University of Florida/Egm4313/s12.team8.dupre/R3.3

Problem Statement
Find the complete solution for $$\displaystyle y''-3y'+2y=4x^2 $$ (3.1) with the initial conditions $$\displaystyle y(0)=1,y'(0)=0 $$. Plot the solution $$\displaystyle y(x) $$.

Solution
First, we set r(x), or the right side of the given equation, equal to 0, to solve for the homogenous equation. We solve this by plugging into the quadratic formula, which gives us: $$\displaystyle \frac{3\pm \sqrt{9-4(1)(2)}}{2(1)} $$ This gives us the solutions for the roots of equation (3.1) to be $$\displaystyle \lambda _{1}=2,\lambda _{2}=1 $$ Using these roots and the fact that they are both real and positive, we know that the homogenous equation of (3.1) is equal to: $$\displaystyle y_{h}=c_{1}e^{2x}+c_{2}e^{x} $$ (3.2) The next step in this solution is to solve for the particular solution of (3.1). Using the Basic Rule and (3.2), we find that $$\displaystyle y_{p}(x)=c_{0}+c_{1}x+c_{2}x^2 $$ (3.3) We also need the first and second derivatives of (3.3), and solving for these gives us: $$\displaystyle y_{p}'(x)=c_{1}+2c_{2}x $$  (3.4) and $$\displaystyle y_{p}''(x)=2c_{2} $$   (3.5) To solve for the unknown c constants in (3.3),(3.4), and (3.5), we plug these equations into (3.1), giving us: $$\displaystyle 2c_{2}-3(c_{1}+2c_{2}x)+2(c_{0}+c_{1}x+c_{2}x^2)=(0*1)+(0*x)+4x^2 $$ Simplifying this equation... $$\displaystyle 2c_{2}-3c_{1}-6c_{2}x+2c_{0}+2c_{1}x+2c_{2}x^2=4x^2 $$   (3.6) Creating equations for the non-coefficient terms, along with the coefficients of x and x squared terms, respectively, we get: $$\displaystyle 2c_{0}-3c_{1}+2c_{2}=0 $$   (3.7) $$\displaystyle 0(c_{0})+2c_{1}-6c_{2}=0 $$  (3.8) $$\displaystyle 0(c_{0})+0(c_{1})+2c_{2}=4$$  (3.9) In matrix form of Ac=d, these would appear as: $$\displaystyle \begin{bmatrix} 2 & -3 & 2\\ 0& 2 & -6\\  0& 0 & 2 \end{bmatrix}\begin{Bmatrix} c_{0}\\c_{1} \\c_{2} \end{Bmatrix}=\begin{Bmatrix} 0\\ 0\\ 4 \end{Bmatrix}$$ Using this, we can solve for all three c constants, as shown: $$\displaystyle 2c_{2}=4 $$ which leads to $$\displaystyle c_{2}=2 $$ $$\displaystyle 2c_{1}-(6)(2)=0 $$ which leads to  $$\displaystyle 2c_{1}=12 $$ and finally $$\displaystyle c_{1}=6 $$ $$\displaystyle 2c_{0}-(3)(6)+(2)(2)=0 $$ which leads to $$\displaystyle 2c_{0}=14 $$ and finally $$\displaystyle c_{0}=7 $$ Plugging these constants back into (3.3) gives us the final particular solution of (3.1): $$\displaystyle y_{p}(x)=2x^2+6x+7 $$   (3.10) To find the general solution of (3.1), we combine the particular (3.10) and homogenous (3.2) equations to get: $$\displaystyle y(x)=2x^2+6x+7+c_{1}e^{2x}+c_{2}e^{x} $$   (3.11) To solve for the final two constants, we use the given initial conditions in (3.11) and the derivative of that, which is: $$\displaystyle y'_{p}=4x+6+2c_{1}e^{2x}+c_{2}e^{x} $$   (3.12) Using the initial conditions in (3.11) and (3.12) allows us to obtain the following: $$\displaystyle y_{p}(0)=1=7+6(0)+2(0)^2+c_{1}e^{0}+c_{2}e^{0} $$ which leads to $$\displaystyle 1=7+c_{1}+c_{2} $$ and finally gives us $$\displaystyle -6=c_{1}+c_{2} $$    (3.13) $$\displaystyle y'_{p}(0)=0=6+4(0)+2c_{1}e^{0}+c_{2}e^{0} $$ Which leads to $$\displaystyle 0=6+c_{1}+c_{2} $$ And finally comes out to $$\displaystyle -6=2c_{1}+c_{2} $$    (3.14) Solving for $$\displaystyle c_{1} $$ in equation (3.13) and plugging that into equation (3.14) allows us to solve for $$\displaystyle c_{2} $$: $$\displaystyle -6=2(-c_{2}-6)+c_{2} $$ $$\displaystyle -6=-2c_{2}+12+c_{2} $$ $$\displaystyle c_{2}=-6 $$    (3.15) Plugging (3.15) into (3.13) allows us to solve for the remaining constant: $$\displaystyle -6=c_{1}-6 $$ $$\displaystyle 0=c_{1} $$    (3.16) Plugging (3.15) and (3.16) into (3.11) gives us our final general solution of: $$\displaystyle y(x)=2x^2+6x+7-6e^{x} $$    (3.17)

Plot
The plot of equation (3.17) is shown below: