α2k+axekx.

We have still to handle the case when k is the double root of the equation  f(k)=0  and thus  f(k)=0.  Putting  r:=k  into (7), the right hand side reduces to  Af′′(k)ekx=2Aekx; this equals to R(x)=αekx  when choosing A:=α2.  So we have the particular solution

y:=α2x2ekx

of the given inhomogeneous equation.

5:  Suppose that in (1) the right hand side R(x) is a sum of several functions,

d2ydx2+adydx+by=R1(x)+R2(x)++Rn(x), (8)

and one can find a particular solution yi(x) for each of the equations

d2ydx2+adydx+by=Ri(x).

Then evidently the sum y1(x)+y2(x)++yn(x) is a particular solution of the equation (8).

References

  • 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset III.1.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

Title on inhomogeneous second-order linear ODE with constant coefficients
Canonical name OnInhomogeneousSecondorderLinearODEWithConstantCoefficients
Date of creation 2014-03-05 16:25:57
Last modified on 2014-03-05 16:25:57
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 14
Author pahio (2872)
Entry type Derivation
Classification msc 34A05