User:Egm6321.f12.team4/Report6

Statement
Find $$y_{xxxxx}$$ in terms of the derivatives of $$y$$ with respect to t where,

Solution
From the above given equations, we can conclude that, Find $$y_{xxxxx}$$ in terms of the derivatives of $$y$$ with respect to t where,

Solving this equations we get,

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Rui Che

Problem R*6.2 – Solving an Euler (homogeneous) L2-ODE-VC
sec31-3

Statement
Solve the Euler (homogeneous) linear second order ordinary differential equation with varying coefficients (L2-ODE-VC)
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$$ x^2y''-2xy'+2y=0 $$     (6.2.1)
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using Method 2 (sec31-4 ) with the trial solution $$ y=x^r $$, such that $$ r $$ is a constant, with the boundary conditions $$ y(1) = 3 y(2) = 4 $$. Plot the results over the domain of the L2-ODE-VC.

Solution
In Method 2 (see link above), one must begin with $$ y=x^r $$ and take the proper derivatives.


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$$ y'=rx^{r-1} $$     (6.2.2)
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$$ y''=r(r-1)x^{r-2} $$     (6.2.3)
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Once the derivatives are computed, they must then be substituted back into equation 6.2.1. This allows one to determine a characteristic equation and solve for the roots of the equation.


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$$ (x^2)r(r-1)x^{r-2}-(2x)rx^{r-1}+2x^r=0 $$     (6.2.4)
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Adding exponents of like bases, one can reduce equation 6.2.4 to


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$$ r(r-1)x^r-2rx^r+2x^r=0 $$     (6.2.5)
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Dividing out the $$ x^r $$ from each term, one can see that the characteristic equation becomes


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$$ r(r-1)-2r+2=0 $$     (6.2.6)
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Equation 6.2.6 simplifies to


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$$ r^2-3r+2=0 $$     (6.2.7)
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Equation 6.2.7 can then be solved to get the two roots, $$ r_1=1 $$ and $$ r_2=2 $$. These roots are then substituted into the general solution


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$$ y(x)=c_1x^{r_1}+c_2x^{r_2} $$     (6.2.8)
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where $$ c_1 $$ and $$ c_2 $$ are constants that can be solved for via boundary conditions. Plugging the roots into equation 6.2.8 and using the boundary conditions in the problem statement, one can see that


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$$ y(x)=c_1x^1+c_2x^2 $$     (6.2.9)
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$$ y(1)=c_1(1)+c_2(1)=3 $$     (6.2.10)
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$$ y(2)=c_1(2)+c_2(4)=4 $$     (6.2.11)
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Multiplying equation 6.2.10 by 2 and subtracting equation 6.2.11 from it, one can see that $$ c_1 = 4 $$ and $$ c_2 = -1 $$. Thus


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$$ y(x)=4x-x^2 $$     (6.2.12)
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Equation 6.2.12 is plotted over the domain ?????. This plot is shown below.

Author and References

 * Solved and typed by -- Kaitlin Harris
 * Reviewed by -- Rui Che

Statement
Trial solution (5)p.30-4 in Method 2 can be written as:

while the combined trial solutions (3)-(4)p.30-4 in Method 1 can be written as:

Solution
In Method 2, trial solution (6.3.0.1) is substituted directly into the following Euler Ln-ODE-VC:

while in Method 1, (6.3.0.2) is first substituted into (6.3.1.1) to replace x with t. Then a trial solution for y with respect to t (6.3.0.3) is substituted into equation (6.3.1.1).

In Method 2, we inspect a particular term: $$a_ix^iy^{(i)}$$ :

In Method 2, we first substitute (6.3.0.2) into the specific term in the left hand side of (6.3.1.1) and recall that:

So:

We then plug (6.3.0.3) into the above expression to replace $$y$$ with $$e^{rt}$$ :

We then substitute (6.3.0.2) back into (6.3.1.5) to replace t in terms of x:

which is the same expression as (6.3.1.2). Thus Method 1 is equivalent to the direct Method 2.

Author and References

 * Solved and Typed by -- Rui Che
 * Reviewed by -- Jinchao Lu

Statement
Consider the characteristic equation:

where $$\lambda=5$$ ,

(1) Euler L2-ODE-VC:


 * (1.1) Find $$a_2,a_1,a_0$$ such that (6.4.0.1) is characteristic equation of (6.4.0.2).


 * (1.2) 1st homogeneous solution: $$y_1(x)=x^\lambda$$


 * (1.3) Complete solution: find c(x) such that


 * (1.4) Find the 2nd homogeneous solution $$y_2(x)$$

(2) L2-ODE-CC

Repeat the steps.

