User:Eml6321.f12.team4.alphonso

Statement
Draw the polar coordinate lines $$(\xi_1,\xi_2)=(r,\theta)$$ in a 2-D plane emanating from a point not at the origin.

Solution


The polar coordinates may be represented by concentric circles around the point, which is not the origin (0,0)

Statement
Show that $$ L_2(\cdot) $$ is linear

where

$$ L_2(\cdot)=\frac{d^2(\cdot)}{dx^2}+ a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot)$$

Solution
$$ L_2(\cdot)$$ is Linear if and only if (iff)

$$ L_2(\alpha u+\beta v)= \alpha L_2(u)+ \beta L_2(v)\, \forall\alpha,\beta \in R $$

where u and v are functions.

LHS

= $$ L_{2}\left ( \alpha u+\beta v \right )$$

= $$\frac{\mathrm{d^2} }{\mathrm{d} x^2}\left ( \alpha u+\beta v \right )+ a_1(x)\frac{\mathrm{d} }{\mathrm{d} x}\left ( \alpha u+\beta v \right )+a_0(x)\left ( \alpha u+\beta v \right )$$

= $$ \alpha \frac{\mathrm{d^2}u }{\mathrm{d} x^2} + \beta \frac{\mathrm{d^2}v }{\mathrm{d} x^2}+ a_1(x)\alpha\frac{\mathrm{d} u}{\mathrm{d} x}+a_1(x)\beta\frac{\mathrm{d} v}{\mathrm{d} x} + a_0(x)\alpha u +a_0(x)\beta v $$

Separating out the $$ \alpha $$ and $$ \beta $$ terms,

= $$\alpha \left [ \frac{\mathrm{d^{2}u} }{\mathrm{d} x^{2}}+a_{1}(x)\frac{\mathrm{d}u }{\mathrm{d} x}+a_{0}(x)u \right ]+ \beta \left [ \frac{\mathrm{d^{2}v} }{\mathrm{d} x^{2}}+a_{1}(x)\frac{\mathrm{d}v }{\mathrm{d} x}+a_{0}(x)v \right ] $$

= $$\alpha L_2\left ( u \right )+\beta L_2\left ( v \right )$$

=RHS, hence linearity is proved.

Statement
Solve the N1-ODE $$ \frac{d}{dx}\phi(x,y)=75x^{4}+ (\cos y){y}^'= 0 $$ and verify the solution is $$ y(x)= sin^{-1}(k-15x^{5})$$

Solution
The solution to the above problem is to first verify if the N1-ODE meets the 1st and 2nd exactness conditions,

1st Exactness Condition - Verifying if the N1-ODE is affine, that is it can be represented in the form $$ M(x,y)+N(x,y)\,y'=0 $$

By observation, the 1st Exactness Condition is met, where $$ M(x,y)= 75x^4 $$ and $$ N(x,y)= cos y $$

2nd Exactness Condition - Verify that $$ M_{y}(x,y)= 0 $$ and $$ N_{x}(x,y)= 0 $$ which by calculation is also true.

Hence there exist a function $$ \phi \left ( x,y \right )$$. Once we confirmed the existence of $$ \phi \left ( x,y \right )$$, we will now solve to find it's value.

Integrating N1-ODE,

$$ \int 75x^{4}+ (cos y) y' = 0 $$ (2.7.1)

$$\int 75x^{4} + \int (cos y) y' = 0 $$ (2.7.2)

$$\frac{75x^{5}}{5} + sin y + k = 0$$ where k = sum total of the constants of integration. (2.7.3)

$$sin y = k - 15x^{5}$$ (2.7.4)

$$y(x)= sin^{-1}(k - 15x^{5})$$

Which is the required solution.

Author and References

 * Solved and Typed by -- Jason Alphonso

Lecture Notes Reference:

Statement
Explain why it is so challenging to solve for Euler's integrating factor -> h(x,y)

Reasoning
Understanding the complexity of solving for the integrating factor h(x,y)requires an understanding of the origin of the factor.

