User:Egm4313.s12.team8.koester/R3.5

=R3.5 - Particular solution to L2-ODE-CC from series=

Problem Statement
Complete the solution for equation (2) p 7-11 as follows:

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

Step 1
Obtain equations for 2, 3, 4, and 6 from p 7-14

-coefficients of $$ \displaystyle x $$

-coefficients of $$ \displaystyle x^{2} $$

-coefficients of $$ \displaystyle x^{3} $$

-coefficients of $$ \displaystyle x^{5} $$

Use equation (2) from p 7-13:

$$ \ \sum_{j = 0}^{3} \left [ c_{j+2} (j+2) (j+1) - 3c_{j+1} (j+1) + 2c_j\right] x^j - 3c_5(5)x^4 + 2 \left [c_4x^4 + c_5x^5 \right ] = 4x^2 - 6x^5\ $$

For the coefficients of x, let j=1 to get:

$$ \displaystyle (c_3(3)(2)-3c_2(2)+2c_1)x=0 $$

Reducing to:

$$ \displaystyle 6c_3-6c_2+2c_1=0 $$

For the coefficients of x^{2}, let j=2 to get:

$$ \displaystyle (c_4(4)(3)-3c_3(3)+2c_2)x^{2}=4x^{2} $$

Reducing to:

$$ \displaystyle 12c_4-9c_3+2c_2=4 $$

For the coefficients of x^{3}, let j=3 to get:

$$ \displaystyle (c_5(5)(4)-3c_4(4)+2c_3)x^{3}=0 $$

Reducing to:

$$ \displaystyle 20c_5-12c_4+2c_3=0 $$

For the coefficients of x^{5}, let j=5 to get:

$$ \displaystyle 2c_5x^{5}=-6x^{5} $$

Reducing to:

$$ \displaystyle 2c_5=-6 $$

Step 2
Verify all equations by long-hand expansion of the series in equation (4) p 7-12:

$$ \ \sum_{j = 2}^{5} c_j(j)(j-1)(x^{j-2}) - 3\sum_{j = 1}^{5} c_j(j)(x^{j-1}) + 2\sum_{j = 0}^{5} c_jx^j = 4x^2 - 6x^5\ $$

Expanding these series the following terms are obtained:

$$ \displaystyle c_2(2)(1)+c_3(3)(2)x+c_4(4)(3)x^2+c_5(5)(4)x^3-3c_1(1)-3c_2(2)x-3c_3(3)x^2-3c_4(4)x^3-3c_5(5)x^4 $$

$$ \displaystyle +2c_0+2c_1x+2c_2x^2+2c_3x^3+2c_4x^4+2c_5x^5= 4x^2-6x^5  $$

Collecting like coefficients the equation can be rearranged to:

$$ \displaystyle (2c_0-3c_1+2c_2)+(2c_1-6c_2+6c_3)x+(2c_2-9c_3+12c_4)x^2+(2c_3-12c_4+20c_5)x^3+(2c_4-15c_5)x^4+(2c_5)x^5=4x^2-6x^5  $$

Examining this equation it can be seen that the coefficients match those previously found so the equations are verified.

Step 3
Put the system of equations into matrix form:

$$ \ \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 \\ 0 & 0 & 2 & -9 & 12 & 0 \\ 0 & 0 & 0 & 2 & -12 & 20 \\ 0 & 0 & 0 & 0 & 2 & -15 \\ 0 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 4 \\ 0 \\ 0 \\ -6 \end{bmatrix} $$

Step 4
Solve for the coefficients by back substitution.

This is done by first solving for $$ \displaystyle c_5 $$ and then $$ \displaystyle c_4 $$ and eventually all the way back to $$ \displaystyle c_0 $$

$$ \displaystyle 2c_5=-6 $$

$$ \displaystyle c_5=-3 $$

$$ \displaystyle -15c_5+2c_4=0 $$

$$ \displaystyle 2c_4=-45 $$

$$ \displaystyle c_4=-22.5 $$

$$ \displaystyle 20c_5-12c_4+2c_3=0 $$

$$ \displaystyle 2c_3=-210 $$

$$ \displaystyle c_3=-105 $$

$$ \displaystyle 12c_4-9c_3+2c_2=4 $$

$$ \displaystyle 2c_2=-671 $$

$$ \displaystyle c_2=-335.5 $$

$$ \displaystyle 6c_3-6c_2+2c_1=0 $$

$$ \displaystyle 2c_1=-1383 $$

$$ \displaystyle c_1=-691.5 $$

$$ \displaystyle 2c_2-3c_1+2c_0=0 $$

$$ \displaystyle 2c_0=-1403 $$

$$ \displaystyle c_0=-701.75 $$

Step 5
Consider the initial conditions:

$$ \displaystyle y(0)=1 $$

$$ \displaystyle y'(0)=0 $$

Find the solution y(x) and plot it

$$ \displaystyle y(x)= y_p + y_h $$

$$ \displaystyle y_p= c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + c_5x^5 $$

So for our equation:

$$ \displaystyle y_p=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5 $$

$$ \displaystyle y_h= C_1e^{\lambda_1x}+C_2e^{\lambda_2x} $$

where:

$$ \displaystyle \lambda_1=1 $$ and $$ \displaystyle \lambda_2=2 $$ from quadratic equation.

So the overall solution becomes:

$$ \displaystyle y(x)= -701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5+C_1e^{x}+C_2e^{2x} $$

Now use the given initial conditions to solve for the constants:

$$ \displaystyle y(0)=1=-701.75 + C_1 + C_2 $$

$$ \displaystyle y'(0)=0=-691.5 + C_1 + 2C_2 $$

Cleaning up to:

$$ \displaystyle 702.75 = C_1 + C_2 $$

$$ \displaystyle 691.5 = C_1 + 2C_2 $$

Subtracting the first equation from the second the constants can then be solved for.

$$ \displaystyle C_2=-11.25 $$

$$ \displaystyle C_1=714 $$

Yielding the final complete solution of:

$$ \displaystyle y(x)= -701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5+714e^{x}-11.25e^{2x} $$