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 nonzeroto obtain
Becauseis 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 assumesince
is a polynomial of degree one.
and
Substitute these into (1)
Equating coefficients ofgives
Equating constants gives
when
implies
(3)
at
implies
(4)
(3)+(4) givesthen from (3)
The solution is