User:Egm6321.f11.team2/HW1

=Problem 1.1- Total derivative=

Given
The first total time derivative is given by $$\displaystyle \frac {d}{dt}f(Y^1(t),t)=\frac {\partial f(Y^1(t),t)}{\partial S}\dot Y^1 + \frac {\partial f(Y^1(t),t)}{\partial t} $$ with $$\displaystyle \dot Y^1 := \frac {dY^1(t)}{dt}$$

Find
Show that

$$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1 + f_{,SS}(Y^1,t)(\dot Y^1)^2 + 2f_{,St}(Y^1,t)\dot Y^1 + f_{,tt}(Y^1,t) $$

with

Solution
Solved on my own

Chain Rule
According to chain rule , the chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function f ∘ g in terms of the derivatives of f and g.

Example
Given $$u = x^2 + 2y$$ where $$x = r\sin(t)$$ and $$y = \sin^2(t)$$, determine the value of $$\partial u/\partial r$$ and $$\partial u/\partial t$$ using the chain rule.

$$ \begin{align} \frac{\partial u}{\partial r}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \\&= \left(2x\right)\left(\sin(t)\right)+\left(2\right)\left(0\right) \\&=2r\sin^2(t) \end{align} $$

and

$$ \begin{align} \frac{\partial u}{\partial t}&=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t} \\&= \left(2x\right)\left(r\cos(t)\right)+\left(2\right)\left(2\sin(t)\cos(t)\right) \\&= 2\left(r\sin(t)\right)r\cos(t)+4\sin(t)\cos(t) \\&= 2\left(r^2+2\right)\sin(t)\cos(t) \end{align} $$

First total time derivative
$$ \begin{align} S&=Y^1(t) \\ \frac {d}{dt}f(Y^1(t),t)&=\frac {\partial f(S,t)}{\partial S}\frac {\partial S}{\partial t}+\frac {\partial f(S,t)}{\partial t} \\ &= \frac {\partial f(S,t)}{\partial S} \frac {\partial S}{\partial t}+\frac {\partial f(S,t)}{\partial t} \\ &= \frac {\partial f(Y^1(t),t)}{\partial S} \dot Y^1 +\frac {\partial f(Y^1(t),t)}{\partial t} \end{align} $$

Second total time derivative
$$ \displaystyle \frac {d}{dt}(\frac {\partial f(Y^1(t),t)}{\partial S} \dot Y^1 +\frac {\partial f(Y^1(t),t)}{\partial t})= \frac {d}{dt}(\frac {\partial f(S,t)}{\partial S}\frac {\partial S}{\partial t}+\frac {\partial f(S,t)}{\partial t}) $$ $$ =\displaystyle \frac {\partial^2 f(S,t)}{\partial S \partial S}(\frac {\partial S}{\partial t})^2 + \frac {\partial^2 f(S,t)}{\partial S \partial t} \frac {\partial S}{\partial t} + \frac {\partial f(S,t)}{\partial S} \frac {\partial^2 S}{\partial t^2} + \frac {\partial^2 f(S,t)}{\partial S \partial t} \frac{\partial S}{\partial t} + \frac {\partial^2 f(S,t)}{\partial t^2}

$$ $$ =\displaystyle \frac {\partial^2 f(S,t)}{\partial S \partial S}\dot Y^1\dot Y^1 + \frac {\partial^2 f(S,t)}{\partial S \partial t} \dot Y^1 + \frac {\partial f(S,t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(S,t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(S,t)}{\partial t^2} $$ $$ =\displaystyle \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial S}(\dot Y^1)^2 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial f(Y^1(t),t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial t^2} $$ $$ =\displaystyle \frac {\partial f(Y^1(t),t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial S}(\dot Y^1)^2 + 2\frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial t^2} $$ From Equation (1.1) and (1.2),

$$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1 + f_{,SS}(Y^1,t)(\dot Y^1)^2 + 2f_{,St}(Y^1,t)\dot Y^1 + f_{,tt}(Y^1,t) $$

=Problem 1.2 - Derive 1st and 2nd total time derivative, compare with the derivation of the Coriolis force=

