User:Egm4313.s12.team13.fsp.r2

Problem Statement
Find the solution for the general excitation for each of the two systems listed below:

No excitation when $$ r(x)=0\!$$. Solution will then be plotted.

Since the roots of the differential equations have already been given to us, it is simply a matter of transforming it into a general solution from here.

$$ \mathbf{Part\ a}\!$$

$$ \lambda_1=3,\!\qquad\lambda_2= -6\!$$

$$ y(0)=-5,\qquad\! y'(0)=2$$

The roots of the differential equation can be written as

(2.0) $$\color{Red} y(x)=c_1 e^{3x} + c_2 e^{-6x}\!$$

To find the constants, we shall now apply the initial conditions. Since the there are two unknowns in this expression, it will be necessary to take the derivative of equation 2.0 to be able to apply the second initial condition.

(2.1 $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= 3c_1 e^{3x} - 6c_2 e^{-6x}\!$$

When applying the first initial condition, it results in

(2.2) $$ -5=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}-5=c_1+c_2\!$$

Applying the second initial conditions yields

(2.3) $$ 2=3c_1 e^{0} + -6c_2 e^{0}\longrightarrow\!\color{Red}2=3c_1-6c_2\!$$

By solving the system of equations, the constants come out to be

$$\color{Blue}c_1= \frac{-28}{9}\,\qquad c_2=\frac{-17}{9}\,$$

Solution Part a
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-28}{9} e^{3x} + \frac{-17}{9}e^{-6x} }$$



$$ \mathbf{Part\ b}\!$$

$$ \lambda_1=-2,\!\qquad\!\lambda_2=+5\!$$

$$ y(0)=1,\!\qquad y'(0)=0$$

Following the exact same procedure in part b as in part a, the general solution is shown below:

(2.4) $$ \color{Red} y(x)=c_1 e^{-2x} + c_2 e^{5x}\!$$

The derivative of equation 2.4 is

(2.3) $$ \frac{d }{dx}\,$$ $$y(x)= \color{Red}y'= -2c_1 e^{-2x} +5c_2 e^{5x}\!$$

Applying the first initial condition yields

(2.5) $$ 1=c_1 e^{0} + c_2 e^{0}\longrightarrow\!\color{Red}1=c_1+c_2\!$$

Applying the second initial condition yields

(2.6) $$ 0=-2c_1 e^{0} + 5c_2 e^{0}\longrightarrow\!\color{Red}0=-2c_1+5c_2\!$$

Solving the system of equations, the constants are figured out to be

$$\color{Blue}c_1= \frac{-5}{2}\,\qquad c_2=\frac{2}{7}\,$$

Solution Part b
$$\color{Red}\mathbf{\therefore\! \qquad y(x)=\frac{-5}{2} e^{-2x} + \frac{2}{7}e^{5x} }$$



Section 3 Notes