Solution(1.1)
Given,


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$$ a_2x^2y''+a_1 xy'+a_1y=0 $$     (6.4.1.1)
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We start by transforming the variable x to t, by choosing $$ x=e^t $$


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$$ \frac{dy}{dx}= (\frac{d}{dt} \frac{dt}{dx})y $$     (6.4.1.2)
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Since $$\frac{dt}{dx} = e^{-t} $$


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$$ \frac{dy}{dx}= (e^{-t} \frac{d}{dt})y $$     (6.4.1.3)
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To calculate $$ y'' $$,


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$$ \frac{d}{dx}(\frac{dy}{dx})= (\frac{d}{dt} \frac{dt}{dx})(\frac{dy}{dx}) $$     (6.4.1.4)
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$$ \frac{d}{dx}(\frac{dy}{dx})= (e^{-t} \frac{d}{dt})(e^{-t} \frac{d}{dt})y $$     (6.4.1.5)
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Differentiating by parts,
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$$ \frac{d}{dx}(\frac{dy}{dx})= (e^{-2t} \frac{dy}{dt}) + (e^{-2t} \frac{d^2y}{dt^2}) $$     (6.4.1.6)
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Substituting (6.4.1.4) & (6.4.1.6) into (6.4.1.1),


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$$ x^2 e^{-2t}a_2(y_{tt}-y_t) + a_1xe^{-t}y_t + a_0y =0 $$     (6.4.1.7)
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Substituting $$ x=e^t $$ & $$ y=e^{rt} $$ the characteristic equation becomes,


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$$ a_2r^2 + (-a_2+a_1)r + a_0 =0 $$     (6.4.1.8)
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For the characteristic equation to be $$ (\lambda -5)^2=0$$, the following equations should be same,


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$$ a_2r^2 + (-a_2+a_1)r + a_0 =0 $$
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$$ r^2 - 10r +25 =0 $$
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Comparing the above,


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$$ a_2 = 1 $$
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$$a_1 = -9 $$
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$$a_0= 25 $$
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Solution(1.2)
The Euler L2-ODE VC becomes,

Substituting $$ y = x^\lambda $$ ,

which gives, $$ (\lambda -5)^2=0 $$,

Hence, the first homogeneous solution is,

Solution(1.3)
The variation of parameters:

Substitute it into the Euler L2-ODE-VC:

We have:

It can be further reorganized as:

where $$k_1, k_2$$ are constants.

Solution(1.4)
Substitute (6.4.3.7) and expression for $$y_1(x)$$ into (6.4.0.3), we have:

Then it is obvious that the second term is the first homogeneous solution and the first term is the second homogeneous solution.

Solution(2.1)
Now,we consider the Euler L2-ODE CC,

Start by choosing $$ y= e^rx $$,

Substituting (6.4.2.2) and (6.4.2.3) into (6.4.2.1)

which can be rewritten as,

Comparing with the characteristic equation,

we get ,

Solution(2.2)
The first homogeneous solutions can be found from as we know the value of r from the characteristic equation,

Solution(2.3)
From the variation of parameters,

Substituting in Eq. (6.4.2.1)

which gives

From this we get,

Hence the complete solution is ,

Solution(2.4)
From (6.4.2.9) we can identify the second term $$e^{5x} $$ as the first homogeneous solution. The second homogeneous solution is thus $$ x e^{5x} $$

Author and References

 * Solved and Typed by -- Pushkar Mishra and Rui Che
 * Reviewed by -- Pushkar Mishra

Problem R*6.5-Particular solution by using variation parameters method
sec11-3 sec11-4 sec11-5 sec12-2 sec32-2

Statement
Recall p.12-2 R*3.3 Use the same idea of variation of constants(parameters) to find the particular solution $$y_P(x)$$ after knowing the homogeneous solution $$y_H(x)$$, i.e. consider the following solution $$y(x)=A(x)y_H(x)$$, with $$A(x)$$ being the unknown to be found.

Solution
Recall

Derive Eq(6.5.1) we obtain

Derive Eq(6.5.3) we obtain

Substitute Eq(6.5.5) into Eq(6.5.4)

Substitute Eq(6.5.1) and Eq(6.5.6) into Eq(6.5.2), we obtain

Rearrange Eq(6.5.7)

Integrate Eq(6.5.8)

Finally, we obtain the particular solution is

where

End

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Pushkar Mishra

Problem R6.6 – Use of special IFM to solve a Nonhomogeneous L2-ODE-CC with general excitation f(t)
sec35-5 sec33-1

Statement
Solve the nonhomogeneous L2-ODE-CC
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$$ a_2y''+a_1y'+a_0y=f(t) $$     (6.6.1)
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Solving PDE's to determine integrating factor h(x,y)

The non homogenous L2-ODE-CC with an arbitrary excitation factor f(t). The PDE's as described above are complex and difficult to solve (?)

