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7.2 Nonhomogeneous linear equations  (Page 3/9)

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Keep in mind that there is a key pitfall to this method. Consider the differential equation y ″ + 5 y ′ + 6 y = 3 e −2 x . Based on the form of r ( x ) , we guess a particular solution of the form y p ( x ) = A e −2 x . But when we substitute this expression into the differential equation to find a value for A , we run into a problem. We have

y p ′ ( x ) = −2 A e −2 x

and

y p ″ = 4 A e −2 x ,

so we want

y ″ + 5 y ′ + 6 y = 3 e −2 x 4 A e −2 x + 5 ( −2 A e −2 x ) + 6 A e −2 x = 3 e −2 x 4 A e −2 x − 10 A e −2 x + 6 A e −2 x = 3 e −2 x 0 = 3 e −2 x ,

which is not possible.

Looking closely, we see that, in this case, the general solution to the complementary equation is c 1 e −2 x + c 2 e −3 x . The exponential function in r ( x ) is actually a solution to the complementary equation, so, as we just saw, all the terms on the left side of the equation cancel out. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by x . Using the new guess, y p ( x ) = A x e −2 x , we have

y p ′ ( x ) = A ( e −2 x − 2 x e −2 x )

and

y p ″ ( x ) = −4 A e −2 x + 4 A x e −2 x .

Substitution gives

y ″ + 5 y ′ + 6 y = 3 e −2 x ( −4 A e −2 x + 4 A x e −2 x ) + 5 ( A e −2 x − 2 A x e −2 x ) + 6 A x e −2 x = 3 e −2 x −4 A e −2 x + 4 A x e −2 x + 5 A e −2 x − 10 A x e −2 x + 6 A x e −2 x = 3 e −2 x A e −2 x = 3 e −2 x .

So, A = 3 and y p ( x ) = 3 x e −2 x . This gives us the following general solution

y ( x ) = c 1 e −2 x + c 2 e −3 x + 3 x e −2 x .

Note that if x e −2 x were also a solution to the complementary equation, we would have to multiply by x again, and we would try y p ( x ) = A x 2 e −2 x .

Problem-solving strategy: method of undetermined coefficients

  1. Solve the complementary equation and write down the general solution.
  2. Based on the form of r ( x ) , make an initial guess for y p ( x ).
  3. Check whether any term in the guess for y p ( x ) is a solution to the complementary equation. If so, multiply the guess by x . Repeat this step until there are no terms in y p ( x ) that solve the complementary equation.
  4. Substitute y p ( x ) into the differential equation and equate like terms to find values for the unknown coefficients in y p ( x ).
  5. Add the general solution to the complementary equation and the particular solution you just found to obtain the general solution to the nonhomogeneous equation.

Solving nonhomogeneous equations

Find the general solutions to the following differential equations.

  1. y ″ − 9 y = −6 cos 3 x
  2. x ″ + 2 x ′ + x = 4 e − t
  3. y ″ − 2 y ′ + 5 y = 10 x 2 − 3 x − 3
  4. y ″ − 3 y ′ = −12 t
  1. The complementary equation is y ″ − 9 y = 0 , which has the general solution c 1 e 3 x + c 2 e −3 x (step 1). Based on the form of r ( x ) = −6 cos 3 x , our initial guess for the particular solution is y p ( x ) = A cos 3 x + B sin 3 x (step 2). None of the terms in y p ( x ) solve the complementary equation, so this is a valid guess (step 3).
    Now we want to find values for A and B , so substitute y p into the differential equation. We have
    y p ′ ( x ) = −3 A sin 3 x + 3 B cos 3 x and y p ″ ( x ) = −9 A cos 3 x − 9 B sin 3 x ,

    so we want to find values of A and B such that
    y ″ − 9 y = −6 cos 3 x −9 A cos 3 x − 9 B sin 3 x − 9 ( A cos 3 x + B sin 3 x ) = −6 cos 3 x −18 A cos 3 x − 18 B sin 3 x = −6 cos 3 x .

    Therefore,
    −18 A = −6 −18 B = 0 .

