# 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.

#### (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:

$y=e^{{\frac {1}{4}}x}-2e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}\!$ #### (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:

$y=(1-3.5x)e^{-2x}+e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!$ 