on inhomogeneous second-order linear ODE with constant coefficients

Let’s consider solving the ordinary second-order lineardifferential equation

${\displaystyle \frac{d^{2}y}{d{x}^2}}+a{\displaystyle \frac{dy}{dx}}+by=R(x)$(1)

which isinhomogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation), i.e.$R(x)\not\equiv 0$.

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

${\displaystyle \frac{d^{2}y}{d{x}^2}}+a{\displaystyle \frac{dy}{dx}}+by=\mathrm{\hspace{0.33em}0}$(2)

someparticular solution (http://planetmath.org/SolutionsOfOrdinaryDifferentialEquation)of the inhomogeneous equation (1).  A latter one canalways be gotten by means of the variation of parameters, butin many cases there exist simpler ways to find a particularsolution of (1).

$1^{\circ }$:  $R(x)$ is a nonzero constant function$x\mapsto c$.  In this case, apparently  $y=\frac{c}{b}$ isa solution of (1), supposing that  $b\ne 0$.  If  $b=0$ but $a\ne 0$,  a particular solution is $y=\frac{c}{a}x$. If  $a=b=0$,  a solution is gotten via two consecutiveintegrations.

$2^{\circ }$:  $R(x)$ is a polynomial function of degree$n\ge 1$.  Now (1) has as solution a polynomial which can befound by using indetermined coefficients.  If  $b\ne 0$, the polynomial is of degree $n$ and is uniquely determined. If  $b=0$  and  $a\ne 0$,  the degree of the polynomialis $n+1$ and its constant term is arbitrary. If $a=b=0$  the polynomial is of degree $n+2$ and isgotten via two integrations.

$3^{\circ }$:  Let $R(x)$ in (1) be of the form$\alpha \sin nx+\beta \cos nx$ with $\alpha$, $\beta$, $n$constants.  We try to find a solution of the same form and putinto (1) the expression

$y:=A\sin nx+B\cos nx.$(3)

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

$$[(b-n^2)A-anB]\sin nx+[anA+(b-n^2)B]\cos nx.$$

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

$$(b-n^2)A-anB=\alpha \mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}anA+(b-n^2)B=\beta .$$

These determine uniquely the values of $A$ and $B$ provided thatthe determinant

does not vanish.  Then we obtain the particular solution (3). The determinant vanishes only if  $a=0$ and  $b=n^2$, inwhich case the differential equation (1) reads

${\displaystyle \frac{d^{2}y}{d{x}^2}}+n^{2}y=\alpha \sin nx+\beta \cos nx.$(4)

Unless we have  $\alpha =\beta =0$, the equation (4) has nosolution of the form (3), since

${\displaystyle \frac{d^2}{d{x}^2}}(A\sin nx+B\cos nx)+n^2(A\sin nx+B\cos nx)=\mathrm{\hspace{0.33em}0}$(5)

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

$$\frac{d^2}{d{x}^2}(Ax\cos nx-Bx\sin nx)+n^2(Ax\cos nx-Bx\sin nx)=-2nA\sin nx-2nB\cos nx.$$

The right hand side coincides with the right hand side of(4) iff  $-2nA=\alpha$  and  $-2nB=\beta$, and thus (4)has the solution

$$y:=-\frac{\alpha }{2n}x\cos nx+\frac{\beta }{2n}x\sin nx.$$

$4^{\circ }$:  Let $R(x)$ in (1) now be$\alpha e^{kx}$ where $\alpha$ and $k$ are constants.  Denotethe left hand side of (1) briefly$\frac{d^{2}y}{d{x}^2}+a\frac{dy}{dx}+by=:F(y)$.  We seekagain a solution of the same form $A{e}^{kx}$ as $R(x)$.

First we have

$$F(A{e}^{kx})=A\underset{f(k)}{\underbrace{(k^2+ak+b)}}{e}^{kx}=Af(k)e^{kx}.$$

Thus $A$ can be determined from the condition  $Af(k)=\alpha$.  If  $f(k)\ne 0$,  i.e. $k$ is not a root of thecharacteristic equation  $f(r)=0$  corresponding thehomogeneous equation (2), then we obtain theparticular solution

$$y:=\frac{\alpha }{f(k)}{e}^{kx}=\frac{\alpha }{k^2+ak+b}{e}^{kx}$$

of the inhomogeneous equation (1).

If  $f(k)=0$, then $e^{kx}$ and $A{e}^{kx}$ satisfy thehomogeneous equation  $F(y)=0$.  Now we may start from theidentity

$$F(A{e}^{rx})=Af(r)e^{rx}$$

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

$F(Ax{e}^{rx})=A{e}^{rx}[f^{\prime }(r)+xf(r)],$(6)

and differentiating anew,

$F(A{x}^2{e}^{rx})=A{e}^{rx}[f^{\prime \prime }(r)+2x{f}^{\prime }(r)+x^{2}f(r)].$(7)

If $k$ is a simple root of the equation  $f(r)=0$,  i.e.if  $f(k)=0$  but  $f^{\prime }(k)\ne 0$,  then  $r:=k$ makes the right hand side of (6) to $A{f}^{\prime }(k)e^{kx}$, whichequals to  $R(x)=\alpha e^{kx}$  by choosing $A:=\frac{\alpha }{f^{\prime }(k)}$.  Then we have found theparticular solution

$$y:=\frac{\alpha }{f^{\prime }(k)}x{e}^{kx}=\frac{\alpha }{2k+a}x{e}^{kx}.$$

We have still to handle the case when $k$ is the double root ofthe equation  $f(k)=0$  and thus  $f^{\prime }(k)=0$.  Putting $r:=k$  into (7), the right hand side reduces to $A{f}^{\prime \prime }(k)e^{kx}=2A{e}^{kx}$; this equals to $R(x)=\alpha e^{kx}$ when choosing $A:=\frac{\alpha }{2}$.  So we have theparticular solution

$$y:=\frac{\alpha }{2}{x}^2{e}^{kx}$$

of the given inhomogeneous equation.

$5^{\circ }$:  Suppose that in (1) the right hand side$R(x)$ is a sum of several functions,

${\displaystyle \frac{d^{2}y}{d{x}^2}}+a{\displaystyle \frac{dy}{dx}}+by=R_1(x)+R_2(x)+\mathrm{\dots }+R_n(x),$(8)

and one can find a particular solution $y_i(x)$ for each ofthe equations

$$\frac{d^{2}y}{d{x}^2}+a\frac{dy}{dx}+by=R_i(x).$$

Then evidently the sum $y_1(x)+y_2(x)+\mathrm{\dots }+y_n(x)$ is aparticular solution of the equation (8).