If consists of the sum of two functions and we say that is the solution based on and the solution based on . Then, using a superposition principle, we can say that the particular integral is
Typical forms of the particular integral
In order to find the particular integral, we need to 'guess' its form, with some coefficients left as variables to be solved for. This takes the form of the first derivative of the complementary function. Below is a table of some typical functions and the solution to guess for them.
Function of x
Form for y
If a term in the above particular integral for y appears in the homogeneous solution, it is necessary to multiply by a sufficiently large power of x in order to make the solution independent. If the function of x is a sum of terms in the above table, the particular integral can be guessed using a sum of the corresponding terms for y.
Find a particular integral of the equation
The right side t cos t has the form
with n = 2, α = 0, and β = 1.
Since α + iβ = i is a simple root of the characteristic equation
we should try a particular integral of the form
Substituting yp into the differential equation, we have the identity
Comparing both sides, we have
which has the solution
We then have a particular integral
Consider the following linear nonhomogeneous differential equation:
This is like the first example above, except that the nonhomogeneous part () is not linearly independent to the general solution of the homogeneous part (); as a result, we have to multiply our guess by a sufficiently large power of x to make it linearly independent.
Here our guess becomes:
By substituting this function and its derivative into the differential equation, one can solve for A:
So, the general solution to this differential equation is:
Find the general solution of the equation:
is a polynomial of degree 2, so we look for a solution using the same form,
Plugging this particular function into the original equation yields,
Solving for constants we get:
To solve for the general solution,
where is the homogeneous solution , therefore, the general solution is:
^ abRalph P. Grimaldi (2000). "Nonhomogeneous Recurrence Relations". Section 3.3.3 of Handbook of Discrete and Combinatorial Mathematics. Kenneth H. Rosen, ed. CRC Press. ISBN 0-8493-0149-1.
^Zill, Dennis G., Warren S. Wright (2014). Advanced Engineering Mathematics. Jones and Bartlett. p. 125. ISBN 978-1-4496-7977-4.CS1 maint: multiple names: authors list (link)
^ abDennis G. Zill (14 May 2008). A First Course in Differential Equations. Cengage Learning. ISBN 978-0-495-10824-5.
Boyce, W. E.; DiPrima, R. C. (1986). Elementary Differential Equations and Boundary Value Problems (4th ed.). John Wiley & Sons. ISBN 0-471-83824-1.
Riley, K. F.; Bence, S. J. (2010). Mathematical Methods for Physics and Engineering. Cambridge University Press. ISBN 978-0-521-86153-3.
Tenenbaum, Morris; Pollard, Harry (1985). Ordinary Differential Equations. Dover. ISBN 978-0-486-64940-5.
de Oliveira, O. R. B. (2013). "A formula substituting the undetermined coefficients and the annihilator methods". Int. J. Math. Educ. Sci. Technol. 44 (3): 462–468. Bibcode:2013IJMES..44..462R. doi:10.1080/0020739X.2012.714496. S2CID 55834468.