# equation $y^{\prime\prime}=f(x)$

A simple special case of the second order linear differential equation with constant coefficients is

 $\displaystyle\frac{d^{2}y}{dx^{2}}\;=\;f(x)$ (1)

where $f$ is continuous.  We obtain immediately  $\displaystyle\frac{dy}{dx}=C_{1}+\!\int\!f(x)\,dx$,

 $\displaystyle y\;=\;C_{1}x+C_{2}+\!\int\!\left(\int f(x)\,dx\right)dx.$ (2)

A particular solution $y(x)$ of (1) satisfying the initial conditions

 $y(x_{0})=y_{0},\quad y^{\prime}(x_{0})=y_{0}^{\prime}$

is obtained more simply by integrating (1) twice between the limits (http://planetmath.org/DefiniteIntegral) $x_{0}$ and $x$, thus getting

 $y(x)=y_{0}+y_{0}^{\prime}\!\cdot\!(x\!-\!x_{0})+\!\int_{x_{0}}^{x}\!\left(\int% _{x_{0}}^{x}f(x)\,dx\right)dx.$

But here, the two first addends are the first terms of the Taylor polynomial of $y(x)$, expanded by the powers of $x-x_{0}$, whence the double integral is the corresponding remainder term (http://planetmath.org/RemainderVariousFormulas)

 $\int_{x_{0}}^{x}y^{\prime\prime}(x)(x\!-\!t)\,dt\;=\;\int_{x_{0}}^{x}f(t)(x\!-% \!t)\,dt.$

Hence the particular solution can be written with the simple integral as

 $\displaystyle y(x)\;=\;y_{0}+y_{0}^{\prime}\!\cdot\!(x\!-\!x_{0})+\int_{x_{0}}% ^{x}f(t)(x\!-\!t)\,dt.$ (3)

The result may be generalised for the $n^{\mathrm{th}}$ order (http://planetmath.org/ODE) differential equation

 $\displaystyle\frac{d^{n}y}{dx^{n}}\;=\;f(x)$ (4)

with corresponding $n$ initial conditions:

 $\displaystyle y(x)\,=\,y_{0}\!+\!y_{0}^{\prime}\!\cdot\!(x\!-\!x_{0})\!+\!% \frac{y_{0}^{\prime\prime}}{2!}(x\!-\!x_{0})^{2}\!+\ldots+\!\frac{y_{0}^{(n-1)% }}{(n\!-\!1)!}(x\!-\!x_{0})^{n-1}\!+\!\frac{1}{(n\!-\!1)!}\int_{x_{0}}^{x}f(t)% (x\!-\!t)^{n-1}\,dt.$ (5)
Title equation $y^{\prime\prime}=f(x)$ EquationYFx 2013-03-22 18:35:33 2013-03-22 18:35:33 pahio (2872) pahio (2872) 10 pahio (2872) Topic msc 34A30 msc 34-01