User:Egm4313.s12.team10.R1

To return to the Team 10 site click here = Problem 1: Equation of motion for a spring-dashpot system in parallel =

Given
Refer to Lecture slide [1-5] for problem statement. A spring-dashpot system in parallel, with a mass and an applied force f(t)



Find
The equation of motion of the system.

Solution
Free Body Diagram:



Kinematics:
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {y}={y_k}+{y_c} $$
 * $$\displaystyle {y}={y_k}+{y_c} $$


 * style= |
 * }

Kinetics:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{f_k}=k\overline{y} $$
 * $$\displaystyle \overline{f_k}=k\overline{y} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{f_c}=c\overline{{y}'} $$
 * $$\displaystyle \overline{f_c}=c\overline{{y}'} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m\overline{a}=m\overline{{y}''} $$
 * $$\displaystyle m\overline{a}=m\overline{{y}''} $$


 * style= |
 * }

Final Solution:

By summing forces in the Y direction we obtain:
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{f(t)}=m\overline{{y}''} +c\overline{{y}'}+k\overline{y} $$
 * $$\displaystyle \overline{f(t)}=m\overline{{y}''} +c\overline{{y}'}+k\overline{y} $$


 * style= |
 * }

= Problem 2: Equation of motion for a spring-dashpot system in series=

Given
Refer to Lecture slide [1-5] for problem statement.



Find
The equation of motion of the spring mass-dashpot.

Solution
Free Body Diagram



Kinematics:
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {y}={y_k}+{y_c} $$
 * $$\displaystyle {y}={y_k}+{y_c} $$


 * style= |
 * }

Kinetics:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m\overline{a}=m\ddot{y}=m\overline{y}''$$
 * $$\displaystyle m\overline{a}=m\ddot{y}=m\overline{y}''$$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=k\overline{y}$$
 * $$\displaystyle \overline{F_k}=k\overline{y}$$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=C\overline{y'}$$
 * $$\displaystyle \overline{F_c}=C\overline{y'}$$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \sum \overline{F_y}=m\overline{a} $$
 * $$\displaystyle \sum \overline{F_y}=m\overline{a} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \sum\overline{F_y}=-\overline{F_c}-\overline{F_k}+\overline{F}_{r}=m\overline{a} $$
 * $$\displaystyle \sum\overline{F_y}=-\overline{F_c}-\overline{F_k}+\overline{F}_{r}=m\overline{a} $$


 * style= |
 * }

Final Solution:
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m\overline{y''}+\overline{y'}c+\overline{y}k=\overline{r(t)} $$
 * $$\displaystyle m\overline{y''}+\overline{y'}c+\overline{y}k=\overline{r(t)} $$


 * style= |
 * }

= Problem 3: Equation of motion for a spring-dashpot system in series =

Given
Refer to Lecture slide [1-6] for problem statement.



Find
The equation of motion of the spring mass-dashpot.

Solution
Free Body Diagram



Kinematics:
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {y}={y_k}+{y_c} $$
 * $$\displaystyle {y}={y_k}+{y_c} $$


 * style= |
 * }

Kinetics:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=k\overline{y} $$
 * $$\displaystyle \overline{F_k}=k\overline{y} $$

$$
 * $$\displaystyle
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=c\overline{{y}'} $$
 * $$\displaystyle \overline{F_c}=c\overline{{y}'} $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{y}=\overline{y_k}+\overline{y_c} $$
 * $$\displaystyle \overline{y}=\overline{y_k}+\overline{y_c} $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{{y}}=\overline{{y_k}}+\overline{{y_c}''} $$
 * $$\displaystyle \overline{{y}}=\overline{{y_k}}+\overline{{y_c}''} $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline=\overline{F_k}=\overline{F_c} $$
 * $$\displaystyle \overline=\overline{F_k}=\overline{F_c} $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m\overline{{y}''}+\overline{F_I}=\overline{F}(t) $$
 * $$\displaystyle m\overline{{y}''}+\overline{F_I}=\overline{F}(t) $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle k\overline{y_k}=c\overline{{y_c}'} $$
 * $$\displaystyle k\overline{y_k}=c\overline{{y_c}'} $$

|| $$\displaystyle $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \overline{{y_c}'}=\frac{k}{c}\overline $$
 * $$\displaystyle \overline{{y_c}'}=\frac{k}{c}\overline $$

|| $$\displaystyle $$
 * }

Final Solution:
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$

|| $$\displaystyle $$
 * }

= Problem 4: Equation Derivation =

Given
Refer to Lecture slide [2-2] for problem statement. The components of the circuit can be related by the following formulas:




 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$

|| $$\displaystyle (Eq. 4.1) $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$

|| $$\displaystyle (Eq. 4.2) $$
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$

|| $$\displaystyle (Eq. 4.3) $$
 * }

Find
Derive equations (3) and (4) from equation (2).

Equation (3)

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle I=C\frac{dV_c}{dt}$$
 * $$\displaystyle I=C\frac{dV_c}{dt}$$

$$
 * $$\displaystyle (Eq. 4.1a)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{I}{C}=\frac{dV_c}{dt} $$
 * $$\displaystyle \frac{I}{C}=\frac{dV_c}{dt} $$

$$
 * $$\displaystyle (Eq. 4.2a)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {V}'=\frac{dV}{dt} $$
 * $$\displaystyle {V}'=\frac{dV}{dt} $$

$$
 * $$\displaystyle (Eq. 4.3a)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}(I)=\frac{d}{dt}(C\frac{dV_c}{dt})=C\frac{d^2V_c}{dt}={I}' $$
 * $$\displaystyle \frac{d}{dt}(I)=\frac{d}{dt}(C\frac{dV_c}{dt})=C\frac{d^2V_c}{dt}={I}' $$

$$
 * $$\displaystyle (Eq. 4.4a)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}({I}')=\frac{d}{dt}(C\frac{d^2V_c}{dt^2})=C\frac{d^3V_c}{dt^3}={I}'' $$
 * $$\displaystyle \frac{d}{dt}({I}')=\frac{d}{dt}(C\frac{d^2V_c}{dt^2})=C\frac{d^3V_c}{dt^3}={I}'' $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.5a)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$

$$ Taking the derivative of equation (2) yields
 * <p style="text-align:left;">$$\displaystyle (Eq. (2) )
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}(V)=\frac{d}{dt}(LC\frac{d^2V_c}{dt^2})+\frac{d}{dt}(RC\frac{dV_c}{dt})+\frac{d}{dt}(V_c) $$
 * $$\displaystyle \frac{d}{dt}(V)=\frac{d}{dt}(LC\frac{d^2V_c}{dt^2})+\frac{d}{dt}(RC\frac{dV_c}{dt})+\frac{d}{dt}(V_c) $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.6a)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{dV}{dt}=LC\frac{d^3V_c}{dt^3}+RC\frac{d^2V_c}{dt^2}+\frac{dV_c}{dt} $$
 * $$\displaystyle \frac{dV}{dt}=LC\frac{d^3V_c}{dt^3}+RC\frac{d^2V_c}{dt^2}+\frac{dV_c}{dt} $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.7a)
 * }

Substituting in from Eq. 4.3a,4.4a, and 4.5a yields:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. (3))
 * }

Equation (4)

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle Q=CV_c $$
 * $$\displaystyle Q=CV_c $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.1b)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}(Q)=\frac{d}{dt}(CV_c)=\frac{dV_c}{dt}C={Q}' $$
 * $$\displaystyle \frac{d}{dt}(Q)=\frac{d}{dt}(CV_c)=\frac{dV_c}{dt}C={Q}' $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.2b)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}({Q}')=\frac{d}{dt}(C\frac{dV_C}{dt})=C\frac{d^2V_c}{dt^2}={Q}'' $$
 * $$\displaystyle \frac{d}{dt}({Q}')=\frac{d}{dt}(C\frac{dV_C}{dt})=C\frac{d^2V_c}{dt^2}={Q}'' $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.3b)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{Q}{C}=V_c\ $$
 * $$\displaystyle \frac{Q}{C}=V_c\ $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. 4.4b)
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. (2))
 * }

Substituting in from equations 4.2b,4.3b, and 4.4b yields.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$

$$
 * <p style="text-align:left;">$$\displaystyle (Eq. (4) )
 * }

= Problem 5: Find the general solution. Check solution by substitution. =

Given
Refer to Lecture slide [2-5] for problem statement.

