second-order linear ODE with constant coefficients

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

${\displaystyle \frac{d^{2}y}{d{x}^2}}+a{\displaystyle \frac{dy}{dx}}+by=\mathrm{\hspace{0.33em}0}$(1)

which ishomogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation)and the coefficients $a,b$ of which are constants.  Asmentionned in the entry“finding another particular solution of linear ODE”, a simple substitutionmakes possible to eliminate from it the addend containing firstderivative of the unknown function.  Therefore weconcentrate upon the case  $a=0$.  We have two casesdepending on the sign of  $b=\pm k^2$.

$1^{\circ }$.  $b>0$.  We will solve the equation

${\displaystyle \frac{d^{2}y}{d{x}^2}}+k^{2}y=\mathrm{\hspace{0.33em}0}.$(2)

Multiplicating both addends by the expression $2\frac{dy}{dx}$ it becomes

$$2\frac{dy}{dx}\frac{d^{2}y}{d{x}^2}+2k^{2}y\frac{dy}{dx}=\mathrm{\hspace{0.33em}0},$$

where the left hand side is the derivative of${\left(\frac{dy}{dx}\right)}^2+k^2{y}^2$.  The latter one thus has a constant valuewhich must be nonnegative; denote it by $k^2{C}^2$.  We then have the equation

${\left({\displaystyle \frac{dy}{dx}}\right)}^2=k^2(C^2-y^2).$(3)

After taking the square root and separating the variables it reads

$$\frac{dy}{\pm \sqrt{C^2-y^2}}=kdx.$$

Integrating (see the table of integrals) this yields

$$\arcsin \frac{y}{C}=k(x-x_0)$$

where $x_0$ is another constant.  Consequently, the generalsolution of the differential equation (2) may be written

$y=C\sin k(x-x_0)$(4)

in which $C$ and $x_0$ are arbitrary real constants.

If one denotes  $C\cos k{x}_0=C_1$  and $-C\sin k{x}_0=C_2$,then (4) reads

$y=C_1\sin kx+C_2\cos kx.$(5)

Here, $C_1$ and $C_2$ are arbitrary constants.  Because both$\sin kx$ and $\cos kx$ satisfy the given equation (2) and arelinearly independent, its general solution can be written as (5).

$2^{\circ }$.  .  An analogical treatment ofthe equation

${\displaystyle \frac{d^{2}y}{d{x}^2}}-k^{2}y=\mathrm{\hspace{0.33em}0}.$(6)

yields for it the general solution

$y=C_1{e}^{kx}+C_2{e}^{-kx}$(7)

(note that one can eliminate the square root from the equation$y\pm \sqrt{y^2+C}=C^{\prime }{e}^{kx}$ and its “inverted equation”$y\mp \sqrt{y^2+C}=-\frac{C}{C^{\prime }}{e}^{-kx}$).  The linear independenceof the obvious solutions $e^{\pm kx}$ implies also the linearindependence of $\cosh kx$ and $\sinh kx$ and thus allows us togive the general solution also in the alternative form

$y=C_1\sinh kx+C_2\cosh kx.$(8)

Remark.  The standard method for solving ahomogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation)ordinary second-order linear differential equation (1) withconstant coefficients is to use in it thesubstitution

$y=e^{rx}$(9)

where $r$ is a constant; see the entry “second order lineardifferential equation with constant coefficients”.  This methodis possible to use also for such equations of higher order.