Any differential equation of the form
(1) is a second order differential equations and there is a standard technique for solving any equation of this sort. We solve the homogeneous equation
(2) first for the 'complementary' solution
We assume a solution of the form
and substitute this into (2). We extract the non zero factor – since no exponential is zero for any finite x -
to obtain a quadratic equation. We solve this equation to obtain solutions
and
and then the general solution is
Having found the complementary solution
we now find a particular solution
to (1) by assuming a solution 'similar' to
then the general solution to (1) is given by![]()
Example: Solve the equation
(2) subject to
and
at![]()
![]()
Substitution of the above expressions into
gives![]()
We can factor out the nonzero
to obtain![]()
Because
is non zero we can divide by it to obtain
and we factorise this expression to obtain
and solve to obtain
and
The complementary solution is then
![]()
To find a particular solution we assume
since
is a polynomial of degree one.
and
Substitute these into (1)
![]()
Equating coefficients of
gives![]()
Equating constants gives![]()
![]()
when
implies
(3)
at
implies
(4)
(3)+(4) gives
then from (3)![]()
The solution is![]()