Determine integrating factor using trial solution

Trial solution for the integrating factor $$ h(t)=e^{\alpha t}$$


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$$ \int e^{\alpha t} \left [ a_2y''+a_1y'+a_0 \right]dt =\int e^{\alpha t}f(t)dt $$     (6.6.2)
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LHS: The integration is in the form


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$$ \int e^{\alpha t} \left [ a_2y+a_1y'+a_0\right ]dt= e^{\alpha t} \left [{\bar a_2}y+{\bar a_1}y'+{\bar a_0}\right] $$     (6.6.3)
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2.1) $$ (\bar{a_1},\bar{a_0})$$ in terms of $$(a_1,a_0,a_2) $$

2.2)$$ $$ Find the quadratic equation for $$ \alpha $$

2.3) Obtain a reduced order equation.

2.4) Use the IFM to solve,

Recall,

2.5) Show that $$ \alpha \beta = \frac{a_0}{a_2} $$ and $$ \alpha + \beta = \frac{a_1}{a_2} $$, thus making $$ \alpha $$ and $$ \beta $$ solutions of the characteristic equation
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$$ (\lambda - \alpha)(\lambda - \beta) = \lambda ^2 - (\alpha + \beta )\lambda + \alpha \beta $$     (6.6 2.5.1)
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2.6) Deduce the particular solution $$ y_p(x) $$ for a general excitation $$ f(t) $$

2.7) Verify result with table of particular solutions for $$f(t)=t \exp(bt)$$

2.8) Solve the nonhomogeneous L2-ODE-CC (1)sec32-5

with the following excitation:

Gaussian distribution:

$$f(t)=exp(-t^2)$$

For the coefficients $$(a_0,a_1,a_2)$$, consider two different characteristic equations:

2.8.1)   $$(r+1)(r-2)=0$$

2.8.2)   $$(r-4)^2=0$$

2.9) For each case in 2.8.1 and 2.8.2, determine the fundamental period of undamped free vibration. Plot the homogeneous soln $$y_H(t)$$ for about 5 periods, the particular solution $$y_P(t)$$ for the excitation (3)-(4) p33-5 for the same time interval, and the complete solution $$y(t)$$, assuming zero initial conditions.

2.1) Find $$ (\bar{a_1},\bar{a_0})$$ in terms of $$(a_1,a_0,a_2)$$
By observation we know to avoid having the term $$ y^{'''}$$ the value of $$ \bar a_2=0 $$

Differentiating with respect to 't' the RHS of (6.6.3) we obtain the following,


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$$ \frac{\mathrm{d}e^{\alpha t}[a_1y'+a_0 y]}{\mathrm{d} t} $$ (6.6.4)
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$$ \frac{\mathrm{d} \left [e^{\alpha t} \bar a_1\ y'\right ]}{\mathrm{d} t}+ \frac{\mathrm{d} \left [e^{\alpha t} \bar a_0\ y\right ]}{\mathrm{d} t} $$ (6.6.5)
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$$ \bar a_1\left [ e^{\alpha t} y''+ y' e^{\alpha t} {\alpha} \right ] + \bar a_0\left [e^{\alpha t} y'+ y e^{\alpha t}{\alpha} \right] $$     (6.6.6)
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$$ \bar a_1e^{\alpha t}.y'' + \left [\bar a_1 e^{\alpha t}{\alpha} + \bar a_0 e^{\alpha t} \right ].y' + \bar a_0 e^{\alpha t}{\alpha}.y $$ (6.6.7)
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Compare coefficients of $$ y^{''}$$, $$ y^{'}$$, and $$ y $$ in equation (6.6.7) with the LHS of equation (6.6.2), hence by comparison,


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$$ \bar a_1=a_2, \bar a_0.{\alpha}= a_0 $$     (6.6.8)
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and


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$$ \bar a_1.{\alpha}+ \bar a_0 = a_1 $$     (6.6.9)
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which by using (6.6.8) can also be written as


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$$ \bar a_0 = a_1 - {\alpha}a_2 $$     (6.6.10)
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which are the values of $$ (\bar a_1, \bar a_0) $$ in terms of $$ (a_0, a_1, a_2)$$

2.2) Find the quadratic equation for $$ \alpha $$
As seen equation (6.6.8),


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$$ \bar a_0.{\alpha}= a_0, or,  \bar a_0=\frac{a_0}{\alpha} $$     (6.6.11)
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Equating the LHS terms from (6.6.10) and (6.6.11)


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$$ a_1 - {\alpha}a_2=\frac{a_0}{\alpha} $$     (6.6.12)
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$$ a_1.{\alpha}- a_2.{\alpha}^2 = a_0 $$     (6.6.13)
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$$ a_2.{\alpha}^2 - a_1.{\alpha}+ a_0 = 0 $$     (6.6.14)
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which is the quadratic equation in terms of $$ {\alpha} $$

2.3) Obtain a reduced order equation
Equating the RHS from equation (6.6.2) and (6.6.3),