Solve problems 4 and 5 from page 59 of Advanced Engineering Mathematics

Problem (4)

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$


 * style= |
 * }

Problem (5)

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$


 * style= |
 * }

Find
General solution

Problem (4)
Given Equation
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$

Constant Values
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle a = 4 $$
 * $$\displaystyle a = 4 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle b = (\pi^2 + 4) $$
 * $$\displaystyle b = (\pi^2 + 4) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda = \frac{dy}{dx} $$
 * $$\displaystyle \lambda = \frac{dy}{dx} $$

Substituting Lambda for y Values
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda^2 + f\lambda + (\pi^2 + 4) = 0 $$
 * $$\displaystyle \lambda^2 + f\lambda + (\pi^2 + 4) = 0 $$

Solving For Omega
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega^2 = \pi^2 $$
 * $$\displaystyle \omega^2 = \pi^2 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \omega = \pi $$
 * $$\displaystyle \omega = \pi $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega = -2\, \pm \pi i $$
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega = -2\, \pm \pi i $$

Solving For y1 and y2
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$


 * style= |
 * }

The general equation is:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^\frac{-ax}{2}(A\cos\omega x + B\sin\omega x)$$
 * $$\displaystyle y = e^\frac{-ax}{2}(A\cos\omega x + B\sin\omega x)$$


 * style= |
 * }

Solving for our equation, we get:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$

Pluging y", y' and y into Initial Equation
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$


 * style= |
 * }

Therefore:


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$


 * style= |
 * }

is the solution

Problem (5)

 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$


 * style= |
 * }

The ODE above can be simplified by using the variables below.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle a = 2\pi $$
 * $$\displaystyle a = 2\pi $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle b = \pi ^2 $$
 * $$\displaystyle b = \pi ^2 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' + a y' + by = 0 $$
 * $$\displaystyle y'' + a y' + by = 0 $$


 * style= |
 * }

The equation is now in the general format for linear ODE’s.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{\lambda x} $$
 * $$\displaystyle y = e^{\lambda x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = \lambda e^{\lambda x} $$
 * $$\displaystyle y' = \lambda e^{\lambda x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$

Solving For Lambda Values
 * style= |
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$


 * style= |
 * }

This means that the characteristic equation has a doubleroot.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_2 = u y_1 $$
 * $$\displaystyle y_2 = u y_1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle u = c_2 x + c_1, c_1 = 0, c_2 = 1 $$
 * $$\displaystyle u = c_2 x + c_1, c_1 = 0, c_2 = 1 $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y = e^{-\pi x}(c_1 + c_2 x) $$
 * $$\displaystyle y = e^{-\pi x}(c_1 + c_2 x) $$


 * style= |
 * }

This is the general solution to the ordinary differential equation.


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y' = -\pi c_1 e^{-\pi x}+c_2e^{-\pi x}-\pi c_2 x e^{-\pi x} $$
 * $$\displaystyle y' = -\pi c_1 e^{-\pi x}+c_2e^{-\pi x}-\pi c_2 x e^{-\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle y'' = \pi ^2 c_1 e^{-\pi x}-\pi c_2e^{-\pi x}+\pi ^2 c_2 x e^{-\pi x} - \pi c_2e^{-\pi x} $$
 * $$\displaystyle y'' = \pi ^2 c_1 e^{-\pi x}-\pi c_2e^{-\pi x}+\pi ^2 c_2 x e^{-\pi x} - \pi c_2e^{-\pi x} $$


 * style= |
 * }


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle e^{-\pi x} [\pi ^2 c_1 -\pi c_2 + \pi ^2 c_2 x - \pi - 2\pi^2 c_1 + 2\pi c_2 - 2\pi ^2 c_2 x + c_1 \pi ^2 + \pi ^2 c_2x] = 0 $$
 * $$\displaystyle e^{-\pi x} [\pi ^2 c_1 -\pi c_2 + \pi ^2 c_2 x - \pi - 2\pi^2 c_1 + 2\pi c_2 - 2\pi ^2 c_2 x + c_1 \pi ^2 + \pi ^2 c_2x] = 0 $$


 * style= |
 * }

All the terms inside the bracket cancel out and equal zero, so the general solution holds.

= Problem 6: =

Given
Refer to Lecture slide [2-2] for problem statement.

Problem (A)

 * {| style="width:100%" border="0"

$$\displaystyle y'' = g = const. $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (B)

 * {| style="width:100%" border="0"

$$\displaystyle mV' = mg-bV^2$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (C)

 * {| style="width:100%" border="0"

$$\displaystyle h' = -k \sqrt h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (D)

 * {| style="width:100%" border="0"

$$\displaystyle m{Y''} + K{Y} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (E)

 * {| style="width:100%" border="0"

$$\displaystyle {Y''} + \omega_o^2 Y = \cos \omega {t} $$, $$\displaystyle \omega_o \approx \omega $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (F)

 * {| style="width:100%" border="0"

$$\displaystyle LI''+RI'+\frac{I}{C}=E $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (G)

 * {| style="width:100%" border="0"

$$\displaystyle {E}{I}{y^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (H)

 * {| style="width:100%" border="0"

$$\displaystyle L\theta''+g\sin{\theta}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Find
Determine the order of the differential equations for the first 8 figures above. For each equation, determine if it's linear and whether the superposition principle can be applied.

