User:Egm6321.f10.team4.Auerbach

= Problem 5 statement= From Meeting 28, p 28-2 to Meeting 29, p 29-1.

Given
A spring-mass-damper system with an applied force:

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Find
5.1 PDEs for the integrating factor $$\displaystyle h(t,y)$$

5.2 Trial solution for the integrating factor $$\displaystyle h(t,y)$$


 * 5.2.1 $$\displaystyle \bar{a}_1$$ and $$\displaystyle \bar{a}_2$$ in terms of $$\displaystyle a_2,a_1,a_0$$


 * 5.2.2 Quadratic equation for $$\displaystyle \alpha $$


 * 5.2.3 Reduced order equation using integrating factor


 * 5.2.4 y by using the integrating factor method (IFM)


 * 5.2.5 Show $$\displaystyle \alpha \beta =\frac{\alpha _{0}}{\alpha _{2}} $$ and $$\displaystyle \alpha + \beta =\frac{\alpha _{1}}{\alpha _{2}} $$


 * 5.2.6 Deduce an expression for the particular solution $$\displaystyle y_P $$ for a general excitation $$\displaystyle f(t) $$


 * 5.2.7 Verify with the table of particular solutions


 * 5.2.8 Solve the linear 2nd order ordinary differential equation with constant coefficients (L2_ODE_CC) with $$f(t)=exp(-t^2) $$. Find the coefficients $$\displaystyle a_2,a_1,a_0$$ such that the L2_ODE_CC accepts the following as characteristic equations:


 * 5.2.8.1 $$\displaystyle (r+1)(r-2)=0 $$


 * 5.2.8.2 $$\displaystyle (r-4)^4=0 $$

Solution
5.1

The equation of motion for a spring-mass-damper system can be written as:

Using an integrating factor $$\displaystyle h(t,y)$$, equation 5-1 becomes

If we then set $$\displaystyle p=y'$$, equation 5-2 can be written in a form that satisfies the first condition of exactness:

The second condition of exactness requires the following 2 conditions must be met:


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$$\displaystyle {g_{pp}}=f_{tp}+pf_{yp}+2f_{y} $$ $$
 * $$\displaystyle (Eq. 5-3)
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$$\displaystyle {f_{tt}}+2pf_{ty}+p^2f_{yy}=g_{tp}+pg_{yp}-g_{y} $$ $$
 * $$\displaystyle (Eq. 5-4)
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Taking the partial derivatives in equation 5-3,


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$$\displaystyle g_{pp}=0 $$
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$$\displaystyle f_{tp}=0 $$
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$$\displaystyle f_{yp}=0 $$
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$$\displaystyle f_{y}=h_ya_0 $$
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Equation 5-3 reduces to


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$$\displaystyle 0=0+p0+2h_ya_0 $$
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so,


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$$\displaystyle h_y=0 $$
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Taking the partial derivatives in equation 5-4,


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$$\displaystyle f_{tt}=h_{tt}a_2 $$
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$$\displaystyle f_{ty}=h_{ty}a_2=0 $$
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$$\displaystyle f_{yy}=h_{yy}a_2=0 $$
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$$\displaystyle g_{tp}=h_{t}a_1 $$
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$$\displaystyle g_{yp}=h_{y}a_1=0 $$
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$$\displaystyle g_{y}=h_{y}a_1p+h_{y}a_0y+ha_0-h_yf(t)=ha_0 $$
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Equation 5-4 reduces to


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$$\displaystyle h_{tt}a_2+2p0+p^20=h_{t}a_1+p0-ha_0 $$
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or


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$$\displaystyle h_{tt}a_2-h_{t}a_1+ha_0=0 $$
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5.2

Starting off with the trial solution for the intergration factor $$\displaystyle h(t)=exp(\alpha t)$$, equation 5-1 becomes

If we integrate both sides with respect to $$\displaystyle t$$

Assuming the result will be a reduction of order

5.2.1

Differentiating the LHS of equation 5-7 and the LHS of equation 5-6 and comparing:

Examining like $$\displaystyle y''$$ terms:

Thus

Examining like $$\displaystyle y'$$ terms:

Thus

Examining like $$\displaystyle y$$ terms:

Thus

5.2.2

Using equations 5-10 and 5-11

or

5.2.3

Starting with equation 5-7

Dividing both sides by $$\displaystyle \bar{a}_1$$

5.2.4

Using what was previous discussed in lecture 10,

where

Thus, using the equation presented in 10-3

5.2.5

From earlier definitions:

5.2.6