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$$ e^{\alpha t} \left [{\bar a_2}.y''+{\bar a_1}.y'+{\bar a_0}.y \right] =\int e^{\alpha t}f(t)dt $$     (6.6.15)
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We know the value of $$ \bar a_2=0 $$, hence the equation is reduced to an order of 1


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$$ e^{\alpha t} \left [{\bar a_1}.y'+{\bar a_0}.y \right] =\int e^{\alpha t}f(t)dt $$     (6.6.16)
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$$ \bar a_1.y'+\bar a_0.y =e^{-\alpha t}\int e^{\alpha t}f(t)dt $$     (6.6.17)
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which is now a L1-ODE-CC

2.4) Finding the Solution y(t)
Rearranging the equation (3)33-3 we get,

Hence, applying equation 6.6.19 and (4)33-3 into equation (1)11-5 we get,

2.5) Proof of Coefficients
Given the quadratic equation for $$ \alpha $$ from part 2.2 above as
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$$ a_2 \alpha ^2 - a_1 \alpha + a_0 = 0 $$     (6.6 2.5.2)
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one can divide the entire equation by $$ a_2 $$ to obtain a coefficient of $$ 1 $$ on the $$ \alpha ^2 $$ term. One can then equate this result to the right hand side of equation 6.6. 2.5.3 in the problem statement such that
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$$ \alpha ^2 - \frac{a_1}{a_2} \alpha + \frac{a_0}{a_2} = 0 = \lambda ^2 - (\alpha + \beta )\lambda + \alpha \beta $$     (6.6. 2.5.3)
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By simple observation, one can see that, in the above equation, $$ \alpha + \beta = \frac{a_1}{a_2} $$ and $$ \alpha \beta = \frac{a_0}{a_2} $$.

2.6) Particular solution
Given the equation
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$$ y(t)=e^{-\beta t} \int ^t (\frac{e^{(\beta - \alpha)s}}{\bar {a_1}} \int ^s (e^{\alpha t} f(t)dt))ds + e^{-\beta t} \int ^t kds $$
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one can deduce that the particular solution $$ y_p(t) $$ is the portion of the above equation that includes
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$$ y_p(t)=e^{-\beta t} \int ^t (\frac{e^{(\beta - \alpha)s}}{\bar {a_1}} \int ^s (e^{\alpha t} f(t)dt))ds $$
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The remaining portion would be the homogeneous solution.

2.7) Verification of result with table of particular solution
Recall

And

Substitute Eq(2.7.2) into Eq(2.7.1), we obtain

Finally,we obtain

where

Compare to the table

They match.

End

2.8.1) Characteristic equation $$(r+1)(r-2)=0$$
- Recall

And

We obtain

Substitute Eq(2.8.1.2) and Eq(2.8.1.3) into Eq(2.8.1.1), we obtain

NOTE: Wolfram|Alpha was used to determine the following integrals:

integrate e^(-t-t^2)dt

http://www.wolframalpha.com/input/?i=integrate+e^%28-t-t^2%29dt

integrate e^(3s)1/2 e^(1/4) sqrt(π) erf(1/2+s)ds

http://www.wolframalpha.com/input/?i=integrate+e^%283s%291%2F2+e^%281%2F4%29+sqrt%28%CF%80%29+erf%281%2F2%2Bs%29ds

2.8.2) Characteristic equation $$(r+1)(r-2)=0$$
Similar to 2.8.1 above

we have

Recall

And

Substitute Eq(2.8.2.1) and Eq(2.8.2.3) into Eq(2.8.2.2), we obtain

NOTE: Wolfram|Alpha was used to determine the following integrals:

integrate e^(4t-t^2)dt

http://www.wolframalpha.com/input/?i=integrate+e^%284t-t^2%29dt

integrate -1/2 e^4 sqrt(π) erf(2-t) dt

http://www.wolframalpha.com/input/?i=integrate+-1%2F2+e^4+sqrt%28%CF%80%29+erf%282-t%29+dt

Author and References

 * Solved and Typed by -- Jinchao Lu & Jason Alphonso & Kaitlin Harris
 * Reviewed by -- Seong Hyeon Hong

Problem R*6.7 – Comparison with the method in King 2003 p.8
sec35-4

Statement
Show that (1) p 34.6 agrees with King 2003 p.8 i.e.


