# University of Florida/Egm4313/s12.team11.perez.gp/R3.9
#### Problem Statement

Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

#### Given
#### (K 2011 pg.85 #13)

$8y''-6y'+y=6\cosh x\!$ (1)
Initial conditions are:
$y(0)=0.2,y'(0)=0.05\!$

#### Solution

The general solution of the homogeneous ordinary differential equation is
$8y''-6y'+y=0\!$
We can use this information to determine the characteristic equation:
$8\lambda ^{2}-6\lambda +1=0\!$
And proceeding to find the roots,
$4\lambda (2\lambda -1)-1(2\lambda -1)=0\!$
Thus, $(4\lambda -1)(2\lambda -1)=0\!$.
Solving for the roots, we find that $\lambda ={\frac {1}{4}},{\frac {1}{2}},\!$
where the general solution is
$y_{k}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}\!$.
The solution of $y_{p}\!$ of the non-homogeneous ordinary differential equation is
$x={\frac {e^{x}+e^{-x}}{2}}\!$.
Using the Sum rule as described in Section 2.7, the above function translates into the following:
$y_{p}=y_{p1}+y_{p2}\!$, where Table 2.1 tells us that:
$y_{p1}=Ae^{x}\!$ and $y_{p2}=Be^{x}\!$.
Therefore, $y_{p}=Ae^{x}+Be^{-x}\!$.
Now, we can substitute the values ($y_{p},y_{p}',y_{p}''\!$) into (1) to get:
$8(Ae^{x}+Be^{-x})-6(Ae^{x}-Be^{-x})+Ae^{x}+Be^{-x}=6({\frac {e^{x}+e^{-x}}{2}})\!$
$=3Ae^{x}+15Be^{-x}\!$
$=3(e^{x}+e^{-x})\!$
Now that we have this equation, we can equate coefficients to find that:
$3A=3\!$
$\therefore A=1\!$
$B={\frac {3}{15}}={\frac {1}{5}}\$
and thus, $y_{p}=e^{x}+{\frac {1}{5}}e^{-x}\!$
We find that the general solution is in fact:
$y=y_{k}+y_{p}\!$
$y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+3e^{-x}\!$
whereas the general solution of the given ordinary differential equation is actually:
$y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+{\frac {1}{5}}e^{-x}\!$
Solving for the initial conditions given and first plugging in $y(0)=0.2\!$, we get that:
$0.2=c_{1}e^{{\frac {1}{4}}(0)}+c_{2}e^{{\frac {1}{2}}(0)}+e^{0}+3e^{(0)}\!$
$0.2=c_{1}e^{(0)}+c_{2}e^{(0)}+e^{(0)}+3e^{(0)}\!$
$0.2=c_{1}+c_{2}+1+{\frac {1}{5}}\!$
$\therefore c_{1}+c_{2}=-1\!$. (2)
And now we can determine the first order ODE :
$y'={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(x)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(x)}+e^{x}-{\frac {1}{5}}e^{x}\!$
The second initial condition that was given to us, $y'(0)=0.05\!$ can now be plugged in:
$0.05={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(0)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!$
$0.05={\frac {1}{4}}c_{1}e^{(0)}+{\frac {1}{2}}c_{2}e^{(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!$
$0.05={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}+1-{\frac {1}{5}}\!$
${\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}=-0.75\!$
$\therefore c_{1}+2c_{2}=-3\!$ (3)
Once we solve (2) and (3), we can get the values:
$c_{1}=1,c_{2}=-2\!$.
And once we substitute these values, we get the following solution for this IVP:
$y=e^{{\frac {1}{4}}x}-2e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}\!$

#### Given
#### (K 2011 pg.85 #14)

$y''+4y'+4y=e^{-2x}sin2x\!$ (1)
Initial conditions are:
$y(0)=1,y'(0)=-1.5\!$

#### Solution

The general solution of the homogeneous ordinary differential equation is
$y''+4y'+4y=0\!$
We can use this information to determine the characteristic equation:
$\lambda ^{2}+4\lambda +4=0\!$
And proceeding to find the roots,
$(\lambda +2)(\lambda +2)=0\!$
Solving for the roots, we find that $\lambda =-2,-2\!$
where the general solution is:
$y_{k}=c_{1}e^{-2x}+c_{2}e^{-2x}x\!$, or:
$y_{k}=(c_{1}+c_{2}x)e^{-2x}\!$
Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:
$y_{p}=e^{-2x}(Kxcos2x+Mxsin2x)\!$, since the solution of $y_{k}\!$ is a double root of the characteristic equation.
We can then derive to get $y_{p}'\!$:
$y_{p}'=-2e^{-2x}(Kxcos2x+Mxsin2x)+e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!$
$y_{p}'=(-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-}2xxsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx\!$
Deriving once again to solve for $y_{p}''\!$, we get the following:
$y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!$
$y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-4e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)+e^{-2x}(-4Ksin2x-4Kxcos2x+4Mcos2x-4Mxsin2x)\!$
$y_{p}''=(-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x\$
Now, we can substitute the values ($y_{p},y_{p}',y_{p}''\!$) into (1) to get:
$(-4K+4M)e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx)+4(e^{-2x}(Kxcos2x+Mxsin2x))=e^{2x}sin2x\!$
$\therefore (-3K+4M)e^{-2x}cos2x+(-3M-4K)e^{-2x}sin2x=e^{-2x}sin2x\!$
Now that we have this equation, we can equate coefficients to find that:
$-3K+4M=0\!$ and $-4K-3M=1\!$
and finally discover that:
$M=-{\frac {3}{25}}\!$ and $K=-{\frac {4}{25}}\!$.
Plugging in these values in $y_{p}\!$, we find that:
$y_{p}=e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$
And finally, we arrive at the general solution of the given ordinary differential equation:
$y=y_{k}+y_{p}\!$
$y=(c_{1}+c_{2}x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$
Solving for the initial conditions given and first plugging in $y(0)=1\!$, we get that:
$1=(c_{1}+c_{2}(0))e^{-2(0)}+e^{-2(0)}(-{\frac {4}{25}}(0)cos2(0)-{\frac {3}{25}}(0)sin2(0))\!$
$1=(c_{1}+c_{2}(0))e^{0}\!$
$\therefore c_{1}=1\!$
The second initial condition that was given to us, $y'(0)=-1.5\!$ can now be plugged in:
$y'=-{\frac {1}{5}}e^{-}2x(10c_{1}+10c_{2}x-5c_{2}+(3-14x)sin2x+(4-2x)cos(2x))\!$
$-1.5=-{\frac {1}{5}}(10c_{1}-5c_{2}+4)\!$
$\therefore c_{2}=-3.5\!$
And once we substitute these values, we get the following solution for this IVP:
$y=(1-3.5x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$