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)

${\displaystyle 8y''-6y'+y=6\cosh x\!}$ (1)

Initial conditions are:

${\displaystyle y(0)=0.2,y'(0)=0.05\!}$

Solution

The general solution of the homogeneous ordinary differential equation is

${\displaystyle 8y''-6y'+y=0\!}$

We can use this information to determine the characteristic equation:

${\displaystyle 8\lambda ^{2}-6\lambda +1=0\!}$

And proceeding to find the roots,

${\displaystyle 4\lambda (2\lambda -1)-1(2\lambda -1)=0\!}$

Thus, ${\displaystyle (4\lambda -1)(2\lambda -1)=0\!}$.

Solving for the roots, we find that ${\displaystyle \lambda ={\frac {1}{4}},{\frac {1}{2}},\!}$

where the general solution is

${\displaystyle y_{k}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}\!}$.

The solution of ${\displaystyle y_{p}\!}$ of the non-homogeneous ordinary differential equation is

${\displaystyle x={\frac {e^{x}+e^{-x}}{2}}\!}$.

Using the Sum rule as described in Section 2.7, the above function translates into the following:

${\displaystyle y_{p}=y_{p1}+y_{p2}\!}$, where Table 2.1 tells us that:

${\displaystyle y_{p1}=Ae^{x}\!}$ and ${\displaystyle y_{p2}=Be^{x}\!}$.

Therefore, ${\displaystyle y_{p}=Ae^{x}+Be^{-x}\!}$.

Now, we can substitute the values (${\displaystyle y_{p},y_{p}',y_{p}''\!}$) into (1) to get:

${\displaystyle 8(Ae^{x}+Be^{-x})-6(Ae^{x}-Be^{-x})+Ae^{x}+Be^{-x}=6({\frac {e^{x}+e^{-x}}{2}})\!}$

${\displaystyle =3Ae^{x}+15Be^{-x}\!}$

${\displaystyle =3(e^{x}+e^{-x})\!}$

Now that we have this equation, we can equate coefficients to find that:

${\displaystyle 3A=3\!}$

${\displaystyle \therefore A=1\!}$

${\displaystyle B={\frac {3}{15}}={\frac {1}{5}}\ }$

and thus, ${\displaystyle y_{p}=e^{x}+{\frac {1}{5}}e^{-x}\!}$

We find that the general solution is in fact:

${\displaystyle y=y_{k}+y_{p}\!}$

${\displaystyle y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+3e^{-x}\!}$

whereas the general solution of the given ordinary differential equation is actually:

${\displaystyle y=c_{1}e^{\frac {1}{4}}x+c_{2}e^{\frac {1}{2}}x+e^{x}+{\frac {1}{5}}e^{-x}\!}$

Solving for the initial conditions given and first plugging in ${\displaystyle y(0)=0.2\!}$, we get that:

${\displaystyle 0.2=c_{1}e^{{\frac {1}{4}}(0)}+c_{2}e^{{\frac {1}{2}}(0)}+e^{0}+3e^{(0)}\!}$

${\displaystyle 0.2=c_{1}e^{(0)}+c_{2}e^{(0)}+e^{(0)}+3e^{(0)}\!}$

${\displaystyle 0.2=c_{1}+c_{2}+1+{\frac {1}{5}}\!}$

${\displaystyle \therefore c_{1}+c_{2}=-1\!}$. (2)

And now we can determine the first order ODE :

${\displaystyle y'={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(x)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(x)}+e^{x}-{\frac {1}{5}}e^{x}\!}$

The second initial condition that was given to us, ${\displaystyle y'(0)=0.05\!}$ can now be plugged in:

${\displaystyle 0.05={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}(0)}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!}$

${\displaystyle 0.05={\frac {1}{4}}c_{1}e^{(0)}+{\frac {1}{2}}c_{2}e^{(0)}+e^{(0)}-{\frac {1}{5}}e^{(0)}\!}$

${\displaystyle 0.05={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}+1-{\frac {1}{5}}\!}$

${\displaystyle {\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}=-0.75\!}$

${\displaystyle \therefore c_{1}+2c_{2}=-3\!}$ (3)

Once we solve (2) and (3), we can get the values:

${\displaystyle c_{1}=1,c_{2}=-2\!}$.