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$$ y_P(x) = \int^{x} f(s) [\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}] ds $$ (6.7.1)
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with,


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$$ W(s) := u_1u_2' - u_2u_1' $$
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Discuss the feasibility of the following choices for the variation of parameters,


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$$ y(x) = U (x) \pm u_1(x) $$
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$$ y(x) = U (x) / u_1(x) $$
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$$ y(x) = u_1(x) / U(x) $$
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 * }

Solution
From (1)p 34.6,


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$$ y_P(x)= u_1(x)[\int\frac{1}{h(x)}[\int h(x)\frac{f(x)}{u_1(x)}dx]dx] $$     (6.7.2)
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 * }

From the hint, we observe that, (4)35.4,


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$$ (\frac{u_2}{u_1})'=\frac{1}{h(x)} $$     (6.7.3)
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 * }

Further differentiating,


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$$ (\frac{u_1u_2'-u_2u_1'}{u_1^2})=\frac{1}{h(x)} $$     (6.7.4)
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 * }

Integrating Eq. (6.7.2) by parts,


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$$ \int u dv = uv - \int vdu $$     (6.7.5)
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 * }

We select,


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$$ u = [\int h(x)\frac{f(x)}{u_1(x)}dx] $$     (6.7.6)
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 * }


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$$ dv = \frac{1}{h(x)} = (\frac{u_2}{u_1})' $$     (6.7.7)
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 * }

Therefore,


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$$ y_P(x)= u_1(x)[\int\frac{1}{h(x)}dx[\int h(s)\frac{f(s)}{u_1(s)}ds] - \int[\int\frac{dx}{h(x)}][\frac{h(s)f(s)}{u_1(s)}]ds] $$     (6.7.8)
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 * }

As $$ \int\frac{1}{h(x)}dx = \frac{u_2}{u_1}$$ and using (6.7.4),


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$$ y_P(x)= \not{u_1(x)}[\frac{u_2(x)}{\not{u_1(x)}}[\int \frac{{u_1(s)}^{\not2}}{u_1(s)u_2'(s)-u_2(s)u_1'(s)}\frac{f(s)}{\not{u_1(s)}}ds]] - u_1(x)[\int[\frac{u_2(s)}{\not{u_1(s)}}][\frac{\not{u_1(s)^2} f(s)}{(u_1(s)u_2'(s)-u_2(s)u_1'(s){\not{u_1(s)}}}]ds]$$     (6.7.9)
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 * <p style="text-align:right">
 * }


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

$$ y_P(x)= [\int \frac{{u_1(s)}u_2(x)}{u_1(s)u_2'(s)-u_2(s)u_1'(s)}f(s)ds] -[\int [\frac{u_2(s){u_1(x)}}{(u_1(s)u_2'(s)-u_2(s)u_1'(s)}f(s)]ds] $$     (6.7.10)
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 * }

which can be rewritten as,


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$$ y_P(x)= [\int \frac{{u_1(s)}u_2(x)-u_2(s)u_1(x)}{u_1(s)u_2'(s)-u_2(s)u_1'(s)}f(s)ds] $$     (6.7.11)
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 * }

which is in the form given in King 2003 p8.

Feasibility of Choices

1.)$$ y(x)= U(x) \pm u_1(x) $$


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$$ [y(x)= U(x) \pm u_1(x)] \times a_0(x) $$     (6.7.12)
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 * }


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$$ [y'(x)= (U'(x) \pm u_1'(x))] \times a_1(x) $$     (6.7.13)
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 * <p style="text-align:right">
 * }


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

$$ [y(x)= (U(x) \pm u_1''(x))] \times a_2(x) $$     (6.7.14)
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 * <p style="text-align:right">
 * }

Adding (6.7.12), (6.7.13),(6.7.14)


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$$ a_0(x)y+a_1(x)y'+y= a_0(x) U(x) + a_1(x) U'(x) + a_2(x) U(x) + (a_0(x) u_1(x) + a_1(x) u_1'(x) + u_1''(x)) $$     (6.7.15)
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 * }

Since, $$(a_0(x) u_1(x) + a_1(x) u_1'(x) + u_1''(x) = 0 $$,


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$$ a_0(x)y+a_1(x)y'+y= a_0(x) U(x) + a_1(x) U'(x) + a_2(x) U(x) = f(x) $$     (6.7.15)
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 * <p style="text-align:right">
 * }

2.)$$ y(x)= U(x) / u_1(x) $$


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

$$ y'= \frac{u_1(x)U'(x)-U(x)u_1'(x)}{u_1(x)^2} $$     (6.7.16)
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 * <p style="text-align:right">
 * }


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

$$ y= \frac{u_1^2(u_1(x)U(x)+U'(x)u_1'(x)-U'(x)u_1'(x)-U(x)u_1''(x))-(u_1(x)U'(x)-U(x)u_1'(x))2u_1(x)}{u_1(x)^4} $$     (6.7.17)
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 * <p style="text-align:right">
 * }

3.) $$ y(x)= u_1(x) / U(x) $$


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

$$ y'= \frac{u_1'(x)U(x)-U'(x)u_1(x)}{U(x)^2} $$     (6.7.18)
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 * <p style="text-align:right">
 * }


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

$$ y= \frac{U(x)^2(U(x)u_1(x)+u_1'(x)U'(x)-u_1'(x)U'(x)-u_1(x)U''(x))-(U(x)u_1'(x)-u_1(x)U'(x))2U(x)}{U(x)^4} $$     (6.7.19)
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 * <p style="text-align:right">
 * }

From the above, we can see that our choices of trial functions does not reduce the order of non-homogeneous L2-ODE VC, hence they are not feasible.

