User:Egm6321.f11.team1.zheng.zx/personal place/HW5

= R*5.3 Find the expressions for X(x)in terms of $$\cos Kx,\ \sin Kx,\ \cosh Kx,\ \sinh Kx$$ =

Given
$$ \displaystyle X^{(4)}-K^4X=0 $$

$$ \displaystyle K:= \frac{\omega^2m}{EI} $$

Assume:

$$ \displaystyle X_{(x)}=e^{(rx)} $$

characteristic equation:

$$ \displaystyle r^4=K^4 $$

then:

$$ \displaystyle r_1=K,\ r_2=-K,\ r_3=iK,\ r_4=-iK,\ i:=\sqrt{-1} $$

Find
the expressions for $$X_{(x)}$$ in terms of:

$$ \displaystyle \cos Kx, \ \sin Kx,\ \cosh Kx,\ \sinh Kx $$

Solution

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\displaystyle X_{(x)}=c_1e^{Kx}+c_2e^{-Kx}+c_3e^{iKx}+c_4e^{-iKx} $$ $$
 * $$\displaystyle (Equation\;5.3.1 )
 * }
 * }

$$ \displaystyle \begin{align} A_1: &=c_1e^{Kx}+c_2e^{-Kx}\ \in \mathbb{R} \\ &= c_1(\frac{e^{Kx}+e^{-Kx}}{2}+\frac{e^{Kx}-e^{-Kx}}{2}) + c_2(\frac{e^{Kx}+e^{-Kx}}{2}-\frac{e^{Kx}-e^{-Kx}}{2})\\ &= c_1(\cosh Kx+ \sinh Kx) + c_2(\cosh Kx- \sinh Kx)\\ &= (c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx \end{align} $$

We can rewrite it to:
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\displaystyle A_1=b_1\cosh Kx + b_2\sinh Kx $$ $$
 * $$\displaystyle (Equation\;5.3.2 )
 * }
 * }

$$ \displaystyle \begin{align} A_2: &=c_3e^{iKx}+c_4e^{-iKx} \\ &=c_3(\cos Kx +i\sin Kx)+c_4(\cos Kx -i\sin Kx) \\ &= (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx \ \in \mathbb{R} \end{align} $$

We can rewrite it to:
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\displaystyle A_2=b_3\cos Kx- b_4\sin Kx $$ $$
 * $$\displaystyle (Equation\;5.3.3 )
 * }
 * }

So, add up the two terms from equation 5.3.2 and equation 5.3.3:
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\displaystyle X_{x}=A_1+A_2=b_1\cosh Kx + b_2\sinh Kx + b_3\cos Kx- b_4\sin Kx $$ $$
 * $$\displaystyle (Equation\;5.3.4 )
 * }
 * }

Where:

$$ \displaystyle b_1=c_1+c_2,\ b_2=c_1-c_2,\ b_3=c_3+c_4,\ b_4=-i(c_3-c_4) $$

= R5.6 Show that the trial solution in Method 2 is equivalent to the combined trial solutions in Method 1 =

Given
Show the equivalence of two methods using Euler L3-ODE-VC as an example:
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\displaystyle a_3x^3y^{(3)}+a_2x^2y^{(2)}+a_1xy'+a_0y=0,\ y'=y^{(1)},\ y=y^{(0)} $$ $$
 * $$\displaystyle (Equation\;5.6.1)
 * }
 * }

Find
That the trial solution in Method 2 is equivalent to the combined trial solutions in Method 1.

Solution
1.Method I Stage 1,

$$ \displaystyle x=e^t $$

and we have derived in class that:
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\displaystyle y'=e^{-t}y_t $$ $$
 * $$\displaystyle (Equation\; 5.6.2)
 * }
 * }


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\displaystyle y^{(2)}=e^{-2t}(y_{tt}-y_t) $$ $$
 * $$\displaystyle (Equation\; 5.6.3)
 * }
 * }


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\displaystyle y^{(3)}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t) $$ $$
 * $$\displaystyle (Equation\; 5.6.4)
 * }
 * }

plug equations 2,3,4 into 1, we obtain:
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\displaystyle a_3e^{3t}\cdot e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+a_2e^{2t}\cdot e^{-2t}(y_{tt}-y_t)+a_1e^t\cdot e^{-t}y_t+a_0y=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.5)
 * }
 * }

Rearranging:
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\displaystyle a_3y_{ttt}+(a_2-3a_3)y_{tt}+(2a_3-a_2+a_1)y_t+a_0y=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.6)
 * }
 * }

Stage 2,

$$ \displaystyle y=e^{rt},\ r=constant $$

substitute into equation 5.6.6, we get:
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\displaystyle a_3r^3e^{rt}+(a_2-3a_3)r^2e^{rt}+(a_1-a_2+2a_3)re^{rt}+a_0e^{rt}=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.7)
 * }
 * }

Finally:

2.Method II

$$ \displaystyle y=x^r,\ r=constant $$

substitute into equation 1:
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\displaystyle a_3x^3r(r-1)(r-2)x^{r-3}+a_2x^2r(r-1)x^{r-2}+a_1xrx^{r-1}+a_0x^r=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.8)
 * }
 * }


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\displaystyle a_3r(r-1)(r-2)x^r+a_2r(r-1)x^r+a_1rx^r+a_0x^r=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.9)
 * }
 * }

rearranging:
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\displaystyle a_3(r^3-3r^2+2r)+a_2(r^2-r)+a_1r+a_0=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.10)
 * }
 * }

And, finally:

So, using whatever method I or method II, we can derive the exact same characteristic equation

thus, they are just the same.

= References =