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

Problem Statement

Find the complete solution for $y^{″}-3y^{\prime }+2y=4x^2$ (3.1) with the initial conditions $y(0)=1,y^{\prime }(0)=0$. Plot the solution $y(x)$.

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:

$\frac{3\pm \sqrt{9-4(1)(2)}}{2(1)}$

This gives us the solutions for the roots of equation (3.1) to be

${\lambda }_1=2,{\lambda }_2=1$

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:

$y_h=c_1{e}^{2x}+c_2{e}^x$ (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

$y_p(x)=c_0+c_{1}x+c_2{x}^2$ (3.3)

We also need the first and second derivatives of (3.3), and solving for these gives us:

$y_{p^{\prime }}(x)=c_1+2c_{2}x$ (3.4) and $y_{p^{″}}(x)=2c_2$ (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:

$2c_2-3(c_1+2c_{2}x)+2(c_0+c_{1}x+c_2{x}^2)=(0*1)+(0*x)+4x^2$

Simplifying this equation...

$2c_2-3c_1-6c_{2}x+2c_0+2c_{1}x+2c_2{x}^2=4x^2$ (3.6)

Creating equations for the non-coefficient terms, along with the coefficients of x and x squared terms, respectively, we get:

$2c_0-3c_1+2c_2=0$ (3.7)

$0(c_0)+2c_1-6c_2=0$ (3.8)

$0(c_0)+0(c_1)+2c_2=4$ (3.9)

In matrix form of Ac=d, these would appear as:

$\left[\begin{array}{ccc}2& -3& 2\\ 0& 2& -6\\ 0& 0& 2\end{array}\right]\left\{\begin{array}{c}{c}_0\\ c_1\\ c_2\end{array}\right\}=\left\{\begin{array}{c}0\\ 0\\ 4\end{array}\right\}$

Using this, we can solve for all three c constants, as shown:

$2c_2=4$ which leads to $c_2=2$

$2c_1-(6)(2)=0$ which leads to $2c_1=12$ and finally $c_1=6$

$2c_0-(3)(6)+(2)(2)=0$ which leads to $2c_0=14$ and finally $c_0=7$

Plugging these constants back into (3.3) gives us the final particular solution of (3.1):

$y_p(x)=2x^2+6x+7$ (3.10)

To find the general solution of (3.1), we combine the particular (3.10) and homogenous (3.2) equations to get:

$y(x)=2x^2+6x+7+c_1{e}^{2x}+c_2{e}^x$ (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:

$y{\text{'}}_p=4x+6+2c_1{e}^{2x}+c_2{e}^x$ (3.12)

Using the initial conditions in (3.11) and (3.12) allows us to obtain the following:

$y_p(0)=1=7+6(0)+2(0{)}^2+c_1{e}^0+c_2{e}^0$

which leads to $1=7+c_1+c_2$

and finally gives us $-6=c_1+c_2$ (3.13)

$y{\text{'}}_p(0)=0=6+4(0)+2c_1{e}^0+c_2{e}^0$

Which leads to $0=6+c_1+c_2$

And finally comes out to $-6=2c_1+c_2$ (3.14) Solving for $c_1$ in equation (3.13) and plugging that into equation (3.14) allows us to solve for $c_2$:

$-6=2(-c_2-6)+c_2$

$-6=-2c_2+12+c_2$

$c_2=-6$ (3.15)

Plugging (3.15) into (3.13) allows us to solve for the remaining constant:

$-6=c_1-6$

$0=c_1$ (3.16)

Plugging (3.15) and (3.16) into (3.11) gives us our final general solution of:

$y(x)=2x^2+6x+7-6e^x$     (3.17)

Plot

The plot of equation (3.17) is shown below: