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

Problem Statement

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

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

is of the form

$y_p(x)=\sum _{j=0}^n{c}_j{x}^j$

with n=5, i.e,

$y_p(x)=\sum _{j=0}^5{c}_j{x}^j$.

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 $r(x)$ , you can find what $y_p(x)$ you should choose to use.

From (4.1), we know that:

$r(x)=4x^2-6x^5$ (4.2)

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

The $y_p(x)$ for $4x^2$ is:

$k_2{x}^2+k_1{x}^1+k_0$ (4.3)

And the $y_p(x)$ for $6x^5$ is:

$K_5{x}^5+K_4{x}^4+K_3{x}^3+K_2{x}^2+K_1{x}^1+K_0$ (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 $y_p(x)$. The solution for this is as follows:

$k_2{x}^2+k_1{x}^1+k_0+K_5{x}^5+K_4{x}^4+K_3{x}^3+K_2{x}^2+K_1{x}^1+K_0$

Adding similar variables gives us:

$K_5{x}^5+K_4{x}^4+K_3{x}^3+(K_2+k_2)x^2+(K_1+k_1)x^1+(K_1+k_0)$ (4.5)

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

$K_5=c_5$

$K_4=c_4$

$K_3=c_3$

$(K_2+k_2)=c_2$

$(K_1+k_1)=c_1$

$(K_1+k_0)=c_0$

Our FINAL solution for the $y_p(x)$ of (4.1) is:

$y_p(x)=c_5{x}^5+c_4{x}^4+c_3{x}^3+c_2{x}^2+c_1{x}^1+c_0$ (4.6)

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

$y_p(x)=\sum _{j=0}^5{c}_j{x}^j$