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

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Problem 3.4

## Problem Statement

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

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

is of the form

$\displaystyle y_{p}(x)=\sum _{j=0}^{n}c_{j}x^{j}$ with n=5, i.e,

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

### 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 $\displaystyle r(x)$ , you can find what $\displaystyle y_{p}(x)$ you should choose to use.

From (4.1), we know that:

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

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

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

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

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

$\displaystyle K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x^{1}+K_{0}$ (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 $\displaystyle y_{p}(x)$ . The solution for this is as follows:

$\displaystyle k_{2}x^{2}+k_{1}x^{1}+k_{0}+K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x^{1}+K_{0}$ Adding similar variables gives us:

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

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

$\displaystyle K_{5}=c_{5}$ $\displaystyle K_{4}=c_{4}$ $\displaystyle K_{3}=c_{3}$ $\displaystyle (K_{2}+k_{2})=c_{2}$ $\displaystyle (K_{1}+k_{1})=c_{1}$ $\displaystyle (K_{1}+k_{0})=c_{0}$ Our FINAL solution for the $\displaystyle y_{p}(x)$ of (4.1) is:

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

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

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