User:Egm4313.s12.team14.turano

=R2.1= Given the two roots and the initial conditions:


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

\lambda_1=-2, \lambda_2=+5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1a)
 * }
 * }


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

y(0)=1,y'(0)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1b)
 * }
 * }

Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$\displaystyle r(x) $$

Consider no excitation:


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r(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

Plot the solution.

Characteristic equation:


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(\lambda-\lambda_1)(\lambda-\lambda_2)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Substituting $$\displaystyle(Eq. 1a)$$ into $$\displaystyle(Eq. 3)$$:


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(\lambda+2)(\lambda-5)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }


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\lambda^2-3\lambda-10=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

Non-homogeneous solution:


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y''-3y'-10y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 6)
 * }
 * }

Homogeneous solution:


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y_h(x)=C_1e^{-2x}+C_2e^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
 * }
 * }

Overall solution:


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y(x)=C_1e^{-2x}+C_2e^{5x}+y_p(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
 * }
 * }

No excitation:


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r(x)=0 => y_p(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 9)
 * }
 * }

From intitial conditions:


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y(0)=1=C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 10)
 * }
 * }


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y'(0)=0=-2C_1+5C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 11)
 * }
 * }

Solving for $$\displaystyle C_1 $$ and $$\displaystyle C_2 $$:


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C_1=\frac{5}{7} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

and


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C_2=\frac{2}{7} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

So, the final solution is:

$$\displaystyle y(x)=\frac{5}{7}e^{-2x}+\frac{2}{7}e^{5x} $$



=R2.2=


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y'' - 10y' +25y = r(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Initial conditions:


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y(0) = 1, y'(0) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

No excitation:


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r(x) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Find and plot the solution for $$\displaystyle (Eq. 1) $$

Due to no excitation, $$\displaystyle (Eq. 1) $$ becomes:


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y'' - 10y' +25y = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Substituting $$\displaystyle \lambda $$ into $$\displaystyle (Eq. 4) $$:


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\lambda^2-10\lambda+25=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 5)
 * }
 * }

Factoring, and solving for $$\displaystyle \lambda $$:


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(\lambda-5)^2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 6)
 * }
 * }


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\lambda_2=\lambda_1=\lambda=5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Since $$\displaystyle \lambda $$ is a double root, the general solution:


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y=C_1e^{5x}+C_2xe^{5x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 7)
 * }
 * }

From intitial conditions:


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y(0)=1=C_1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 10)
 * }
 * }


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y'(0)=0=5C_1+C_2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 11)
 * }
 * }

Solving for $$\displaystyle C_1 $$ and $$\displaystyle C_2 $$:


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C_1=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

and


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C_2=-5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

So, the final solution is:

$$\displaystyle y(x)=e^{5x}-5xe^{5x} $$



=R1.6G= G) Deformation of a Beam

$$\displaystyle{EIy''=f(x)}$$

$$\cdot$$ This equation is of SECOND order and LINEAR.

$$\cdot$$ To prove the applicability of Superposition, the following is done:

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


 * $$\displaystyle{EIy_h''=0}$$


 * <p style="text-align:right;">$$\displaystyle (Eq. 1) $$
 * }
 * }

Particular:


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 * $$\displaystyle{EIy_p''=f(x)}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2) $$
 * }
 * }
 * }

Add (Eq. 1) and (Eq. 2):

$$\displaystyle{EIy_h+EIy_p=f(x}$$

It could be said that:

$$\displaystyle{y_{h}+y_{p}=Y}$$

and due to the fact that,

$$\displaystyle{EI(y_h''+y_p)=EIY}$$

=R1.1=

Team14P1R16Diagram1.png‎

$$y=y_k=y_c$$

Spring Force, $$\vec{F}_s=k(y)$$

Dashpot Force, $$\vec{F}_D=c(y')$$

Applied Force, $$\vec{F}(t)$$


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\Sigma\vec{F}=\vec{F}(t)-\vec{F}_s-\vec{F}_D $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 1)
 * }
 * }


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m\vec{a}=\vec{F}_i=my'' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 2)
 * }
 * }

From Newton's 2nd Law:


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\Sigma\vec{F}=m\vec{a} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 3)
 * }
 * }

Substituting $$ (Eq. 1)$$ and $$ (Eq. 2)$$ into $$ (Eq. 3)$$:


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\vec{F}(t)-\vec{F}_s-\vec{F}_D=\vec{F}_i $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Solving for $$\vec{F}(t)$$:

$$\vec{F}(t)=\vec{F}_s+\vec{F}_D+\vec{F}_i$$

Substituting $$\vec{F}_s=ky$$, $$\vec{F}_D=cy'$$, and $$\vec{F}_i=my''$$ into $$ (Eq. 4)$$:

$$\vec{F}(t)=ky+Cy'+my''$$

Putting into standard form:

$$my''+Cy'+ky=\vec{F}(t)$$