User:Egm4313.s12.team8.lucas/R3.9

=R3.9=

Problem Statement
Find the solution the L2-ODE-CC using the given initial values.

Number 13) $$\ 8y''-6y'+y=6coshx$$  y(0)=0.2,  y'(0)=0.05

Number 14) $$\ y''+4y'+4y=e^{-2x}sin2x$$  y(0)=1,  y'(0)=-1.5

Number 13
$$\ 8y''-6y'+y=6coshx$$

$$\ 6coshx=6\frac{e^{x}+e^{-x}}{2}=3e^{x}+3e^{-x}$$

$$\ 8y''-6y'+y=3e^{x}+3e^{-x}$$

The homogeneous solution will take the form:

$$\ y_{h}(x)=c_{1}e^{\lambda_{1}x}+c_{2}e^{\lambda_{2}x}$$

To find the homogeneous solutions, set r(x)=0:

$$\ 8y''-6y'+y=0$$

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

$$\ (2\lambda-1)(4\lambda-1)=0$$

$$\ \lambda_{1}=\frac{1}{2},   \lambda_{2}=\frac{1}{4}$$

$$\ y_{h}(x)=c_{1}e^{x/2}+c_{2}e^{x/4}$$

To find the particular solution, we must find the form $$\ y_{p}(x)$$ will take based on r(x):

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

From table 2.1 p. 82, the form of $$\ y_{p}(x)$$ will be:

$$\ y_{p}(x)=Me^{x}+Ne^{-x}$$

Differentiating $$\ y_{p}(x)$$ gives:

$$\ y'_{p}(x)=Me^{x}-Ne^{-x}$$

Differentiating $$\ y'_{p}(x)$$ gives:

$$\ y''_{p}(x)=Me^{x}+Ne^{-x}$$

Then, we substitute $$\ y_{p}(x)$$, $$\ y'_{p}(x)$$, $$\ y''_{p}(x)$$ into the original ODE:

$$\ 8[Me^{x}+Ne^{-x}]-6[Me^{x}-Ne^{-x}]+[Me^{x}+Ne^{-x}]=r(x)=3e^{x}+3e^{-x}$$

By collecting like terms, we get:

$$\ 3Me^{x}+15Ne^{-x}=3e^{x}+3e^{-x}$$

By equating like terms, we get:

$$\ 3Me^{x}=3e^{x}  =>  M=1$$

$$\ 15Ne^{-x}=3e^{-x}   => N=\frac{1}{5}$$

By substituting M and N into the general form for the particular solution we get:

$$\ y_{p}(x)=e^{x}+\frac{1}{5}e^{-x}$$

To find the final solution to the ODE, we add the homogeneous and particular solutions:

$$\ y(x)=y_{h}(x)+y_{p}(x)$$

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

To find the unknown coefficients, we apply the initial conditions:

$$\ y(0)=0.2, y'(0)=0.05$$

$$\ y(0)=0.2=c_{1}e^{0}+c_{2}e^{0}+e^{0}+\frac{1}{5}e^{0}=c_{1}+c_{2}+1.2$$

$$\ c_{1}+c_{2}=-1$$ (Eqn. 1)
 * 

Then we find the derivative of the general solution:

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

Then we apply initial conditions:

$$\ y'(0)=0.05=\frac{1}{2}c_{1}e^{0}+\frac{1}{4}c_{2}e^{0}+e^{0}-\frac{1}{5}e^{0}=\frac{1}{2}c_{1}+\frac{1}{4}c_{2}+0.8$$

$$\ \frac{1}{2}c_{1}+\frac{1}{4}c_{2}=-0.75$$ (Eqn. 2)
 * 

By substituting Equation 1 into Equation 2, we get:

$$\ 0.5c_{1}+0.25(-1-c_{1})=-0.75  =>  c_{1}=-2$$

$$\ (-2)+c_{2}=-1 =>  c_{2}=1$$

By substituting the known coefficients into the general solution, we get the final solution:

$$\ y(x)=-2e^{x/2}+e^{x/4}+e^{x}+\frac{1}{5}e^{-x}$$

Number 14
$$\ y''+4y'+4y=e^{-2x}sin2x$$

The homogeneous solution will take the form:

$$\ y_{h}(x)=c_{1}e^{\lambda_{1}x}+c_{2}e^{\lambda_{2}x}$$

To find the homogeneous solutions, set r(x)=0:

$$\ y''+4y'+4y=0$$

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

$$\ (\lambda+2)(\lambda+2)=0$$

$$\ \lambda_{1}=-2,   \lambda_{2}=-2$$

Because of the double root at $$\ \lambda=-2$$, the homogeneous solution will take the form:

$$\ y_{h}(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}$$

To find the particular solution, we must find the form $$\ y_{p}(x)$$ will take based on r(x):

$$\ r(x)=e^{-2x}sin2x$$

From table 2.1 p. 82, the form of $$\ y_{p}(x)$$ will be:

$$\ y_{p}(x)=e^{-2x}(Ksin2x+Mcos2x)$$

Differentiating $$\ y_{p}(x)$$ gives:

$$\ y'_{p}(x)=-2e^{-2x}(Ksin2x+Mcos2x)+e^{-2x}(-2Msin2x+2Kcos2x)$$

Differentiating $$\ y'_{p}(x)$$ gives:

$$\ y''_{p}(x)=4e^{-2x}(Ksin2x+Mcos2x)-2e^{-2x}(-2Msin2x+2Kcos2x)-2e^{-2x}(-2Msin2x+2Kcos2x)+e^{-2x}(-4Ksin2x-4Mcos2x)$$

$$\ y''_{p}(x)=4e^{-2x}(Ksin2x+Mcos2x)-4e^{-2x}(-2Msin2x+2Kcos2x)+e^{-2x}(-4Ksin2x-4Mcos2x)$$

Then, we substitute $$\ y_{p}(x)$$, $$\ y'_{p}(x)$$, $$\ y''_{p}(x)$$ into the original ODE:

$$\ [4e^{-2x}(Ksin2x+Mcos2x)-4e^{-2x}(-2Msin2x+2Kcos2x)+e^{-2x}(-4Ksin2x-4Mcos2x)]+4[-2e^{-2x}(Ksin2x+Mcos2x)+e^{-2x}(-2Msin2x+2Kcos2x)]+4[e^{-2x}(Ksin2x+Mcos2x)]=e^{-2x}sin2x$$

By collecting like terms, we get:

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

By equating like terms, we get:

$$\ e^{-2x}cos2x(-4M)=0  =>  M=0$$

$$\ e^{-2x}sin2x(4M-4K)=e^{-2x}sin2x  => K=-\frac{1}{4}$$

By substituting K and M into the general form for the particular solution we get:

$$\ y_{p}(x)=e^{-2}sin2x$$

To find the final solution to the ODE, we add the homogeneous and particular solutions:

$$\ y(x)=y_{h}(x)+y_{p}(x)$$

$$\ y(x)=c_{1}e^{-2x}+c_{2}xe^{-2x}-\frac{1}{4}e^{-2x}sin2x$$

To find the unknown coefficients, we apply the initial conditions:

$$\ y(0)=1, y'(0)=-1.5$$

$$\ y(0)=1=c_{1}e^{0}+c_{2}(0)e^{0}+\frac{1}{4}e^{0}(sin(0))=c_{1} => c_{1}=1$$

Then we find the derivative of the general solution:

$$\ y'(x)=-2c_{1}e^{-2x}+c_{2}e^{-2x}-2c_{2}xe^{-2x}+\frac{1}{2}e^{-2x}sin2x-\frac{1}{2}e^{-2x}cos2x$$

Then we apply initial conditions:

$$\ y'(0)=-1.5=-2c_{1}e^{0}+c_{2}e^{0}-2c_{2}(0)e^{0}+\frac{1}{2}e^{0}sin(0)-\frac{1}{2}e^(0)cos(0)=-2c_{1}+c_{2}-0.5=-1.5$$

$$\ -2c_{1}+c_{2}=-1$$

$$\ -2(1)+c_{2}=-1 => c_{2}=1$$

By substituting the known coefficients into the general solution, we get the final solution:

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