Given
$$\displaystyle \frac {d}{dt}f(Y^1(t),t)=\frac {\partial f(Y^1(t),t)}{\partial S}\dot Y^1 + \frac {\partial f(Y^1(t),t)}{\partial t} $$ with $$\displaystyle \dot Y^1 := \frac {dY^1(t)}{dt}$$

$$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1 + f_{,SS}(Y^1,t)(\dot Y^1)^2 + 2f_{,St}(Y^1,t)\dot Y^1 + f_{,tt}(Y^1,t) $$ with $$ \displaystyle f_{,S}(Y^1,t) :=\frac{\partial f(Y^1,t)}{\partial S}$$ $$ \displaystyle f_{,St}(Y^1,t):=\frac{\partial ^2 f(Y^1,t)}{\partial S \partial t} $$

Find
Derive $$\displaystyle \frac {d}{dt}f(Y^1(t),t)=\frac {\partial f(Y^1(t),t)}{\partial S}\dot Y^1 + \frac {\partial f(Y^1(t),t)}{\partial t} $$ $$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1 + f_{,SS}(Y^1,t)(\dot Y^1)^2 + 2f_{,St}(Y^1,t)\dot Y^1 + f_{,tt}(Y^1,t) $$ And show the similarity with the derivation of the Coriolis force.

Solution
Solved on my own

First total time derivitive
$$\displaystyle Y^1(t)$$ is function of S. $$\displaystyle \frac {d}{dt}f(Y^1(t),t) $$

This equation is following that

$$\displaystyle \frac {d}{dt}f(Y^1(t),t)=\frac {d}{dt}f(S,t)$$ According to the Chain rule, $$ \displaystyle \frac {d}{dt}f(Y^1(t),t)=\frac {\partial f(S,t)}{\partial S}\frac {\partial S}{\partial t}+\frac {\partial f(S,t)}{\partial t}

= \frac {\partial f(S,t)}{\partial S} \frac {\partial S}{\partial t}+\frac {\partial f(S,t)}{\partial t} = \frac {\partial f(Y^1(t),t)}{\partial S} \dot Y^1 +\frac {\partial f(Y^1(t),t)}{\partial t} $$ $$\frac {df(Y^1(t),t)} {dt} = \frac {\partial f(Y^1(t),t)}{\partial S}\dot Y^1 + \frac {\partial f(Y^1(t),t)}{\partial t} $$

Second total derivative
$$ \displaystyle \frac {d}{dt}(\frac {\partial f(Y^1(t),t)}{\partial S} \dot Y^1 +\frac {\partial f(Y^1(t),t)}{\partial t})= \frac {d}{dt}(\frac {\partial f(S,t)}{\partial S}\frac {\partial S}{\partial t}+\frac {\partial f(S,t)}{\partial t}) $$ $$ =\displaystyle \frac {\partial^2 f(S,t)}{\partial S \partial S}\frac {\partial S}{\partial t} \frac {\partial S}{\partial t} + \frac {\partial^2 f(S,t)}{\partial S \partial t} \frac {\partial S}{\partial t} + \frac {\partial f(S,t)}{\partial S} \frac {\partial^2 S}{\partial t^2} + \frac {\partial^2 f(S,t)}{\partial S \partial t} \frac{\partial S}{\partial t} + \frac {\partial^2 f(S,t)}{\partial t^2}

$$ $$ =\displaystyle \frac {\partial^2 f(S,t)}{\partial S \partial S}\dot Y^1\dot Y^1 + \frac {\partial^2 f(S,t)}{\partial S \partial t} \dot Y^1 + \frac {\partial f(S,t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(S,t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(S,t)}{\partial t^2} $$ $$ =\displaystyle \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial S}(\dot Y^1)^2 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial f(Y^1(t),t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial t^2} $$ $$ =\displaystyle \frac {\partial f(Y^1(t),t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial S}(\dot Y^1)^2 + 2\frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial t^2} $$ $$\displaystyle \frac{d^2f}{dt^2}=\displaystyle \frac {\partial f(Y^1(t),t)}{\partial S} \ddot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial S \partial S}(\dot Y^1)^2 + 2\frac {\partial^2 f(Y^1(t),t)}{\partial S \partial t} \dot Y^1 + \frac {\partial^2 f(Y^1(t),t)}{\partial t^2} =f_{,S}(Y^1,t)\ddot Y^1 + f_{,SS}(Y^1,t)(\dot Y^1)^2 + 2f_{,St}(Y^1,t)\dot Y^1 + f_{,tt}(Y^1,t)$$(2.1)