When an N1-ODE is affine (meets the 1st exactness condition)but does not posses 2nd exactness criterion, Euler that proposed the use of a factor h(x,y) which satisfies the following equation

$$ h(x,y)\left [ M(x,y)+N(x,y)y^{'}) \right ]=0 $$ (2.8.1)

$$ (hM)+(hN)y^{'}= 0 $$ (2.8.2)

let $$ hM = \hat{M}$$ and $$ hN = \hat{N}$$ (2.8.3)

Now applying 2nd exactness criterion to the equation above,

$$ \hat{M_y}= \hat{N_x}$$ (2.8.4)

where $$ \hat{M_y}= h_{y}M+ hM_{y}$$ and $$ \hat{N_x}= h_{x}N+ hN_{x}$$ (2.8.5)

After solving as per the 2nd exactness criterion above,

$$ h_xN-h_yM+h(N_x - M_y)=0 $$ (2.8.6)

1. As noted above the values of h_x and h_y are partial derivatives of h wrt to x and y respectively. While h(x,y) is unknown the presence of two partial derivative in one equation adds complexity.

2. While N(x,y) and M(x,y) are known, the product of these terms with h_x and h_y respectively adds complexity.

This complex equation can be solved by makes assumptions, that is driving certain terms to zero and studying the case individually.

Author and References

 * Solved and Typed by -- Jason Alphonso

Lecture Notes Reference:

Problem R*3.4 – Determine if the equation is "exact" or "not exact". If "not exact" find the IFM (h) to make it exact
sec12-5

Statement
Consider the general, non-homogeneous, linear, first order ordinary differential equation with varying coefficients (L1-ODE-VC) given below,
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$$ \left [x^{4}y+10\right ]+ \left ( \frac{1}{2}x^{2} \right )y'=0 $$      (3.4.1)
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Is the above L1-ODE-VC "exact"? If "not exact" find IFM (h) to make it exact.

Solution
While the above L1-ODE-VC is affine, the second exactness criteria needs to be verified, which is,
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$$ M_{y}(x,y)= N_{x}(x,y) $$      (3.4.2)
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Comparing the equation provided in the problem statement(3.4.1) to the general affine equation below,


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$$ M(x,y)+N(x,y)\,y'=0 $$      (3.4.3)
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Hence, by comparison between (3.4.1) and (3.4.3) values of $$ M(x,y)= x^{4}y+10 $$ and $$ N(x,y)= \frac{1}{2}x^{2} $$,

To calculate $$ M_{y}(x,y)$$,


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$$ M_{y}(x,y)= \frac{\partial }{\partial y}\left ( x^ {4}y+10\right )= x^{4} $$      (3.4.4)
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To calculate $$ N_{x}(x,y)$$,


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$$ N_{x}(x,y)= \frac{\partial }{\partial x}\left ( \frac{1}{2}x^{2}\right )= x $$ (3.4.5)
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Therefore since


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$$ M_{y}(x,y)\neq N_{x}(x,y) $$      (3.4.6)
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the L1-ODE-VC is "not exact".

Part 2: To calculate the IFM (h),

Consider the derivation of the term


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$$ h(x,y)\left [ M(x,y)+N(x,y)y^{'}) \right ]=0 $$     (3.4.7)
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$$ (hM)+(hN)y^{'}= 0 $$ (3.4.8)
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let
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$$ hM = \hat{M}$$ and $$ hN = \hat{N}$$ (3.4.9)
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2nd exactness criterion applied to the equation above,


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$$ \hat{M_y}= \hat{N_x}$$ (3.4.10)
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where $$ \hat{M_y}= h_{y}M+ hM_{y}$$ and $$ \hat{N_x}= h_{x}N+ hN_{x}$$ (3.4.11)
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After solving as per the 2nd exactness criterion above,
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$$ h_xN-h_yM+h(N_x - M_y)=0 $$ (3.4.12)
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Now (3.4.12) is a complex PDE that can be solved by using a case in which $$ h_y(x,y)=0 $$, making h -> function strictly of x.