Solution
This is the general format for a linear equation:


 * {| style="width:100%" border="0"

$$\displaystyle y''+p(x)y'+q(x)y=r(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (A)

 * {| style="width:100%" border="0"

$$\displaystyle y'' = g = const. $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

The ODE consists of only additive combinations of the dependent variable y and its derivatives. Therefore, the ODE is linear.


 * {| style="width:100%" border="0"

$$\displaystyle y''$$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogenous Equation:


 * {| style="width:100%" border="0"

$$\displaystyle y_h = 0$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:


 * {| style="width:100%" border="0"

$$\displaystyle y_p = g$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{y} = y_h+y_p$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \overline{y} = g$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_h+y_p = g = \overline{y}$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (B)

 * {| style="width:100%" border="0"

$$\displaystyle mV' = mg-bV^2$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle mV' = mg-bV^2 \Rightarrow mV'+bV^2 = mg$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle V^2 \neq V$$ The general format of a linear equation is not followed since the variable, so this is a nonlinear ODE
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle V'$$ is the highest order derivative. Therefore, this is a first-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogenous Equation:


 * {| style="width:100%" border="0"

$$\displaystyle V_h=m{V_h}'+b{V_h}^2=0$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:


 * {| style="width:100%" border="0"

$$\displaystyle V_p=m{V_p}'+b{V_p}^2=mg$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{V}=V_h+V_p$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \overline{V}=m(V_h+V_p)'+b({V_h}^2+2 V_pV_h+V_p^2)=mg$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle V_h+V_p=m(V_h'+V_p')+b(V_h^2+V_p^2)=mg\neq\overline{V}$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle cannot be applied.

Problem (C)

 * {| style="width:100%" border="0"

$$\displaystyle h' = -k \sqrt h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \sqrt h = h^{1/2} \neq h$$ Therefore, this is a nonlinear ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle h' $$ is the highest order derivative. Therefore, this is a first-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous Equation:


 * {| style="width:100%" border="0"

$$\displaystyle h_h=h_h'+k\sqrt{h_h}=0$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:


 * {| style="width:100%" border="0"

$$\displaystyle h_p=h_p'+k\sqrt{h_p}=r(t) $$, where r(t) is some function of time
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{h}=h_h+h_p=(\overline{h})'+k\sqrt{\overline{h}}=r(t)$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{h}=h_h+h_p$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{h}'=(h_h'+h_p')=(h_h+h_p)'$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle h_h+h_p=(h_h'+h_p')+k\sqrt{h_h}+k\sqrt{h_p}=(\overline{h})'+k(\sqrt(h_h)+\sqrt{h_p})=r(t)$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle h_h+h_p\neq\overline{h}$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle cannot be applied.

Problem (D)

 * {| style="width:100%" border="0"

$$\displaystyle m{Y''} + K{Y} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

This is a linear ODE due to the fact that it consists of only additive combinations of one dependent variable and its derivatives.


 * {| style="width:100%" border="0"

$$\displaystyle y'' $$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous equation:


 * {| style="width:100%" border="0"

$$\displaystyle m{Y_h''} + K{Y_h} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:


 * {| style="width:100%" border="0"

$$\displaystyle m{Y_p''} + K{Y_h} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving, we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{Y} = {Y_h} + {Y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle m({Y_h} + {Y_p}) + K({Y_h}+ {Y_p}) = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle m(\overline{Y''}) + K(\overline{Y})=f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{Y} = {Y_h} + {Y_p''} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{Y} = {Y_h} + {Y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (E)

 * {| style="width:100%" border="0"

$$\displaystyle {Y''} + \omega_o^2 Y = \cos \omega {t} $$, $$\displaystyle \omega_o \approx \omega $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

This is a linear ODE.