Coriolis force
Derivation for the Coriolis velocity and acceleration

$$\frac{d\mathbf{A}}{dt} =  \frac{d'\mathbf{A}}{dt} +\mathbf{A_x'}(\mathbf{\Omega} \times \mathbf{\hat x'}) + \mathbf{A_y'}(\mathbf{\Omega} \times \mathbf{\hat y'}) + \mathbf{A_z'}(\mathbf{\Omega} \times \mathbf{\hat z'})$$
 * $$ = \frac{d'\mathbf{A}}{dt} +\mathbf{\Omega} \times (\mathbf{A_x'}\mathbf{\hat x'} + \mathbf{A_y'}\mathbf{\hat y'} + \mathbf{A_z'}\mathbf{\hat z'})$$
 * $$ = \frac{d'\mathbf{A}}{dt} +\mathbf{\Omega} \times \mathbf{A} $$

$$ \frac{d^2\mathbf{A}}{dt^2} = \frac{d}{dt}(\frac{d'\mathbf{A}}{dt} +\mathbf{\Omega} \times \mathbf{A}) $$
 * $$ = \frac{d}{dt}(\frac{d'\mathbf{A}}{dt}) + \mathbf{\Omega} \times \frac{d\mathbf{A}}{dt} + \frac{d\mathbf{\Omega}}{dt} \times \mathbf{A} $$
 * $$ = \frac{d'^2\mathbf{A}}{dt^2} + \mathbf{\Omega} \times \frac{d'\mathbf{A}}{dt} + \mathbf{\Omega} \times (\frac{d'\mathbf{A}}{dt} +\mathbf{\Omega} \times \mathbf{A}) + \frac{d\mathbf{\Omega}}{dt} \times \mathbf{A} $$
 * $$ = \frac{d'^2\mathbf{A}}{dt^2} + 2\mathbf{\Omega} \times \frac{d'\mathbf{A}}{dt} + \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{A}) + \frac{d\mathbf{\Omega}}{dt} \times \mathbf{A} $$(2.2)

Similarities
Eq 2.1 is $$\displaystyle \frac{d^2f}{dt^2}=f_{,S}(Y^1,t)\ddot Y^1 + f_{,SS}(Y^1,t)(\dot Y^1)^2 + 2f_{,St}(Y^1,t)\dot Y^1 + f_{,tt}(Y^1,t)$$ Eq 2.2 is $$ \frac{d^2\mathbf{A}}{dt^2} = \frac{d'^2\mathbf{A}}{dt^2} + 2\mathbf{\Omega} \times \frac{d'\mathbf{A}}{dt} + \mathbf{\Omega} \times (\mathbf{\Omega} \times \mathbf{A}) + \frac{d\mathbf{\Omega}}{dt} \times \mathbf{A} $$

=Problem 1.3 - Analyze Dimensions=

Given
$$\displaystyle c_3(Y^1,t)\ddot Y^1 + c_2(Y^1,t)(\dot Y^1)^2 + c_1(Y^1,t)\dot Y^1 + c_0(Y^1,t) = 0 $$ (1) Ref: VQ&O 1989 CMAME Eqs.(2.5a) $$\displaystyle c_0(Y^1,t) = -F^1[1-\bar R u^2_{,SS}(Y^1,t)]-F^2u^2_{,S}(Y^1,t)-\frac{T}{R}+M\left[(1-\bar R u^2_{,SS}(Y^1,t))(u^1_{,tt}(Y^1,t)-\bar R u^2_{,Stt}(Y^1,t))+u^2_{,S}(Y^1,t)u^2_{,tt}(Y^1,t)\right]$$ (2) Ref: VQ&O 1989 CMAME Eqs.(2.5b) Where:
 * $$\displaystyle u^2_{,SS}(Y^1,t) := \frac {\partial u^2(Y^1,t)}{(\partial S)^2}$$


 * $$\displaystyle u^1_{,tt}(Y^1,t) := \frac {\partial u^1(Y^1,t)}{(\partial t)^2}$$

Find
Analyze the dimension of all terms in $$c_0(Y^1,t)$$, and provide the physical meaning.

Solution
I solved this problem on my own.

