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

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

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

where $f$ is continuous.  We obtain immediately  ${\displaystyle \frac{dy}{dx}}=C_1+{\displaystyle \int f(x)𝑑x}$,

$y=C_{1}x+C_2+{\displaystyle \int \left(\int f(x)𝑑x\right)𝑑x}.$(2)

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

$$y(x_0)=y_0,y^{\prime }(x_0)=y_0^{\prime }$$

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

$y(x)=y_0+y_0^{\prime }\cdot (x-x_0)+{\displaystyle {\int }_{x_0}^x}f(t)(x-t)𝑑t.$(3)

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

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

with corresponding $n$ initial conditions:

$y(x)=y_0+y_0^{\prime }\cdot (x-x_0)+{\displaystyle \frac{y_0^{\prime \prime }}{2!}}{(x-x_0)}^2+\mathrm{\dots }+{\displaystyle \frac{y_0^{(n-1)}}{(n-1)!}}{(x-x_0)}^{n-1}+{\displaystyle \frac{1}{(n-1)!}}{\displaystyle {\int }_{x_0}^x}f(t){(x-t)}^{n-1}𝑑t.$(5)