Calc 3 Lecture Notes
Section 15.2
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Section 15.2: Non-Homogeneous Equations: Undetermined Coefficients
Big idea: The next level of linear second-order differential equations we will learn to solves
includes a function of the independent variable: ay  t   by  t   cy  t   F  t  .
Big skill: You should be able to solve nonhomogeneous second order differential equations with
constant coefficients using the method of “undetermined coefficients”.
In section 15.1, we found that ay  t   by  t   cy t   0 has solution y  t   ert (or y  t   tert )
for some values r. Equations of this form (with zero on the right-hand side) are called
homogeneous.
In this section, we extend the equation we can solve to: ay  t   by  t   cy t   F t  (for some
given function F  t  ). Equations of this form (with a function on the right-hand side) are called
nonhomogeneous.
The general solution to the nonhomogeneous equation will be the general solution of the
corresponding homogeneous equation plus one or more additional terms that called the
particular solution. This is because if we write the solution as y  t   yh  t   y p  t  , then
ay  t   by  t   cy  t    ayh  t   byh  t   cyh t     ay p t   by p t   cy p t  
 0  F t 
.
Practice:
1. Consider the nonhomogeneous equation y  t   y  t   2 y  t   2e3t .
a. Show that y  t   Aet  Be2t is the general solution of the corresponding
homogeneous equation.
Calc 3 Lecture Notes
Section 15.2
Page 2 of 7
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b. Show that y  t   y p  t   e3t is a particular solution of the nonhomogeneous
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equation.
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c. Show that y  t   Aet  Be 2t  y p  t   Aet  Be 2t  e3t is a general solution of
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the nonhomogeneous equation.
Theorem 2.1: General Solution of a Nonhomogeneous, Linear Second-Order Differential
Equation
Let y  t   c1 y1  t   c2 y2 t  be the general solution of the homogeneous equation
ay  t   by t   cy t   0 , and let y p  t  be any solution of the nonhomogeneous equation
ay  t   by t   cy t   F t  .
Then the general solution of the nonhomogeneous equation is:
y  t   c1 y1  t   c2 y2 t   y p t 
Calc 3 Lecture Notes
Section 15.2
Page 3 of 7
To solve the nonhomogeneous equations in this section, we make use of the “Method of
Undetermined Coefficients,” which basically amounts to “make a guess and find the
multiplicative constants that make the guess work.”
Steps to solve the nonhomogeneous equation ay  t   by  t   cy t   F t  :

Solve the corresponding homogeneous equation ay  t   by  t   cy t   0 .

Make a guess for the particular solution:
o If F  t   A cos t   B sin t  , then guess y p  t   c1 cos t   c2 sin t 
 If r  i is a single root of the characteristic equation, then choose
y p  t   t  c1 cos t   c2 sin t   .

If r  i is a double root of the characteristic equation, then choose
y p  t   t 2  c1 cos t   c2 sin t   .
o If F  t   Aekt , then guess y p  t   c1ekt

If r  k is a single root of the characteristic equation, then choose
y p  t   c1tekt .

If r  k is a double root of the characteristic equation, then choose
y p  t   c1t 2ekt .
o If F  t   Pn  t  (i.e., a polynomial of degree n), then guess y p  t   Qn  t  (i.e.,
another polynomial of degree n).
 If r  0 is a single root of the characteristic equation, then chose
y p  t   tQn  t  .



If r  0 is a double root of the characteristic equation, then chose
y p  t   t 2Qn  t  .
o If F  t  is a product of the above three types of functions, then choose a particular
solution that is a product of the above particular solutions.
o Multiplying by t or t2 is done to guarantee that no term in the particular solution is
a solution of the corresponding homogeneous equation.
Plug in the guess for the particular solution; equate like terms to determine the value(s) of
the undetermined coefficient(s).
Write the general solution as the sum of the homogeneous solution and the particular
solution.
Calc 3 Lecture Notes
Section 15.2
Practice:
2. Solve the equation y  t   y  t   2 y  t   t 2 .
3. Solve the equation y  t   2 y  t   t 2 .
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Calc 3 Lecture Notes
Section 15.2
4. Solve the equation y  t   t 2 .
5. Solve the equation y  t   y  t   2 y  t   2et
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Calc 3 Lecture Notes
Section 15.2
6. Solve the equation y  t   2 y  t   y  t   2et .
7. Solve the equation y  t   2 y  t   y  t   sin  t  .
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Calc 3 Lecture Notes
Section 15.2
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8. Determine the form of the particular solution of the equation
u  2u  5u  et sin  2t   t 2et .
9. Determine the form of the particular solution of the equation
u  4u  2t cos  t   t 2 sin  t  .
10. Determine the form of the particular solution of the equation u  2u  u  t 2  4  2e t .