The dimensions of all the components of $$c_0(Y^1,t)$$: $$[F^1] = F$$ $$[1] = 1$$ $$[\bar R] = L$$ $$[u^1(Y^1,t)] = L$$ $$[u^2(Y^1,t)] = L$$ $$[S] = L$$ $$[t] = S$$ $$[T] = FL$$ $$[R] = L$$ $$[M] = M$$ Where, F = Force L = Length S = Time M = Mass $$u^1(Y^1,t)$$ and $$u^2(Y^1,t)$$ have dimension L because both are measures of the displacement of the guideway. Replacing each term in equation (2) by its dimension yields: $$\displaystyle [c_0(Y^1,t)] = -F[1-L \frac {L}{L^2}]-F \frac {L}{L}-\frac{FL}{L}+M\left[(1-L \frac {L}{L^2})(\frac {L}{S^2}-L \frac {L}{LS^2})+\frac {L}{L} \frac {L}{S^2}\right]$$ $$\displaystyle [c_0(Y^1,t)] = -F-F-F+M \frac {L}{S^2}$$ $$[F]= [M] \cdot [a] = \frac {ML}{S^2}$$ $$\displaystyle [c_0(Y^1,t)] = F$$ The dimension of $$c_0(Y^1,t)$$ shows that it is a force. Specifically the horizontal force acting on the wheel/magnet.

=Problem 1.4 - Curvilinear Coordinate Lines=

Find
Draw the polar coordinate lines, in a 2-D plane emanating from a point, not at the origin.

Solution
''Note that this problem was solved as per Professor Vu-Quoc's example during lecture 6, where he implicitly defined coordinate lines as corresponding to only positive values of the coordinate in question. However, if the coordinates curves were defined by allowing a coordinate to vary to all possible values, as done by Wolfram|Alpha, then the radial coordinate curve would be the entire line containing the origin and P, while the axial coordinate curve would be the circle centered at the origin that passes through P.''

Coordinate lines, also called coordinate curves, are produced by varying one coordinate while holding all others constant. In planar polar coordinates, the coordinate curve associated with the radial coordinate is a ray starting at the point P extending away from the origin along a line containing the origin and P. The coordinate curve associated with the axial coordinate is an arc of a circle centered at the origin, extending in a counter-clockwise direction from P.

In the diagram below, the primed coordinates are cartesian coordinates, x1 is the polar coordinate line emanating from the point P, and x2 is the axial coordinate curve emanating from the point P. The red curves show the line containing the origin and the point P associated with the polar coordinate curve and the origin-centered circle associated with the axial coordinate curve.

=Problem 1.5 - Simplification=

Given
The project equation is as follows : $$\displaystyle \frac{1}{g_{i}(\xi_{i})}\frac{d}{d\xi_{i}}\left[g_{i}(\xi_{i})\frac{dX_{i}(\xi_{i})}{d\xi_{i}}\right]+f_{i}(\xi_{i})X_{i}(\xi_{i})=0$$

Find
Show that the Given equation has the following form: $$\displaystyle y''+\frac{g'(x)}{g(x)}y'+a_{0}(x)y=0$$

Solution
In order to simplify the given equation, we shall first change symbols: $$ \displaystyle \xi_{i}=x,\qquad X_{i}(\xi_{i})=y(x);$$ $$ \displaystyle g_{i}(\xi_{i})=g(x),\qquad f_{i}(\xi_{i})=a_{0}(x).$$ Substitute the above expressions into the given equation: $$ \displaystyle \frac{1}{g(x)}\frac{d}{dx}\left[g(x)\frac{dy(x)}{dx}\right]+a_{0}y(x)=0$$ (5.1) $$ \displaystyle \frac{1}{g(x)}\left[\frac{dg(x)}{dx}\frac{dy(x)}{dx}+g(x)\frac{d^{2}y(x)}{dx^{2}}\right]+a_{0}y(x)=0$$ (5.2) $$ \displaystyle y''+\frac{g'(x)}{g(x)}y'+a_{0}(x)y=0$$ where $$ \frac{dg(x)}{dx}=g'(x), \frac{dy(x)}{dx}=y', \frac{d^{2}y(x)}{dx^{2}}=y''.$$

=Problem 1.6 - Nonlinear 2nd-Order Term in Nonlinear 2nd-Order ODE (N2-ODE)=

Given
In the equation of motion (EOM) of wheel/magnet, (1) p.3-3,

$$\displaystyle c_3(Y^1,t)= M[1-\bar {R}u^2_{,SS}(Y^1,t)] $$

Find
Show $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear with respect to $$\displaystyle Y^1$$

Solution
 EGM 6341 Fall 2011 (PEA1 F11) Team 2 HW1.6 (solved on my own)

Linearty has to satisfy the following condition:

Nonlinearty is the condition not satisfing the above statement.