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$$ h_xN+h(N_x - M_y)=0 $$ (3.4.13)
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$$ \frac{h_x}{h}=-\frac{1}{N}(N_{x}-M_{y})=:n\left ( x \right ) $$ (3.4.14)
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To calculate the value of $$ n\left ( x \right ) $$, and from values of $$N(x,y)$$, $$ M_y(x,y) $$ and $$ N_x(x,y) $$ from (3.4.3),(3.4.4) and (3.4.5) respectively,


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$$ n\left ( x \right )= \frac{1}{\frac{1}{2}x^{2}}\left ( x-x^{4} \right ) $$ (3.4.15)
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$$ \frac{-2}{x}+2 x^{2}=:n(x) $$ (3.4.16)
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Integrating on both sides for equation (3.4.14)wrt x,


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$$log h_x= \left [ \int ^{x} \left ( \frac{-2}{x}+2 x^{2} \right )+ k1 \right ] $$ (3.4.17)
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$$h_x=exp^\left( {-2 log x + \frac{2}{3}x^{3}+k1} \right ) $$ (3.4.18)
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where k1=constant of integration.

$$ h_x $$ is the value of the Euler Integration Factor as required by the problem statement.

Author and References

 * Solved and Typed by -- Jason Alphonso

Problem R*3.10 – To calculate the solutions for a L1-ODE-VC and L1-ODE-CC
sec15-1 sec15-2

Statement
An L1-ODE-CC (Linear First Order -ODE - Constant Coefficients) is given in the form $$ \dot x(t)= ax(t)+bu(t)$$.Using IFM show that the solution is


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$$ x(t)= \left [ exp\left \{ a(t-t_{0}) \right \} \right ]x(t_0)+ \int_{t_0}^{t}\left [ exp\left \{ a(t-\tau ) \right \} \right ]bu(\tau)d\tau $$
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Similarly for an L1-ODE-VC (Linear First Order -ODE - Varying Coefficients) given in the form $$ \dot x(t)= a(t)x(t)+b(t)u(t)$$, show that the solution is in the form


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$$ x(t)= \left [ exp\int_{t_0}^{t} a(\tau)d\tau \right ]x(t_0)+ \int_{t_0}^{t}\left [ exp \int_{\tau}^{t} a(s)ds) \right ]b(\tau)u(\tau))d\tau $$
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Solution
L1-ODE-CC : Putting the equation $$ \dot x(t)= ax(t)+bu(t)$$ in the form,


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$$ -\dot x(t)+ (ax(t)+bu(t)) = 0 $$      (3.10.1)
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$$ N=-1$$, and $$ M=ax(t)+bu(t) $$      (3.10.2)
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The two variables are t = independent variable and x = dependent variable, hence there exist a function h (x,t).

Assuming a case with h (x,t) is a function of t only, such that the equation


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$$ \frac{h_t}{h}= \frac{1}{N}\left [ M_x-N_t \right ] $$      (3.10.3)
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where $$M_x=a$$, $$N_t=0$$,  $$N=-1$$ as calculated from equations above.

Hence equation (3.10.3) can now be written as,


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$$ \frac{h_t}{h}= - a $$ (3.10.4)
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Integrating the RHS term between an arbitrary start time $$ t_0$$ and $$t$$,

$$ log h= -a(t-t_0) $$      (3.10.5)
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$$ h=exp ^{-a(t-t_0)} $$      (3.10.6)
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which is Euler's Integrating factor h

Now multiplying both sides of equation (3.10.1) with the integrating factor $$ h=exp ^{-a(t-t_0)} $$


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$$ exp ^{-a(t-t_0)}\dot x(t)= exp ^{-a(t-t_0)}ax(t)+ exp ^{-a(t-t_0)}bu(t) $$      (3.10.7)
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Rearranging the terms,


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$$ exp ^{-a(t-t_0)}\dot x(t) - a exp ^{-a(t-t_0)}x(t)= exp ^{-a(t-t_0)}bu(t) $$      (3.10.8)
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$$ \frac{\mathrm{d} (exp^{-a(t-t_0)}.x(t))}{\mathrm{d} t}=exp ^{-a(t-t_0)}bu(t) $$      (3.10.9)
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Integrating both sides between $$ t_0$$ and $$t$$,


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$$ \int_{t_0}^{t}d(exp^{-a(t-t_0)}x(t)= \int_{t_0}^{t}exp^{-a(\tau-t)}b u(\tau)d\tau $$      (3.10.10)
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$$ exp^{-a(t-t_0)}x(t)-x(t_o)=\int_{t_0}^{t}exp^{-a(\tau-t)}b u(\tau)d\tau $$      (3.10.11)
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$$ exp^{-a(t-t_0)}x(t)= x(t_o)+\int_{t_0}^{t}exp^{-a(\tau-t)}b u(\tau)d\tau $$      (3.10.12)
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Divide by the term $$ exp^{-a(t-t_0)}$$ provides the solution.


