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)

$8y^{″}-6y^{\prime }+y=6\cosh x$ (1)

Initial conditions are:

$y(0)=0.2,y^{\prime }(0)=0.05$

Solution

The general solution of the homogeneous ordinary differential equation is

$8y^{″}-6y^{\prime }+y=0$

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:

$y_{p1}=A{e}^x$ and $y_{p2}=B{e}^x$.

Therefore, $y_p=A{e}^x+B{e}^{-x}$.

Now, we can substitute the values ($y_p,{y_p}^{\prime },{y_p}^{″}$) into (1) to get:

$8(A{e}^x+B{e}^{-x})-6(A{e}^x-B{e}^{-x})+A{e}^x+B{e}^{-x}=6(\frac{e^x+e^{-x}}{2})$

$=3A{e}^x+15B{e}^{-x}$

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

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 :

$y^{\prime }=\frac{1}{4}{c}_1{e}^{\frac{1}{4}(x)}+\frac{1}{2}{c}_2{e}^{\frac{1}{2}(x)}+e^x-\frac{1}{5}{e}^x$

The second initial condition that was given to us, $y^{\prime }(0)=0.05$ 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)

$y^{″}+4y^{\prime }+4y=e^{-2x}sin2x$ (1)

Initial conditions are:

$y(0)=1,y^{\prime }(0)=-1.5$

Solution

The general solution of the homogeneous ordinary differential equation is

$y^{″}+4y^{\prime }+4y=0$

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 ${y_p}^{\prime }$:

${y_p}^{\prime }=-2e^{-2x}(Kxcos2x+Mxsin2x)+e^{-2x}(Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x)$

${y_p}^{\prime }=(-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-}2xxsin2x+K{e}^{-2x}cos2x+M{e}^{-2x}sinx$

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 ($y_p,{y_p}^{\prime },{y_p}^{″}$) into (1) to get:

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

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

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, $y^{\prime }(0)=-1.5$ can now be plugged in:

$y^{\prime }=-\frac{1}{5}{e}^{-}2x(10c_1+10c_{2}x-5c_2+(3-14x)sin2x+(4-2x)cos(2x))$

$-1.5=-\frac{1}{5}(10c_1-5c_2+4)$

$\therefore c_2=-3.5$

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

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