Given a partial differential equation of a function

${\displaystyle F(x_{1},x_{2},\dots ,x_{n})}$

of n variables, it is sometimes useful to guess solution of the form

${\displaystyle F=F_{1}(x_{1})\cdot F_{2}(x_{2})\cdots F_{n}(x_{n})}$

or

${\displaystyle F=f_{1}(x_{1})+f_{2}(x_{2})+\cdots +f_{n}(x_{n})}$

which turns the partial differential equation (PDE) into a set of ODEs. Usually, each independent variable creates a separation constant that cannot be determined only from the equation itself.

When such a technique works, it is called a separable partial differential equation.

Suppose F(x, y, z) and the following PDE:

${\displaystyle {\frac {\partial F}{\partial x}}+{\frac {\partial F}{\partial y}}+{\frac {\partial F}{\partial z}}=0\qquad \qquad (1)}$

We shall guess

${\displaystyle F(x,y,z)=X(x)+Y(y)+Z(z)\qquad \qquad (2)}$

thus making the equation (1) to

${\displaystyle {\frac {dX}{dx}}+{\frac {dY}{dy}}+{\frac {dZ}{dz}}=0}$

(since ${\displaystyle {\frac {\partial F}{\partial x}}={\frac {dX}{dx}}}$).

Now, since X'(x) is dependent only on x and Y'(y) is dependent only on y (so on for Z'(z)) and that the equation (1) is true for every x, y, z it is clear that each one of the term is constant. More precisely,

${\displaystyle {\frac {dX}{dx}}=c_{1}\quad {\frac {dY}{dy}}=c_{2}\quad {\frac {dZ}{dz}}=c_{3}\qquad \qquad (3)}$

where the constants c₁, c₂, c₃ satisfy

${\displaystyle c_{1}+c_{2}+c_{3}=0\qquad \qquad (4)}$

Eq. (3) is actually a set of three ODEs. In this case they are trivial and can be solved by simple integration, giving:

${\displaystyle F(x,y,z)=c_{1}x+c_{2}y+c_{3}z+c_{4}\qquad \qquad (5)}$

where the integration constant c₄ is determined by initial conditions.

Consider the differential equation

${\displaystyle \nabla ^{2}v+\lambda v={\partial ^{2}v \over \partial x^{2}}+{\partial ^{2}v \over \partial y^{2}}+\lambda v=0.}$

First we seek solutions of the form

${\displaystyle v=X(x)Y(y).\,}$

Most solutions are not of that form, but other solutions are sums of (generally infinitely many) solutions of that form.

Substituting,

${\displaystyle {\partial ^{2} \over \partial x^{2}}[X(x)Y(y)]+{\partial ^{2} \over \partial y^{2}}[X(x)Y(y)]+\lambda X(x)Y(y)=}$

${\displaystyle =X''(x)Y(y)+X(x)Y''(y)+\lambda X(x)Y(y)=0\,}$

Divide throughout by X(x)

${\displaystyle ={X''(x)Y(y) \over X(x)}+{X(x)Y''(y) \over X(x)}+{\lambda X(x)Y(y) \over X(x)}}$

${\displaystyle ={X''(x)Y(y) \over X(x)}+Y''(y)+\lambda Y(y)=0}$

and then by Y(y)

${\displaystyle ={X''(x) \over X(x)}+{Y''(y)+\lambda Y(y) \over Y(y)}=0}$

Now X′′(x)/X(x) is a function of x only, and (Y′′(y)+λY(y))/Y(y) is a function of y only, so for their sum to be equal to zero for all x and y, they must both be constant. Thus,

${\displaystyle {X''(x) \over X(x)}=k=-{Y''(y)+\lambda Y(y) \over Y(y)}}$

where k is the separation constant. This splits up into ordinary differential equations

${\displaystyle {X''(x) \over X(x)}=k}$

${\displaystyle X''(x)-kX(x)=0\,}$

and

${\displaystyle {Y''(y)+\lambda Y(y) \over Y(y)}=-k}$

${\displaystyle Y''(y)+(\lambda +k)Y(y)=0\,}$

which we can solve accordingly. If the equation as posed originally was a boundary value problem, one would use the given boundary values. See that article for an example which uses boundary values.