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$$ x(t)= exp^{a(t-t_0)}x(t_o)+\int_{t_0}^{t}exp^{a(t-\tau)}b u(\tau)d\tau $$      (3.10.13)
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Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reference:

Problem R5.11 – Determine the Path of a Particle Moving with Air Resistance
sec63-8 sec64-9b

Statement
Consider the integral in


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$$ \int_{z(0)=0}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (5.11.1)
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$$ \int \frac{dz}{az^n+b}= \frac{1}{b}z\ _2F_1 \left(1,\frac{1}{n};1+ \frac{1}{n};-a\frac{z^n}{b}\right)+k $$     (5.11.2)
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Let value of n=3, a=2 and b=10

Find,

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

Solution
Substituting the values of n=3, a=2 and b=10 in equation (5.11.2),


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$$ \int \frac{dz}{2z^3+10}= \frac{1}{10}z\ _2F_1 \left(1,\frac{1}{3};1+ \frac{1}{3};-2\frac{z^3}{10}\right)+k $$     (5.11.3)
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$$ \int \frac{dz}{2z^3+10}= \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right)+k $$     (5.11.4)
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$$ \int_{z(0)=0}^{z(t)=z}\frac{dz}{2z^3+10}= \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right) $$     (5.11.5)
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From equation (5.11.1) this equates to -t,


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$$ -t = \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right) $$     (5.11.6)
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$$ t = - \frac{z}{10}\ _2F_1 \left(1,\frac{1}{3};\frac{4}{3};-\frac{z^3}{5}\right) $$     (5.11.6)
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Author and References

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

Problem R5.14 – Using Euler's formula find the expression for X(x) in terms of cosh Kx and sinh Kx
sec27-6 sec30-8

Statement
Find the expressions for


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$$ X(x)= e^{rx} $$     (5.14.1)
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in terms of Cos Kx, Sin kx, Cosh Kx and Sinh Kx

Solution
Euler's formula based on Eq(5) sec27-6


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$$ e^{ikx}=cos kx+ i sin kx $$ (5.14.2)
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Trigonometrical definition of $$ e^{ikx} $$ in terms of cosh kx and sinh kx,


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$$ e^{\pm ikx}=cosh(ikx)\pm sinh(ikx) $$     (5.14.3)
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From vibration analysis problem,


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$$ X^{(4)}- K^{4} x = 0 $$     (5.14.4)
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$$ r_{1,2}=\pm K, r_{3,4}=\pm iK $$ (5.14.5)
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Based on Eq(1) sec30-8


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$$ X(x)= \sum_{i=1}^{4}c_{i}e^{r_{i}x} $$     (5.14.6)
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$$ X(x)= c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}+c_{3}e^{r_{3}x}+c_{4}e^{r_{4}x} $$     (5.14.7)
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Substituting from (5.14.5) for values of $$ r_{1}, r_{2}, r_{3}, r_{4} $$


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$$ X(x)= c_{1}e^{Kx}+c_{2}e^{-Kx}+c_{3}e^{iKx}+c_{4}e^{-iKx} $$     (5.14.8)
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Substituting from Eq. (5.14.2) and (5.14.3),


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$$ X(x)= c_{1}e^{Kx}+c_{2}e^{-Kx} + c_{3} (cos{kx}+i\sin{kx})+c_{4}(cosh{kx}-i\sinh{kx}) $$     (5.14.9)
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$$ X(x)= c_{1}e^{Kx}+c_{2}e^{-Kx} + c_{3}\cos{kx}+c_{4}\cosh{kx} +i (c_{3}\sin{kx}-c_{4}\sinh{kx}) $$     (5.14.10)
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which separates it into real and imaginary parts.