Author and References

 * Solved and Typed by -- Pushkar Mishra
 * Reviewed by -- Kaitlin Harris

Problem R*6.8 – Proof of an Invalid Root
sec35-4

Statement
Explain why


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$$ r_2(x)=\frac{1}{x-1} $$     (6.8.1)
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 * }

is not a valid root for the equation


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$$ (x-1)y''-xy'+y=0 $$     (6.8.2)
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 * }

Solution
If $$ y_2=e^{xr_2(x)} $$ then, with equation 6.8.1


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$$ y_2=exp(\frac{x}{x-1}) $$     (6.8.3)
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 * }

One must take the first and second derivatives of equation 6.8.3, and plug these and equation 6.8.3 back into equation 6.8.2. The derivatives are as follows:
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$$ y'=\frac{d}{dx}[exp(\frac{x}{x-1})] $$     (6.8.4) Using the rules
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 * }
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$$ \frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)} $$ and
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 * }
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$$ \frac{d}{dx}(\frac{g(x)}{h(x)})=\frac{h(x)g'(x)-h'(x)g(x)}{(h(x))^2} $$ then
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 * }
 * {| style="width:100%" border="0"

$$ y'=[\frac{(x-1)(1)-x(1)}{(x-1)^2}]exp(\frac{x}{x-1})=\frac{-1}{(x-1)^2}exp(\frac{x}{x-1}) $$     (6.8.5)
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 * }


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$$ y''=(\frac{-1}{(x-1)^2})(\frac{-1}{(x-1)^2})exp(\frac{x}{x-1})+[\frac{(x-1)^2(0)-(-1)(2(x-1)(1))}{(x-1)^4}](\frac{-1}{(x-1)^2})exp(\frac{x}{x-1})=(\frac{2x-1}{(x-1)^4})exp(\frac{x}{x-1}) $$     (6.8.6)
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 * }

Substituting equations 6.8.3, 6.8.5, and 6.8.6 back into equation 6.8.2 gives
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$$ (x-1)[\frac{2x-1}{(x-1)^4}]exp(\frac{x}{x-1})-x[\frac{-1}{(x-1)^2}]exp(\frac{x}{x-1})+exp(\frac{x}{x-1})=0 $$     (6.8.7)
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 * <p style="text-align:right">
 * }

One can divide out the $$ exp(\frac{x}{x-1}) $$ term and simplify equation 6.8.7 to
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$$ \frac{2x-1}{(x-1)^3}+\frac{x}{(x-1)^2}+1=0 $$     (6.8.8)
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This equation is not valid, as there are no values of $$ x $$ that will make the left hand side of the equation equal to zero. Since $$ r_2(x) \ne constant $$, it is not a valid solution to the homogeneous equation.

Author and References

 * Solved and typed by -- Kaitlin Harris and SeongHyeon Hong
 * Reviewed by -- Pushkar Mishra

Problem R*6.9-2nd homogeneous solution by variation of parameters
sec34-4 sec34-5 sec35-3 sec35-4

Statement
For the L2-ODE-VC(1) p.35-3,

select a valid homogeneous solution, and call it $$u_1$$.

Find the 2nd homogeneous solution $$u_2(x)$$ by variation of parameters, and compare to $$e^{xr_2(x)}$$

Solution
Trial solution:$$y=e^{rx},r=constant$$

Characteristic equation:

WA:

From R*6.8, we know that $$r_2(x)$$ is not a valid root. Therefore,

Recall,

Determine $$a_1(x)$$

Hence,

Substitute Eq(6.9.2) and Eq(6.9.5) into Eq(6.9.3), we obtain Hence,

Substitute Eq(6.9.2) and Eq(6.9.6) into Eq(6.9.4), we obtain

Finally, we obtain

Compare $$ u_2(x)=-x $$ to $$u_2(x)=e^{xr_2(x)}$$, we can know that $$u_2(x)=e^{xr_2(x)}$$ is not a valid solution.

End

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Kaitlin Harris

Problem R6.10 – Determine the Path of a Particle Moving with Air Resistance
sec63-8 sec64-9b

Statement
Consider the integral with particular values,


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$$ \int_{z(0)=50}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (6.10.1)
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 * <p style="text-align:right">
 * }


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$$ a=k/m=2, b=g=10, z(t):=v_y(t) $$
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 * }
 * }


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Initial vertical velocity: $$ z(0)=v_y(0)=50 $$
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 * }
 * }

n=2 and n=3

For each value of n

(1) The vertical velocity z(t) for different values of time (t) (2) Plot the altitude y(t) vs t (3) Find time where projectile returns to ground. i.e. t when y(t)= 0

