Sometimes,
is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the
method of variation of parameters .
To simplify our calculations a little, we are going to divide the differential equation through by
so we have a leading coefficient of 1. Then the differential equation has the form
where
and
are constants.
If the general solution to the complementary equation is given by
we are going to look for a particular solution of the form
In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of
x , rather than constants. We want to find functions
and
such that
satisfies the differential equation. We have
Substituting into the differential equation, we obtain
Note that
and
are solutions to the complementary equation, so the first two terms are zero. Thus, we have
If we simplify this equation by imposing the additional condition
the first two terms are zero, and this reduces to
So, with this additional condition, we have a system of two equations in two unknowns:
Solving this system gives us
and
which we can integrate to find
u and
v .
Then,
is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let’s take this opportunity to review Cramer’s rule, which allows us to solve the system of equations using determinants.
Rule: cramer’s rule
The system of equations
has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by
Using cramer’s rule
Use Cramer’s rule to solve the following system of equations.
Problem-solving strategy: method of variation of parameters
Solve the complementary equation and write down the general solution
Use Cramer’s rule or another suitable technique to find functions
and
satisfying
Integrate
and
to find
and
Then,
is a particular solution to the equation.
Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.
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