Author and References

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

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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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)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{\mathrm{d} \left [e^{\alpha t} \bar a_1\ y'\right ]}{\mathrm{d} t}+ \frac{\mathrm{d} \left [e^{\alpha t} \bar a_0\ y\right ]}{\mathrm{d} t} $$ (6.6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \bar a_1\left [ e^{\alpha t} y''+ y' e^{\alpha t} {\alpha} \right ] + \bar a_0\left [e^{\alpha t} y'+ y e^{\alpha t}{\alpha} \right] $$     (6.6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \bar a_1e^{\alpha t}.y'' + \left [\bar a_1 e^{\alpha t}{\alpha} + \bar a_0 e^{\alpha t} \right ].y' + \bar a_0 e^{\alpha t}{\alpha}.y $$ (6.6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Compare coefficients of $$ y^{''}$$, $$ y^{'}$$, and $$ y $$ in equation (6.6.7) with the LHS of equation (6.6.2), hence by comparison,


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

$$ \bar a_1=a_2, \bar a_0.{\alpha}= a_0 $$     (6.6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

and


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

$$ \bar a_1.{\alpha}+ \bar a_0 = a_1 $$     (6.6.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

which by using (6.6.8) can also be written as


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

$$ \bar a_0 = a_1 - {\alpha}a_2 $$     (6.6.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

which are the values of $$ (\bar a_1, \bar a_0) $$ in terms of $$ (a_0, a_1, a_2)$$

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


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

$$ \bar a_0.{\alpha}= a_0, or,  \bar a_0=\frac{a_0}{\alpha} $$     (6.6.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Equating the LHS terms from (6.6.10) and (6.6.11)


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

$$ a_1 - {\alpha}a_2=\frac{a_0}{\alpha} $$     (6.6.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ a_1.{\alpha}- a_2.{\alpha}^2 = a_0 $$     (6.6.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ a_2.{\alpha}^2 - a_1.{\alpha}+ a_0 = 0 $$     (6.6.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

which is the quadratic equation in terms of $$ {\alpha} $$

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


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

$$ e^{\alpha t} \left [{\bar a_2}.y''+{\bar a_1}.y'+{\bar a_0}.y \right] =\int e^{\alpha t}f(t)dt $$     (6.6.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We know the value of $$ \bar a_2=0 $$, hence the equation is reduced to an order of 1


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

$$ e^{\alpha t} \left [{\bar a_1}.y'+{\bar a_0}.y \right] =\int e^{\alpha t}f(t)dt $$     (6.6.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \bar a_1.y'+\bar a_0.y =e^{-\alpha t}\int e^{\alpha t}f(t)dt $$     (6.6.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

which is now a L1-ODE-CC

Problem R6.10 – 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"

$$ \int_{z(0)=50}^{z(t)}\frac{dz}{az^n+b}= -\int_0^t dt = -t $$     (6.10.1)
 * 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: $$ z(0)=v_y(0)=50 $$
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 * style="width:95%" |
 * }
 * }

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,


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

$$ m.\frac{\mathrm{d} v_y}{\mathrm{d} t}= -k.v_y^{n}- mg $$ (6.10.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$ z(t):= v_y(t)$$, there (6.10.1) can also be written as,


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

$$ mz^{'}+kz^{n}= -mg $$     (6.10.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Rearranging the terms,


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

$$ m.\frac{\mathrm{d} z}{\mathrm{d} t}= -(k.z^n+mg) $$     (6.10.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{m.dz}{k.z^n+mg}= -dt $$     (6.10.4)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \int \frac{m.dz}{k.z^n+mg}= - \int dt $$ (6.10.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \int \frac{dz}{\frac{k}{m}z^n+g}= - \int dt $$ (6.10.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

(i) To calculate

Solving this equation by hand,


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

$$ \int \frac{dz}{2.z^2+10} = - \int dt $$ (6.10.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{10} \int \frac{dz}{\frac{z^2}{5}+1} $$     (6.10.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$ \frac{1}{10} \int \frac{\sqrt {5}du}{u^2+1} $$     (6.10.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} \int \frac{1}{u^2+1} $$     (6.10.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} tan^{-1}(u) + constant(k1) $$     (6.10.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) + k1 = -t + k2 $$ (6.10.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}})-\frac{1}{2 \sqrt {5}} tan^{-1}(\frac{50}{\sqrt{5}}) = -t $$     (6.10.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ t= 0.341247 - \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) $$     (6.10.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$ z = \sqrt{5}tan(1.52610 - 2\sqrt{5}t) $$     (6.10.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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 $$ v_y=\frac{\mathrm{d} y_t}{\mathrm{d} t}$$