Solution
Consider the general equation of a particle with air resistance,


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$$ m.\frac{\mathrm{d} v_y}{\mathrm{d} t}= -k.v_y^{n}- mg $$ (6.10.1)
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where $$ z(t):= v_y(t)$$, there (6.10.1) can also be written as,


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$$ mz^{'}+kz^{n}= -mg $$     (6.10.2)
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 * }

Rearranging the terms,


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$$ m.\frac{\mathrm{d} z}{\mathrm{d} t}= -(k.z^n+mg) $$     (6.10.3)
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 * }


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$$ \frac{m.dz}{k.z^n+mg}= -dt $$     (6.10.4)
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 * }


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$$ \int \frac{m.dz}{k.z^n+mg}= - \int dt $$ (6.10.5)
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 * <p style="text-align:right">
 * }


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$$ \int \frac{dz}{\frac{k}{m}z^n+g}= - \int dt $$ (6.10.6)
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 * }

Case # 1 - The value of n=2, $$ a = k/m = 2 $$ and $$ g=10 $$

(i) To calculate

Solving this equation by hand,


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$$ \int \frac{dz}{2.z^2+10} = - \int dt $$ (6.10.7)
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 * }


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$$ \frac{1}{10} \int \frac{dz}{\frac{z^2}{5}+1} $$     (6.10.8)
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 * }

Consider $$ u =\frac{z}{\sqrt{5}} $$ and hence $$ du= \frac{dz}{\sqrt{5}}$$


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$$ \frac{1}{10} \int \frac{\sqrt {5}du}{u^2+1} $$     (6.10.9)
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 * }


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$$ \frac{1}{2 \sqrt {5}} \int \frac{1}{u^2+1} $$     (6.10.10)
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 * }


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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(u) + constant(k1) $$     (6.10.11)
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 * }


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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) + k1 = -t + k2 $$ (6.10.12)
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 * }


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$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}})-\frac{1}{2 \sqrt {5}} tan^{-1}(\frac{50}{\sqrt{5}}) = -t $$     (6.10.13)
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 * }


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$$ t= 0.341247 - \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) $$     (6.10.14)
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WolframAlpha Calculation -> http://www.wolframalpha.com/input/?i={1%2F%282*5^0.5%29}{tan+inverse+%2850%2F5^0.5%29}

or to interpret z in terms of t


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$$ z = \sqrt{5}tan(1.52610 - 2\sqrt{5}t) $$     (6.10.15)
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Plot of Velocity, z(t) vs time.

(ii) To find and plot altitude y(t) vs t

The velocity in the y direction, $$ v_y $$ is the first derivative of the altitude y(t), that is $$ v_y=\frac{\mathrm{d} y_t}{\mathrm{d} t}$$


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$$ \frac{\mathrm{d} y_t}{\mathrm{d} t} = \sqrt{5}tan(1.52610- 2\sqrt{5}.t) $$ (6.10.16)
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 * }

Integrating with respect to time (t),


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$$ \int_{0}^{y_t}dy_t=\int_{0}^{t}\sqrt{5}tan(1.52610- 2\sqrt{5}.t)dt $$     (6.10.17)
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 * }


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$$ y_t=\int_{0}^{t}\sqrt{5}tan(1.52610- 2\sqrt{5}.t)dt $$     (6.10.18)
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 * }

WolframAlpha Calculation -> http://www.wolframalpha.com/input/?i=integrate+sqrt%285%29+tan%28%28-10+t%2Bsqrt%285%29+tan^%28-1%29%2810+sqrt%285%29%29%29%2Fsqrt%285%29%29+dt

Plot of Altitude (vertical distance), y(t) vs time.

(iii) To find the time (t) when projectile returns to the ground

As seen in the graph above, t ~ 0.25 (units unknown)

Case # 2 - The value of n=3, $$ a = k/m = 2 $$ and $$ g=10 $$


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$$ \int \frac{dz}{2.z^3+10} = - \int dt $$ (6.10.19)
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 * }

we know the initial conditions, as $$ z(t=0)=50 $$


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$$ \int_{50}^{z} \frac{dz}{2.z^3+10} = - \int_{0}^{t} dt $$ (6.10.20)
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 * <p style="text-align:right">
 * }

Th solution for the above integral can found using

WolframAlpha Calculation -> https://www.wolframalpha.com/input/?i=++integrate+\frac{dz}{2.z^3%2B10}

The solution will also be obtained using a numerical method like the composite rule, more specifically the composite rectangular rule ,

The field of integration is considered to be $$ \left [ a,b \right ]= \left [ 50,z \right ]$$ with the interval as $$ h=\frac{z-50}{2} $$

The two sub intervals, $$ [50,\frac{z-50}{2}]$$ and $$ [\frac{z-50}{2},z]$$

The corresponding midpoints of the two intervals respectively are $$ h_1=\frac{z-150}{2}$$ and $$ h_2=\frac{z+50}{2} $$

