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Undetermined coefficients

The method of undetermined coefficients    involves making educated guesses about the form of the particular solution based on the form of r ( x ). When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when r ( x ) has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let’s look at some examples to see how this works.

Undetermined coefficients when r ( x ) Is a polynomial

Find the general solution to y + 4 y + 3 y = 3 x .

The complementary equation is y + 4 y + 3 y = 0 , with general solution c 1 e x + c 2 e −3 x . Since r ( x ) = 3 x , the particular solution might have the form y p ( x ) = A x + B . If this is the case, then we have y p ( x ) = A and y p ( x ) = 0 . For y p to be a solution to the differential equation, we must find values for A and B such that

y + 4 y + 3 y = 3 x 0 + 4 ( A ) + 3 ( A x + B ) = 3 x 3 A x + ( 4 A + 3 B ) = 3 x .

Setting coefficients of like terms equal, we have

3 A = 3 4 A + 3 B = 0 .

Then, A = 1 and B = 4 3 , so y p ( x ) = x 4 3 and the general solution is

y ( x ) = c 1 e x + c 2 e −3 x + x 4 3 .
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In [link] , notice that even though r ( x ) did not include a constant term, it was necessary for us to include the constant term in our guess. If we had assumed a solution of the form y p = A x (with no constant term), we would not have been able to find a solution. (Verify this!) If the function r ( x ) is a polynomial, our guess for the particular solution should be a polynomial of the same degree, and it must include all lower-order terms, regardless of whether they are present in r ( x ).

Undetermined coefficients when r ( x ) Is an exponential

Find the general solution to y y 2 y = 2 e 3 x .

The complementary equation is y y 2 y = 0 , with the general solution c 1 e x + c 2 e 2 x . Since r ( x ) = 2 e 3 x , the particular solution might have the form y p ( x ) = A e 3 x . Then, we have y p ( x ) = 3 A e 3 x and y p ( x ) = 9 A e 3 x . For y p to be a solution to the differential equation, we must find a value for A such that

y y 2 y = 2 e 3 x 9 A e 3 x 3 A e 3 x 2 A e 3 x = 2 e 3 x 4 A e 3 x = 2 e 3 x .

So, 4 A = 2 and A = 1 / 2 . Then, y p ( x ) = ( 1 2 ) e 3 x , and the general solution is

y ( x ) = c 1 e x + c 2 e 2 x + 1 2 e 3 x .
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Find the general solution to y 4 y + 4 y = 7 sin t cos t .

y ( t ) = c 1 e 2 t + c 2 t e 2 t + sin t + cos t

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In the previous checkpoint, r ( x ) included both sine and cosine terms. However, even if r ( x ) included a sine term only or a cosine term only, both terms must be present in the guess. The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of r ( x ) and the associated guesses for y p ( x ) are summarized in [link] .

Key forms for the method of undetermined coefficients
r ( x ) Initial guess for y p ( x )
k (a constant) A (a constant)
a x + b A x + B ( Note : The guess must include both terms even if b = 0 . )
a x 2 + b x + c A x 2 + B x + C ( Note : The guess must include all three terms even if b or c are zero.)
Higher-order polynomials Polynomial of the same order as r ( x )
a e λ x A e λ x
a cos β x + b sin β x A cos β x + B sin β x ( Note : The guess must include both terms even if either a = 0 or b = 0 . )
a e α x cos β x + b e α x sin β x A e α x cos β x + B e α x sin β x
( a x 2 + b x + c ) e λ x ( A x 2 + B x + C ) e λ x
( a 2 x 2 + a 1 x + a 0 ) cos β x + ( b 2 x 2 + b 1 x + b 0 ) sin β x ( A 2 x 2 + A 1 x + A 0 ) cos β x + ( B 2 x 2 + B 1 x + B 0 ) sin β x
( a 2 x 2 + a 1 x + a 0 ) e α x cos β x + ( b 2 x 2 + b 1 x + b 0 ) e α x sin β x ( A 2 x 2 + A 1 x + A 0 ) e α x cos β x + ( B 2 x 2 + B 1 x + B 0 ) e α x sin β x

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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