User:Egm4313.s12.team11.Suh/R3.4

Problem Statement
Use Basic Rule (1) and Sum Rule (3) to show that the appropriate particular solution for $$y''-3y'+2y = 4x^{2}-6x^{5}\!$$ is of the form $$\sum_{j=0}^{n}c_{j}x^{j}\!$$, with n = 5.

Finding the Particular Solution with Basic Rule and Sum Rule
We know that in standard form, for a particular solution, $$y''_{p}+ay'_{p}+by_{p}=r(x)\!$$ $$y_{p}=y_{p1}+y_{p2}\!$$

Using Basic Rule (1) and Sum Rule (3), we know that $$y''_{p}+ay'_{p}+by_{p}=\sum_{i}r_i(x)\!$$

Additionally, we choose $$r_1(x)\rightarrow y_{p1}(x)\!$$ $$r_2(x)\rightarrow y_{p2}(x)\!$$

Solving for Particular Solution 1
Using the Method of Undetermined Coefficients, we find that $$y_{p1}=C_{2}x^{2}+C_{1}x+C_{0}\!$$ $$y'_{p1}=2C_{2}x+C_{1}\!$$ $$y''_{p1}=2C_{2}\!$$

Plugging $$y_{p1}\!$$ into $$y''-3y'+2y=4x^{2}\!$$ gives $$(2C_{2})-3(2C_{2}x+C_{1})+2(C_{2}x^{2}+C_{1}x+C_{0})=4x^{2}\!$$

Solving for coefficients $$2C_{2}=4\!$$ $$-6C_{2}+2C_{1}=0\!$$ $$-3C_{1}+2C_{0}+2C_{2}=0\!$$

Results in $$C_{2}=2\!$$ $$C_{1}=6\!$$ $$C_{0}=7\!$$

$$y_{p1}=2x^{2}+6x+7\!$$

Solving for Particular Solution 2
Using the Method of Undetermined Coefficients, we find that $$y_{p2}=K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!$$ $$y'_{p2}=5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1}\!$$ $$y''_{p2}=20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!$$

Plugging $$y_{p2}\!$$ into $$y''-3y'+2y=-6x^{5}\!$$ gives $$(20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2})-3(5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1})+2(K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=-6x^{5}\!$$

Solving for coefficients results in $$K_{5}=-3\!$$ $$K_{4}=-22.5\!$$ $$K_{3}=-105\!$$ $$K_{2}=-337.5\!$$ $$K_{1}=-697.5\!$$ $$K_{0}=-708.75\!$$

$$y_{p2}=-3x^{5}-22.5x^{4}-105x^{3}-337.5x^{2}-697.5x-708.75\!$$

Combining Particular Solutions
Plugging $$y_{p1}\!$$ and $$y_{p2}\!$$ into $$y_{p}=y_{p1}+y_{p2}\!$$ gives

$$y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$$

Comparing to Summation Form
The particular solution in the summation form is $$y_{p}=\sum_{j=0}^{n}c_{j}x^j\!$$

if n=5, then $$y_{p}=\sum_{j=0}^{5}c_{j}x^j\!$$ $$y'_{p}=\sum_{j=0}^{5}c_{j}*j*x^{(j-1)}\!$$ $$y''_{p}=\sum_{j=0}^{5}c_{j}*j*(j-1)*x^{(j-2)}\!$$

Plugging the particular solution into $$y''-3y'+2y=-6x^{5}\!$$, the final solution gives $$y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$$ (check Problem 3.5 for explanation on how to get this answer)

Result
As you can see, the two particular solutions of $$y_{p}\!$$ are equal. Thus, using the Basic Rule (1) and Sum rule (3) does give you the correct particular solution for $$y''-3y'+2y=4x^{2}-6x^{5}\!$$, in the form of $$y_{p}=\sum_{j=0}^{n}c_{j}x^j\!$$