The functions as defined at these midpoints are,


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

$$ f [\frac{z-150}{2}]= \frac{1}{2(\frac{z-150}{2})^3+10} $$     (6.10.21)
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 * }


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$$ f [\frac{z+50}{2}]= \frac{1}{2(\frac{z+50}{2})^3+10} $$     (6.10.22)
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 * <p style="text-align:right">
 * }

Applying the composite rule to calculate the value of the integral
 * {| style="width:100%" border="0"

$$ I_0=h \left ( f(h_1)+f(h_2) \right ) $$     (6.10.23)
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 * <p style="text-align:right">
 * }


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

$$ I_0=\frac{z-50}{2}\left [ \frac{1}{2(\frac{z-150}{2})^3+10}+ \frac{1}{2(\frac{z+50}{2})^3+10} \right ] = -t $$     (6.10.24)
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 * <p style="text-align:right">
 * }


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

$$ t=\frac{50-z}{2}\left [ \frac{1}{2(\frac{z-150}{2})^3+10}+ \frac{1}{2(\frac{z+50}{2})^3+10} \right ] $$     (6.10.25)
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 * <p style="text-align:right">
 * }

which is the value of t and velocity z(t)

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Jinchao Lu

Problem R7.4 – Determine the Path of a Particle Moving with Air Resistance
sec63-8 sec64-9b

Statement
Consider the integral with particular values,


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

$$ \int_{z(0)=50}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (7.10.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ a=k/m=2, b=g=10, z(t):=v_y(t) $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


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

Initial vertical velocity: $$ z(0)=v_y(0)=50 $$ For each value of n, find the vertical velocity z(t) for different values of time (t), plot this function. Also find and plot the altitude y(t) vs time t, then find the time the projectile returns to the ground.
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

If the explicit function cannot be found, use a numerical method to find z(t) for each given value of t.

(1) n=2

(2) n=3. Use the matlab command "roots" to find the appropriate root z for each given time t, verify with WA. Plot z versus t. Find y(t) by integrating z(t) using the trapezoidal rule.

(3) n=3 Use the matlab function "hypergom" to find the time t for each given value of z in the interval [-10,50]. Plot t versus z. Find y(t) by integrating z(t) using the trapezoidal rule. Compare with part 2

(4) Verify the results in Part 1 and 2 using matlab in 2 steps (a) use the command ode45 to integrate and obtain z(t), (b) use the trapezoidal rule to integrate z(t) to obtain y(t)

(5) The eq. of motion can be written as a system 1st-order ODE's to be integrated using matlab ode45:

Statement
For the given equation below,

here is the first homogeneous solution,

Show that the second homogeneous solution is,

Solution
Rearranging the equation (2)sec7-1 we get,

It is proven in the lecture note that,

where $$u_1(x)$$ and $$u_2(x)$$ are first and second homogeneous equations respectively.

Therefore if we substitute $$u_1(x)$$ for $$P_2(x)$$ and $$u_2(x)$$ for $$Q_2(x)$$ we get,

The last equation was obtained by the website WolframAlpha.

From the result, it is clear that the result is not same as the given equation. But if we look at the first term of the result and other two given solutions as below,

it seems like there will be a high chance of getting the same $$Q_2(x)$$ if we apply,

Therefore, now $$Q_2(x)$$ becomes,

Again the integration was done by the website WolframAlpha.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Jason Alphonso

Statement
Consider the Nonhomogeneous Legendre equation:

Given the 1st homogeneous solution:

Find the final solution y(x) by variation of parameters.

Solution
Set $$y(x)=U(x)u_1(x)=U(x)x$$ and substitute it into the Legendre equation;

Since coefficient for $$U(x)$$ equals to 0, we can assume $$Z(x)=U'(x)$$. Then (6.12.5) becomes

It is obvious that the 1st exactness condition is satisfied:

Since

The second exactness condition does not hold. We have to use IFM to solve this linear L2-ODE-VC. Assume $$h(x,y)$$ ,

Set:

Assume $$h=h(x)$$ ,

Then:

So

By substituting (6.12.13) into (6.12.9), we can have:

where $$k=C_2/C_1$$. Then U can be found by integrating Z with respect to x:

If we set $$k'=-k$$, then:

By substituting $$U(x)$$ back, y can be written as:

This is the final result of the Legendre equation, in which the first term is the particular root, while the second and the third therms are the 2nd and the 1st homogeneous roots respectively.

Author and References

 * Solved and Typed by -- Rui Che
 * Reviewed by -- Jason Alphonso

Team Contributions

 * Seong Hyeon Hong (SH) Solved all problems.
 * Jason Alphonso (JA) Solved all problems.
 * Pushkar Mishra (PM) Solved all problems.
 * Jinchao Lu (JL) Solved all problems.
 * Che Rui (CH) Solved all problems.
 * Kaitlin Harris (KH) Solved all problems.