Differential Equations: Nonhomogeneous Equation Problem

Question

The nonhomogeneous equation t^2y′′−2y=29t^2−1, t>0, has homogeneous solutions y1(t)=t^2, y2(t)=t−1. Find the particular solution to the nonhomogeneous equation that does not involve any terms from the homogeneous solution.

Enter an exact answer. Do not use decimal approximations.

Answer 1

You can use either Variation of Parameters or undetermined coefficients.

 

For the method of undetermined coefficients, the particular solution is

 Yp = At^2 + Bt + C based on the right hand side.

Differentiation:

  Y'p = 2At + B

 Y''p = 2A

 

plug those in and equate the coefficients to the right side....

 

 

For variation of parameters, the particular solution is Yp = U*Y1 + V*Y2 where Y1 and Y2 are

the homogenous (complimentary) solutions given.

U and V satisfy the conditions:

 

     U'*Y1 + V'*y2 = 0

 

     U'*Y1' + V'*Y2' = 29t^2 -1

 

 

plugging in:

 

  U'(t^2) + V'(t-1) = 0  --->  U' = V'(1-t)/t^2 <---- call this equation BETA

 

Substituting this into the second equation:

 

   V'(1-t)/t^2(2t) + V' = 29t^2 - 1

 

   [2(1-t)/t + 1 ]V' = 29t^2 - 1

 

Solving for V' we get:

 

 V' = (29t^3 - t)/(2-t) 

 

     = -29t^2 - 58t + 115  - 230/(2-t) <--- polynomial division; call this equation ALPHA

 

Integrating...

 

 V = -29/3t^3 - 29t^2 + 115t - 230 ln | 2- t |

 

You next need to plug equation ALPHA, the function for V' into equation BETA

to find U'

 

Then you need to integrate it to get U

 

Your particular solution will be The parabola is U*Y1 + V*Y2 as stated in the theorem above in bold