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

$$ \frac{\mathrm{d} y_t}{\mathrm{d} t} = \sqrt{5}tan(1.52610- 2\sqrt{5}.t) $$ (6.10.16)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Integrating with respect to time (t),


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

$$ \int_{0}^{y_t}dy_t=\int_{0}^{t}\sqrt{5}tan(1.52610- 2\sqrt{5}.t)dt $$     (6.10.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ y_t=\int_{0}^{t}\sqrt{5}tan(1.52610- 2\sqrt{5}.t)dt $$     (6.10.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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 $$ g=10 $$


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

$$ \int \frac{dz}{2.z^3+10} = - \int dt $$ (6.10.19)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$ \int_{50}^{z} \frac{dz}{2.z^3+10} = - \int_{0}^{t} dt $$ (6.10.20)
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 * style="width:95%" |
 * <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)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ f [\frac{z+50}{2}]= \frac{1}{2(\frac{z+50}{2})^3+10} $$     (6.10.22)
 * style="width:95%" |
 * style="width:95%" |
 * <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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 * style="width:95%" |
 * <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)
 * style="width:95%" |
 * style="width:95%" |
 * <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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 * style="width:95%" |
 * <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 --

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:

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


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

$$ m.\frac{\mathrm{d} v_y}{\mathrm{d} t}= -k.v_y^{n}- mg $$ (7.4.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$ z(t):= v_y(t)$$, there (6.10.1) can also be written as,


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

$$ mz^{'}+kz^{n}= -mg $$     (7.4.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Rearranging the terms,


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

$$ m.\frac{\mathrm{d} z}{\mathrm{d} t}= -(k.z^n+mg) $$     (7.4.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{m.dz}{k.z^n+mg}= -dt $$     (7.4.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \int \frac{m.dz}{k.z^n+mg}= - \int dt $$ (7.4.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \int \frac{dz}{\frac{k}{m}z^n+g}= - \int dt $$ (7.4.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

(i) To calculate

Solving this equation by hand,


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

$$ \int \frac{dz}{2.z^2+10} = - \int dt $$ (7.4.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{10} \int \frac{dz}{\frac{z^2}{5}+1} $$     (7.4.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$ \frac{1}{10} \int \frac{\sqrt {5}du}{u^2+1} $$     (7.4.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} \int \frac{1}{u^2+1} $$     (7.4.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} tan^{-1}(u) + constant(k1) $$     (7.4.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) + k1 = -t + k2 $$ (7.4.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}})-\frac{1}{2 \sqrt {5}} tan^{-1}(\frac{50}{\sqrt{5}}) = -t $$     (7.4.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ t= 0.341247 - \frac{1}{2 \sqrt {5}} tan^{-1}(\frac{z}{\sqrt{5}}) $$     (7.4.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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


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

$$ z = \sqrt{5}tan(1.52610 - 2\sqrt{5}t) $$     (7.4.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Matlab code to plot z(t) against time t:

Figure: Part 1 / velocity z(t) vs. time



Figure: Part 1 / Vertical displacement y(t) vs. time



Part # 2 - n=3, Use of mathlab command "roots"

Matlab code to plot z(t) against time t:

Part # 3 - n=3, Use the matlab function "hypergom" in interval [-10,50], Plot t vs z, Use trapezoidal rule to calculate y(t)

Matlab code to plot z(t) vs t, and y(t) vs t

Figure: Part 3 / velocity z(t) vs. time



Figure: Part 3 / Vertical displacement y(t) vs. time



'''Part # 4 - Use mathlab command "ode45" to integrate and obtain z(t). Use trapezoidal rule to obtain y(t)'''

Figure: Part 4 / velocity z(t) vs. time, n=2



Figure: Part 4 / Vertical displacement y(t) vs. time, n=2



Figure: Part 4 / velocity z(t) vs. time, n=3



Figure: Part 4 / Vertical displacement y(t) vs. time, n=3



Part # 5 - Integrate a system 1st-order ODE's using matlab ode45

Author and References

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