 * {| style="width:100%" border="0"

$$\displaystyle y'' $$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \omega_o \approx \omega $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous equation:


 * {| style="width:100%" border="0"

$$\displaystyle {Y_h''} + \omega_o^2 Y_h = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:


 * {| style="width:100%" border="0"

$$\displaystyle {Y_p''} + \omega_o^2 Y_p = \cos \omega {t} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving, we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{Y} = {Y_h} + {Y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle ({Y_h} + {Y_p}) + \omega_o^2 ({Y_h} + {Y_p}) = \cos\omega{t} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle (\overline{Y''}) + \omega_o^2 (\overline{Y}) = \cos\omega{t} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (F)

 * {| style="width:100%" border="0"

$$\displaystyle LI''+RI'+\frac{I}{C}=E $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

This is a linear ODE.

$$\displaystyle I'' $$ is the highest order derivative. Therefore, this is a second-order ODE.

Homogeneous equation:


 * {| style="width:100%" border="0"

$$\displaystyle I_h=L{I_h}''+R{I_h}'+\frac{I_h}{C}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:


 * {| style="width:100%" border="0"

$$\displaystyle I_p=L{I_p}''+R{I_p}'+\frac{I_p}{C}=E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving, we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{I}=L\overline{I}''+R\overline{I}'+\frac{\overline{I}}{C}=E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{I}=L({I_p+I_h})''+R({I_p+I_h})'+\frac{I_p+I_h}{C}=E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle I_p+I_h=L(I_h+I_p)+R(I_h'+I_p')+\frac{(I_p+I_h)}{C}=0+E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle (I_h+I_p)=(I_h+I_p)'' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle (I_h'+I_p')=(I_h+I_p)'$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{I}=I_p+I_h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (G)
This is a linear ODE.


 * {| style="width:100%" border="0"

$$\displaystyle {E}{I}{y^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle y^{iv} $$ is the highest order derivative. Therefore, this is a fourth-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous equation:


 * {| style="width:100%" border="0"

$$\displaystyle {y_h} = {E}{I}{y_h^{IV}} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:


 * {| style="width:100%" border="0"

$$\displaystyle {y_p} = {E}{I}{y_p^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{y} = {y_h} + {y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \overline{y} = {E}{I}({y_h}+{y_p})^{IV} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

If the superposition principle applies:


 * {| style="width:100%" border="0"

$$\displaystyle {E} {I}\overline{y}^{IV} = {E}{I}({y_h}+{y_p})^{IV} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Equation 1:


 * {| style="width:100%" border="0"

$$\displaystyle {E}{I}{y_h^{IV}} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Equation 2:


 * {| style="width:100%" border="0"

$$\displaystyle {E}{I}{y_p^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Equation 1 + Equation 2:


 * {| style="width:100%" border="0"

$$\displaystyle {E}{I}({y_h^4}+{y_p^{IV}}) = 0 + f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Linearity of Derivative:


 * {| style="width:100%" border="0"

$$\displaystyle -(y^{IV} + x^{IV}) = (y+x)^{IV} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle -{E}{I}({y_h}+{y_p})^{IV} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle -{E}{I}(\overline{y})^{IV} = f(x) = EI({y_h} + {y_p})^{IV} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (H)

 * {| style="width:100%" border="0"

$$\displaystyle L\theta''+g\sin{\theta}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \sin\theta \neq \theta$$ Therefore, this is a nonlinear ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \theta'' $$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }


 * {| style="width:100%" border="0"

$$\displaystyle L\theta''+g\sin{\theta}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous Equation:


 * {| style="width:100%" border="0"

$$\displaystyle \theta_h=L\theta_h''+g\sin{\theta_h}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:


 * {| style="width:100%" border="0"

$$\displaystyle \theta_p=L\theta_p''+g\sin{\theta_p}=r(t) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:


 * {| style="width:100%" border="0"

$$\displaystyle \overline{\theta}=\theta_p+\theta_h=L\overline\theta''+g\sin{\overline{\theta}}=r(t) $$
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 * }


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$$\displaystyle \overline{\theta}=L(\theta_h+\theta_p)''+g\sin({\theta_p+\theta_h}) $$
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 * }


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$$\displaystyle \theta_p+\theta_h=L(\theta_p+\theta_h)+g\sin{\theta_h}+g\sin{\theta_p}=r(t)+0 $$
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$$\displaystyle \theta_p+\theta_h=L(\theta_p+\theta_h)''+g\sin{\theta_h}+g\sin{\theta_p}=r(t) $$
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 * }


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$$\displaystyle \theta_p+\theta_h=L(\overline{\theta})''+g\sin{\theta_h}+g\sin{\theta_p}=r(t) $$
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$$\displaystyle \overline{\theta}\neq\theta_p+\theta_h $$
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Therefore, the superposition principle cannot be applied.