User:Egm6321.f12.team6/report2

=Report 2=

Problem Statement

 * According to the Legendre differential operator, verify that
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$$ \displaystyle L_2(y_H^1)=L_2 (y_H^2)=0 $$
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 * style="width:95%" |
 * (1.1)
 * }

Given

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$$ \displaystyle y_H^1(x)=x $$
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 * (1.3)
 * }
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$$ \displaystyle y_H^2(x)= \frac{x}{2}\log\left(\frac{1+x}{1-x} \right)-1 $$
 * style="width:95%" |
 * style="width:95%" |
 * (1.4)
 * }
 * $$ \displaystyle \frac{d}{dx} log(u)=\frac{1}{u} \frac{du}{dx} $$
 * $$ \displaystyle n=1 $$

Nomenclature

 * $$ \displaystyle \frac{d}{dx}y(x)=:y' $$
 * $$ \displaystyle \frac{d^2}{dx^2}y(x)=:y'' $$

Part 1: Verification of L2(yh1)

 * Step 1.1 - Given Equation 1.2
 * $$ \displaystyle L_2(y):=(1-x^2)y''-2xy'+n(n+1)y=0 $$
 * where
 * $$ \displaystyle L_2(y_H^1):=y_H^1(x)=x $$
 * Step 1.2 - Solve for, $$ \displaystyle y' $$, and $$ \displaystyle y'' $$
 * $$ \displaystyle y_H^1(x)=y(x)=x $$
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$$ \displaystyle y'=1 $$
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 * (1.5)
 * }
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$$ \displaystyle y''=0 $$
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 * style="width:95%" |
 * (1.6)
 * }
 * Step 1.3 - Substitute, $$ \displaystyle y, y', y'', n $$, into the Legendre differential equation, and solve.
 * $$ \displaystyle L_2(y_H^1):=(1-x^2)(0)-2x(1)+2x = 0 $$
 * $$ \displaystyle -2x + 2x = 0 $$
 * $$ \displaystyle 0 = 0 $$

Part 2: Verification of L2(yh2)

 * Step 2.1 - Given Equation 1.2
 * $$ \displaystyle L_2(y):=(1-x^2)y''-2xy'+n(n+1)y=0 $$
 * where
 * $$ \displaystyle L_2(y_H^2):=y_H^2(x)= \frac{x}{2}\log\left(\frac{1+x}{1-x} \right)-1$$
 * Step 2.2 - Solve for, $$ \displaystyle y' $$, using the chain rule, quotient rule and the product rule, and simplify.
 * $$ \displaystyle y'=\frac{x}{2}\left(\frac{1-x}{1+x}\right)\left[\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}\right]+\frac{1}{2}\log \left(\frac{1+x}{1-x}\right) $$
 * $$ \displaystyle y'=\frac{x}{(1+x)(1-x)}+\frac{1}{2}\log \left(\frac{1+x}{1-x}\right) $$


 * Step 2.3 - Solve for, $$ \displaystyle y $$, using the quotient rule, chain rule and the product rule'', then simplify.
 * $$ \displaystyle y''=\frac{1}{2}\left[\frac{1-x}{1+x}\left(\frac{2}{(1-x)(1-x)}\right)\right]+\frac{1-x^2-x(-2x)}{(1-x^2)^2} $$
 * $$ \displaystyle y''=\frac{1}{1-x^2}+\frac{1-x^2+2x^2}{(1-x^2)^2}$$


 * Step 2.4 - Substitute, $$ \displaystyle y, y', y'', n $$, into the Legendre differential equation, and solve.
 * $$ \displaystyle L_2(y_H^2)=0 $$
 * $$ \displaystyle (1-x^2)\frac{2}{(1-x^2)^2}-2x\left[\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]+2\left[\frac{x}{2}\log\left(\frac{1+x}{1-x} \right) -1\right] =0 $$
 * $$ \displaystyle \frac{2}{1-x^2} - x\log \left(\frac{1+x}{1-x}\right) - \frac{2x^2}{1-x^2} + x\log \left(\frac{1+x}{1-x}\right) - 2 = 0$$
 * $$ \displaystyle \frac{2}{1-x^2} - \frac{2x^2}{1-x^2} - 2 = 0 $$
 * $$ \displaystyle 2 \left( \frac{1}{1-x^2} - \frac{x^2}{1-x^2} \right) - 2 = 0 $$
 * $$ \displaystyle 2 \left( \cancelto{1} \frac{1-x^2}{1-x^2}\right) - 2 = 0 $$
 * $$ \displaystyle 2 - 2 = 0 $$
 * $$ \displaystyle 0 = 0 $$