In the given equation, the linearity of $$\displaystyle u^2_{,SS}(Y^1,t)$$ is assumed.

The results show:

Thus,

This proves that $$\displaystyle c_3(Y^1,t)\ddot Y^1$$ is nonlinear wiht respect to $$\displaystyle Y^1.$$

=Problem 1.7- Linear ODE=

Given
An ODE of 2nd order with variable coefficients is read as $$\displaystyle L_2(y)= \frac{d^2(y)}{dx^2}+a_1(x)\frac{d(y)}{dx}+a_0(x)(y) $$ where $$\displaystyle a_0(x), a_1(x)$$ are variable coefficients

Find
show that $$\displaystyle L_2(\cdot)=\frac{d^2(\cdot)}{dx^2}+a_1(x)\frac{d(\cdot)}{dx}+a_0(x)(\cdot) $$   is a linear operator.

Definition of Linear
If $$\displaystyle L_{2} $$ is a linear operator, the it must satisfy $$\displaystyle L_{2}(\alpha u+\beta v)=\alpha L_{2}(u)+\beta L_{2}(v)$$ Where $$\displaystyle u,v $$ are the independent variables, and $$\displaystyle \alpha, \beta \in \mathbb{R}.$$

$$\displaystyle L_{2}(\alpha u+\beta v)=\frac{d^{2}(\alpha u+\beta v)}{dx^{2}}+a_{1}(x)\frac{d(\alpha u+\beta v)}{dx}+a_{0}(x)(\alpha u+\beta v)$$

$$\displaystyle L_{2}(\alpha u+\beta v)=[\frac{d^{2}(\alpha u)}{dx^{2}}+a_{1}(x)\frac{d(\alpha u)}{dx}+a_{0}(x)(\alpha u)]+[\frac{d^{2}(\beta v)}{dx^{2}}+a_{1}(x)\frac{d(\beta v)}{dx}+a_{0}(x)(\beta v)]$$

$$\displaystyle L_{2}(\alpha u+\beta v)=\alpha[\frac{d^{2}(u)}{dx^{2}}+a_{1}(x)\frac{d(u)}{dx}+a_{0}(x)(u)]+\beta[\frac{d^{2}(v)}{dx^{2}}+a_{1}(x)\frac{d(v)}{dx}+a_{0}(x)(v)]$$

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

$$\displaystyle \therefore L_{2}(\cdot)$$ is a linear operator.

=Problem 1.8 - Integration Constants, Boundary Conditions, and Initial Conditions=

Given
The solution for a Linear 2nd Order Ordinary Differential Equation takes the form $$\displaystyle y(x)=c_{1}y^1_{H}(x)+c_{2}y^2_{H}(x)+y_{P}(x)$$ (1)

Find
Find the integration constants $$c_{1}$$ and $$c_{2}$$ in terms of the boundary conditions
 * $$\displaystyle y(a)=\alpha, \ y(b)=\beta$$

and the initial conditions
 * $$\displaystyle y(a)=\alpha, \ y'(a)=\beta$$

Solution
The integration constants in terms of the boundary conditions are: $$\displaystyle y(a)=\alpha=c_{1}y^1_{H}(a)+c_{2}y^2_{H}(a)+y_{P}(a)$$ (2) $$\displaystyle y(b)=\beta=c_{1}y^1_{H}(b)+c_{2}y^2_{H}(b)+y_{P}(b)$$ (3)

To solve for $$c_{1}$$ in terms of the boundary conditions $$c_{2}$$ must be removed. This can be done by multiplying equation (2) by $$y^2_{H}(b)$$ and equation (3) by $$y^2_{H}(a)$$ to obtain: $$\displaystyle y^2_{H}(b)\alpha=c_{1}y^1_{H}(a)y^2_{H}(b)+c_{2}y^2_{H}(a)y^2_{H}(b)+y_{P}(a)y^2_{H}(b)$$ (4) $$\displaystyle y^2_{H}(a)\beta=c_{1}y^1_{H}(b)y^2_{H}(a)+c_{2}y^2_{H}(b)y^2_{H}(a)+y_{P}(b)y^2_{H}(a)$$ (5)