And once we substitute these values, we get the following solution for this IVP:

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

(K 2011 pg.85 #14)

${\displaystyle y''+4y'+4y=e^{-2x}sin2x\!}$ (1)

Initial conditions are:

${\displaystyle y(0)=1,y'(0)=-1.5\!}$

Solution

The general solution of the homogeneous ordinary differential equation is

${\displaystyle y''+4y'+4y=0\!}$

We can use this information to determine the characteristic equation:

${\displaystyle \lambda ^{2}+4\lambda +4=0\!}$

And proceeding to find the roots,

${\displaystyle (\lambda +2)(\lambda +2)=0\!}$

Solving for the roots, we find that ${\displaystyle \lambda =-2,-2\!}$

where the general solution is:

${\displaystyle y_{k}=c_{1}e^{-2x}+c_{2}e^{-2x}x\!}$, or:

${\displaystyle y_{k}=(c_{1}+c_{2}x)e^{-2x}\!}$

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:

${\displaystyle y_{p}=e^{-2x}(Kxcos2x+Mxsin2x)\!}$, since the solution of ${\displaystyle y_{k}\!}$ is a double root of the characteristic equation.

We can then derive to get ${\displaystyle y_{p}'\!}$:

${\displaystyle y_{p}'=-2e^{-2x}(Kxcos2x+Mxsin2x)+e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!}$

${\displaystyle y_{p}'=(-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-}2xxsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx\!}$

Deriving once again to solve for ${\displaystyle y_{p}''\!}$, we get the following:

${\displaystyle y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)-2e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)\!}$

${\displaystyle y_{p}''=4e^{-2x}(Kxcos2x+Mxsin2x)-4e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)+e^{-2x}(-4Ksin2x-4Kxcos2x+4Mcos2x-4Mxsin2x)\!}$

${\displaystyle y_{p}''=(-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x\ }$

Now, we can substitute the values (${\displaystyle y_{p},y_{p}',y_{p}''\!}$) into (1) to get:

${\displaystyle (-4K+4M)e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx)+4(e^{-2x}(Kxcos2x+Mxsin2x))=e^{2x}sin2x\!}$

${\displaystyle \therefore (-3K+4M)e^{-2x}cos2x+(-3M-4K)e^{-2x}sin2x=e^{-2x}sin2x\!}$

Now that we have this equation, we can equate coefficients to find that:

${\displaystyle -3K+4M=0\!}$ and ${\displaystyle -4K-3M=1\!}$

and finally discover that:

${\displaystyle M=-{\frac {3}{25}}\!}$ and ${\displaystyle K=-{\frac {4}{25}}\!}$.

Plugging in these values in ${\displaystyle y_{p}\!}$, we find that:

${\displaystyle y_{p}=e^{-2x}(-{\frac {4}{25}}xcos2x-{\frac {3}{25}}xsin2x)\!}$

And finally, we arrive at the general solution of the given ordinary differential equation:

${\displaystyle y=y_{k}+y_{p}\!}$

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

Solving for the initial conditions given and first plugging in ${\displaystyle y(0)=1\!}$, we get that:

${\displaystyle 1=(c_{1}+c_{2}(0))e^{-2(0)}+e^{-2(0)}(-{\frac {4}{25}}(0)cos2(0)-{\frac {3}{25}}(0)sin2(0))\!}$

${\displaystyle 1=(c_{1}+c_{2}(0))e^{0}\!}$

${\displaystyle \therefore c_{1}=1\!}$

The second initial condition that was given to us, ${\displaystyle y'(0)=-1.5\!}$ can now be plugged in:

${\displaystyle y'=-{\frac {1}{5}}e^{-}2x(10c_{1}+10c_{2}x-5c_{2}+(3-14x)sin2x+(4-2x)cos(2x))\!}$

${\displaystyle -1.5=-{\frac {1}{5}}(10c_{1}-5c_{2}+4)\!}$

${\displaystyle \therefore c_{2}=-3.5\!}$

And once we substitute these values, we get the following solution for this IVP:

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