# University of Florida/Egm4313/s12.team8.dupre/R3.4

Problem 3.4

## Problem Statement

Use the Basic Rule and the Sum Rule to show that the appropriate particular solution to

${\displaystyle \displaystyle y''-3y'+2y=4x^{2}-6x^{5}}$ (4.1)

is of the form

${\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x^{j}}$

with n=5, i.e,

${\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x^{j}}$.

### Solution

The Basic rule and Sum rule allow us to choose our particular y forms from table 2.1 on page 82 in the Kreyszig Advanced Engineering Mathematics book. These rules state that, using your given ${\displaystyle \displaystyle r(x)}$ , you can find what ${\displaystyle \displaystyle y_{p}(x)}$ you should choose to use.

From (4.1), we know that:

${\displaystyle \displaystyle r(x)=4x^{2}-6x^{5}}$ (4.2)

Referring to table 2.1, and knowing that the Basic rule tells us that we can match up the form of our ${\displaystyle \displaystyle r(x)}$ to one in the table, and use the ${\displaystyle \displaystyle y_{p}(x)}$ accordingly, we find the following two equations:

The ${\displaystyle \displaystyle y_{p}(x)}$ for ${\displaystyle \displaystyle 4x^{2}}$ is:

${\displaystyle \displaystyle k_{2}x^{2}+k_{1}x^{1}+k_{0}}$ (4.3)

And the ${\displaystyle \displaystyle y_{p}(x)}$ for ${\displaystyle \displaystyle 6x^{5}}$ is:

${\displaystyle \displaystyle K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x^{1}+K_{0}}$ (4.4)

The sum rule tells us that we are able to add these equations, (4.3) and (4.4), to obtain a final solution for ${\displaystyle \displaystyle y_{p}(x)}$. The solution for this is as follows:

${\displaystyle \displaystyle k_{2}x^{2}+k_{1}x^{1}+k_{0}+K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x^{1}+K_{0}}$

${\displaystyle \displaystyle K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+(K_{2}+k_{2})x^{2}+(K_{1}+k_{1})x^{1}+(K_{1}+k_{0})}$ (4.5)

Since these k and K values are just constants, we can set:

${\displaystyle \displaystyle K_{5}=c_{5}}$

${\displaystyle \displaystyle K_{4}=c_{4}}$

${\displaystyle \displaystyle K_{3}=c_{3}}$

${\displaystyle \displaystyle (K_{2}+k_{2})=c_{2}}$

${\displaystyle \displaystyle (K_{1}+k_{1})=c_{1}}$

${\displaystyle \displaystyle (K_{1}+k_{0})=c_{0}}$

Our FINAL solution for the ${\displaystyle \displaystyle y_{p}(x)}$ of (4.1) is:

${\displaystyle \displaystyle y_{p}(x)=c_{5}x^{5}+c_{4}x^{4}+c_{3}x^{3}+c_{2}x^{2}+c_{1}x^{1}+c_{0}}$ (4.6)

And, as was the original problem statement, equation (4.6) is of the form:

${\displaystyle \displaystyle y_{p}(x)=\sum _{j=0}^{5}c_{j}x^{j}}$