Subtract equation (5) from equation (4) and solve for $$c_{1}$$.
 * $$\displaystyle c_{1}=\frac {y^2_{H}(b)\alpha-y^2_{H}(a)\beta-y_{P}(a)y^2_{H}(b)+y_{P}(b)y^2_{H}(a)}{y^1_{H}(a)y^2_{H}(b)-y^1_{H}(b)y^2_{H}(a)}$$ (6)

Similarly, find $$c_{2}$$ by multiplying equation (2) by $$y^1_{H}(b)$$ and equation (3) by $$y^1_{H}(a)$$ to obtain: $$\displaystyle y^1_{H}(b)\alpha=c_{1}y^1_{H}(a)y^1_{H}(b)+c_{2}y^2_{H}(a)y^1_{H}(b)+y_{P}(a)y^1_{H}(b)$$ (7) $$\displaystyle y^1_{H}(a)\beta=c_{1}y^1_{H}(b)y^1_{H}(a)+c_{2}y^2_{H}(b)y^1_{H}(a)+y_{P}(b)y^1_{H}(a)$$ (8)

Subtract equation (8) from equation (7) and solve for $$c_{2}$$.
 * $$\displaystyle c_{2}=\frac {y^1_{H}(b)\alpha-y^1_{H}(a)\beta-y_{P}(a)y^1_{H}(b)+y_{P}(b)y^1_{H}(a)}{y^2_{H}(a)y^1_{H}(b)-y^2_{H}(b)y^1_{H}(a)}$$ (9)

The integration constants in terms of the initial conditions are: $$\displaystyle y(a)=\alpha=c_{1}y^1_{H}(a)+c_{2}y^2_{H}(a)+y_{P}(a)$$ (10) $$\displaystyle y'(a)=\beta=c_{1}y'_{1H}(a)+c_{2}y'_{2H}(a)+y'_{P}(a)$$ (11) Where $$\displaystyle y'_{1H}=\frac{d(y1_{H}(x))}{dx}$$ $$\displaystyle y'_{2H}=\frac{d(y2_{H}(x)}{dx}$$ Find $$c_{1}$$ similar to the way $$c_{1}$$ was found using the boundary conditions. Multiply equation (10) by $$y'_{2H}(a)$$ and equation (11) by $$y^2_{H}(a)$$ to obtain: $$\displaystyle y'_{2H}(a)\alpha=c_{1}y^1_{H}(a)y'_{2H}(a)+c_{2}y^2_{H}(a)y'_{2H}(a)+y_{P}(a)y'_{2H}(a)$$  (12) $$\displaystyle y^2_{H}(a)\beta=c_{1}y'_{1H}(a)y^2_{H}(a)+c_{2}y'_{2H}(a)y^2_{H}(a)+y'_{P}(a)y^2_{H}(a)$$  (13) Subtract equation (13) from equation (12) and solve for $$c_{1}$$.
 * $$\displaystyle c_{1}=\frac {y'_{2H}(a)\alpha-y^2_{H}(a)\beta-y_{P}(a)y'_{2H}(a)+y'_{P}(a)y^2_{H}(a)}{y^1_{H}(a)y'_{2H}(a)-y'_{1H}(a)y^2_{H}(a)}$$ (14)

Find $$c_{2}$$ similar to the way $$c_{1}$$ was found. Multiply equation (10) by $$y'_{1H}(a)$$ and equation (11) by $$y^1_{H}(a)$$ to obtain: $$\displaystyle y'_{1H}(a)\alpha=c_{1}y^1_{H}(a)y'_{1H}(a)+c_{2}y^2_{H}(a)y'_{1H}(a)+y_{P}(a)y'_{1H}(a)$$ (15) $$\displaystyle y^1_{H}(a)\beta=c_{1}y'_{1H}(a)y^1_{H}(a)+c_{2}y'_{2H}(a)y^1_{H}(a)+y'_{P}(a)y^1_{H}(a)$$ (16) Subtract equation (16) from equation (15) and solve for $$c_{2}$$.
 * $$\displaystyle c_{2}=\frac {y'_{1H}(a)\alpha-y^1_{H}(a)\beta-y_{P}(a)y'_{1H}(a)+y'_{P}(a)y^1_{H}(a)}{y^2_{H}(a)y'_{1H}(a)-y'_{2H}(a)y^1_{H}(a)}$$ (17)

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