    This gives A = 1 3 and B = 0 , so y p ( x ) = ( 1 3 ) cos 3 x (step 4).
    Putting everything together, we have the general solution
    y ( x ) = c 1 e 3 x + c 2 e −3 x + 1 3 cos 3 x .
  2. The complementary equation is x ″ + 2 x ′ + x = 0 , which has the general solution c 1 e − t + c 2 t e − t (step 1). Based on the form r ( t ) = 4 e − t , our initial guess for the particular solution is x p ( t ) = A e − t (step 2). However, we see that this guess solves the complementary equation, so we must multiply by t , which gives a new guess: x p ( t ) = A t e − t (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by t again, which gives x p ( t ) = A t 2 e − t (step 3 again). Now, checking this guess, we see that x p ( t ) does not solve the complementary equation, so this is a valid guess (step 3 yet again).
    We now want to find a value for A , so we substitute x p into the differential equation. We have
    x p ( t ) = A t 2 e − t , so x p ′ ( t ) = 2 A t e − t − A t 2 e − t

    and x p ″ ( t ) = 2 A e − t − 2 A t e − t − ( 2 A t e − t − A t 2 e − t ) = 2 A e − t − 4 A t e − t + A t 2 e − t .
    Substituting into the differential equation, we want to find a value of A so that
    x ″ + 2 x ′ + x = 4 e − t 2 A e − t − 4 A t e − t + A t 2 e − t + 2 ( 2 A t e − t − A t 2 e − t ) + A t 2 e − t = 4 e − t 2 A e − t = 4 e − t .

    This gives A = 2 , so x p ( t ) = 2 t 2 e − t (step 4). Putting everything together, we have the general solution
    x ( t ) = c 1 e − t + c 2 t e − t + 2 t 2 e − t .
  3. The complementary equation is y ″ − 2 y ′ + 5 y = 0 , which has the general solution c 1 e x cos 2 x + c 2 e x sin 2 x (step 1). Based on the form r ( x ) = 10 x 2 − 3 x − 3 , our initial guess for the particular solution is y p ( x ) = A x 2 + B x + C (step 2). None of the terms in y p ( x ) solve the complementary equation, so this is a valid guess (step 3). We now want to find values for A , B , and C , so we substitute y p into the differential equation. We have y p ′ ( x ) = 2 A x + B and y p ″ ( x ) = 2 A , so we want to find values of A , B , and C such that
    y ″ − 2 y ′ + 5 y = 10 x 2 − 3 x − 3 2 A − 2 ( 2 A x + B ) + 5 ( A x 2 + B x + C ) = 10 x 2 − 3 x − 3 5 A x 2 + ( 5 B − 4 A ) x + ( 5 C − 2 B + 2 A ) = 10 x 2 − 3 x − 3 .

    Therefore,
    5 A = 10 5 B − 4 A = −3 5 C − 2 B + 2 A = −3 .

    This gives A = 2 , B = 1 , and C = −1 , so y p ( x ) = 2 x 2 + x − 1 (step 4). Putting everything together, we have the general solution
    y ( x ) = c 1 e x cos 2 x + c 2 e x sin 2 x + 2 x 2 + x − 1 .
  4. The complementary equation is y ″ − 3 y ′ = 0 , which has the general solution c 1 e 3 t + c 2 (step 1). Based on the form r ( t ) = −12 t , our initial guess for the particular solution is y p ( t ) = A t + B (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by t , which gives a new guess: y p ( t ) = A t 2 + B t (step 3). Checking this new guess, we see that none of the terms in y p ( t ) solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for A and B , so we substitute y p into the differential equation. We have y p ′ ( t ) = 2 A t + B and y p ″ ( t ) = 2 A , so we want to find values of A and B such that
    y ″ − 3 y ′ = −12 t 2 A − 3 ( 2 A t + B ) = −12 t − 6 A t + ( 2 A − 3 B ) = −12 t .

    Therefore,
    −6 A = −12 2 A − 3 B = 0 .

    This gives A = 2 and B = 4 / 3 , so y p ( t ) = 2 t 2 + ( 4 / 3 ) t (step 4). Putting everything together, we have the general solution
    y ( t ) = c 1 e 3 t + c 2 + 2 t 2 + 4 3 t .
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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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