second-order linear ODE with constant coefficients

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

 $\displaystyle\frac{d^{2}y}{dx^{2}}+a\frac{dy}{dx}+by\;=\;0$ (1)

which is homogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation) and the coefficients $a,b$ of which are constants.  As mentionned in the entry “finding another particular solution of linear ODE”, a simple substitution makes possible to eliminate from it the addend containing first derivative of the unknown function.  Therefore we concentrate upon the case  $a=0$.  We have two cases depending on the sign of  $b=\pm k^{2}$.

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

 $\displaystyle\frac{d^{2}y}{dx^{2}}+k^{2}y\;=\;0.$ (2)

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

 $2\frac{dy}{dx}\frac{d^{2}y}{dx^{2}}+2k^{2}y\frac{dy}{dx}\;=\;0,$

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 value which must be nonnegative; denote it by $k^{2}C^{2}$.  We then have the equation

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

After taking the square root and separating the variables it reads

 $\frac{dy}{\pm\sqrt{C^{2}\!-y^{2}}}\;=\;k\,dx.$

Integrating (see the table of integrals) this yields

 $\arcsin\frac{y}{C}\;=\;k(x\!-\!x_{0})$

where $x_{0}$ is another constant.  Consequently, the general solution of the differential equation (2) may be written

 $\displaystyle y\,\;=\;C\,\sin k(x\!-\!x_{0})$ (4)

in which $C$ and $x_{0}$ are arbitrary real constants.

If one denotes  $C\cos kx_{0}=C_{1}$  and $-C\sin kx_{0}=C_{2}$, then (4) reads

 $\displaystyle y\,\;=\;C_{1}\sin kx+C_{2}\cos kx.$ (5)

Here, $C_{1}$ and $C_{2}$ are arbitrary constants.  Because both $\sin kx$ and $\cos kx$ satisfy the given equation (2) and are linearly independent, its general solution can be written as (5).

$2^{\circ}$.  $b<0$.  An analogical treatment of the equation

 $\displaystyle\frac{d^{2}y}{dx^{2}}-k^{2}y\;=\;0.$ (6)

yields for it the general solution

 $\displaystyle y\,\;=\;C_{1}e^{kx}+C_{2}e^{-kx}$ (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 independence of the obvious solutions $e^{\pm kx}$ implies also the linear independence of $\cosh kx$ and $\sinh kx$ and thus allows us to give the general solution also in the alternative form

 $\displaystyle y\,\;=\;C_{1}\sinh kx+C_{2}\cosh kx.$ (8)

Remark.  The standard method for solving a homogeneous (http://planetmath.org/HomogeneousLinearDifferentialEquation) ordinary second-order linear differential equation (1) with constant coefficients is to use in it the substitution

 $\displaystyle y\;=\;e^{rx}$ (9)

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

References

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

Title second-order linear ODE with constant coefficients SecondorderLinearODEWithConstantCoefficients 2014-03-01 17:02:54 2014-03-01 17:02:54 pahio (2872) pahio (2872) 8 pahio (2872) Derivation msc 34A05