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Homesecond-order linear ODE with constant coefficients

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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
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 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).

## Mathematics Subject Classification

34A05*no label found*

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