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# on inhomogeneous second-order linear ODE with constant coefficients

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

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

which is inhomogeneous, i.e. $R(x)\not\equiv 0$.

For obtaining the general solution of (1) we have to add to the general solution of the corresponding homogeneous equation

$\displaystyle\frac{d^{2}y}{dx^{2}}+a\frac{dy}{dx}+by\;=\;0$ | (2) |

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

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

$2^{\circ}$: $R(x)$ is a polynomial function of degree
$n\geq 1$. Now (1) has as solution a polynomial^{} which can be
found by using indetermined coefficients. If $b\neq 0$,
the polynomial is of degree $n$ and is uniquely determined.
If $b=0$ and $a\neq 0$, 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.

$3^{\circ}$: 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 put into (1) the expression

$\displaystyle y\;:=\;A\sin{nx}+B\cos{nx}.$ | (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

$(b-n^{2})A-anB\;=\;\alpha\quad\mbox{and}\quad anA+(b-n^{2})B\;=\;\beta.$ |

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

$\left|\begin{matrix}b\!-\!n^{2}&-an\\ an&b\!-\!n^{2}\end{matrix}\right|\;=\;a^{2}n^{2}\!+\!(b\!-\!n^{2})^{2}$ |

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

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

Unless we have $\alpha=\beta=0$, the equation (4) has no solution of the form (3), since

$\displaystyle\frac{d^{2}}{dx^{2}}(A\sin{nx}+B\cos{nx})+n^{2}(A\sin{nx}+B\cos{% nx})\;=\;0$ | (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

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

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. Denote the left hand side of (1) briefly $\frac{d^{2}y}{dx^{2}}+a\frac{dy}{dx}+by\;=:\;F(y)$. We seek again a solution of the same form $Ae^{{kx}}$ as $R(x)$.

First we have

$F(Ae^{{kx}})\;=\;A\underbrace{(k^{2}\!+\!ak\!+\!b)}_{{f(k)}}e^{{kx}}\;=\;Af(k)% e^{{kx}}.$ |

Thus $A$ can be determined from the condition $Af(k)=\alpha$. If $f(k)\neq 0$, i.e. $k$ is not a root of the
characteristic equation^{} $f(r)=0$ corresponding the
homogeneous equation (2), then we obtain the
particular 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 $Ae^{{kx}}$ satisfy the homogeneous equation $F(y)=0$. Now we may start from the identity

$F(Ae^{{rx}})\;=\;Af(r)e^{{rx}}$ |

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

$\displaystyle F(Axe^{{rx}})\;=\;Ae^{{rx}}[f^{{\prime}}(r)\!+\!xf(r)],$ | (6) |

and differentiating anew,

$\displaystyle F(Ax^{2}e^{{rx}})\;=\;Ae^{{rx}}[f^{{\prime\prime}}(r)\!+\!2xf^{{% \prime}}(r)\!+\!x^{2}f(r)].$ | (7) |

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

$y\;:=\;\frac{\alpha}{f^{{\prime}}(k)}xe^{{kx}}\;=\;\frac{\alpha}{2k\!+\!a}xe^{% {kx}}.$ |

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

$\displaystyle\frac{d^{2}y}{dx^{2}}+a\frac{dy}{dx}+by\;=\;R_{1}(x)+R_{2}(x)+% \ldots+R_{n}(x),$ | (8) |

and one can find a particular solution $y_{i}(x)$ for each of the equations

$\frac{d^{2}y}{dx^{2}}+a\frac{dy}{dx}+by\;=\;R_{i}(x).$ |

Then evidently the sum $y_{1}(x)+y_{2}(x)+\ldots+y_{n}(x)$ 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).

## Mathematics Subject Classification

34A05*no label found*

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