 * Step 2.5 - From Equation 1.7 and 1.10, we can verify Equation 1.1, that is;
 * {| style="width:100%" border="0"|-

$$ \displaystyle L_2(y_H^1)=L_2 (y_H^2)=0 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * For the L1-ODE-CC
 * $$\displaystyle p'+p=x$$
 * Verify that
 * $$\displaystyle p(x)=k_1e^{-x}+x-1$$
 * is the solution for this equation.

Given

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


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$$\displaystyle p'+p=x$$
 * (2.1)
 * }
 * {| style="width:100%" border="0"


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$$\displaystyle p(x)=k_1e^{-x}+x-1$$
 * (2.2)
 * }

Nomenclature

 * L1:First Order Linear Equation
 * ODE: Ordinary Differential Equation
 * CC: Constant Coefficients
 * $$\displaystyle p'=\frac{dp}{dx}$$

Solution

 * The first order derivative of Equation 2.2 can be expressed as:
 * $$\displaystyle p'(x)=(k_1e^{-x}+x-1)'=-k_1e^{-x}+1$$
 * {| style="width:100%" border="0"


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$$\displaystyle p'(x)=-k_1e^{-x}+1$$
 * (2.3)
 * }
 * Thus, the left-hand side of Equation 2.1 becomes
 * $$\displaystyle p'+p=(-k_1e^{-x}+1)+(k_1e^{-x}+x-1)$$
 * $$\displaystyle =k_1e^{-x}+k_1e^{-x}+1-1+x$$
 * $$\displaystyle =x$$
 * Which is equal to the right-hand side of Equation 2.1.
 * Therefore, we can conclude that
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$$\displaystyle p(x)=k_1e^{-x}+x-1$$ is a solution for equation $$\displaystyle p'+p=x$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * For N1-ODE
 * $$\displaystyle M(x,y)+N(x,y)\,y'=0$$
 * Show that
 * a) This equation is affine in $$\displaystyle y'$$
 * b) This equation is in general an N1-ODE
 * c) This equation is not the most general N1-ODE. Give an example of a more general N1-ODE

Nomenclature

 * N1:Nonlinear First Order Equation
 * ODE: Ordinary Differential Equation
 * $$\displaystyle y'=\frac{dy}{dx}$$

Part a): Proving Affinity of Equation 3.1

 * Denote that


 * In general,to prove $$\displaystyle G(y',y,x)$$ is linear in $$\displaystyle y'$$
 * One just need to show that


 * Now we divide Equation 3.1 into two parts,
 * For the first part, since $$\displaystyle M(x,y)$$ is not a function of $$\displaystyle y'$$, one can just consider $$\displaystyle M(x,y)$$ as a constant with respect to $$\displaystyle y'$$
 * Clearly, for a constant $$\displaystyle M(x,y)$$
 * $$\displaystyle G_M(\alpha u+\beta v,y,x)=M(x,y)$$
 * $$\displaystyle \alpha\,G_M(u,y,x)+\beta\,G_M(v,y,x)=\alpha\,M(x,y)+\beta\,M(x,y)=(\alpha+\beta)\,M(x,y) $$
 * $$\displaystyle (\alpha+\beta)\,M(x,y) \neq M(x,y),\forall \alpha,\beta\in \mathbb{R} $$
 * Therefore, the first part is nonlinear in $$\displaystyle y'$$ because
 * $$\displaystyle G_M(\alpha u+\beta v,y,x)\neq \alpha\,G_M(u,y,x)+\beta\,G_M(v,y,x)$$
 * For the second part, we have
 * $$\displaystyle G_N(\alpha u+\beta v,y,x)=N(x,y)\,(\alpha u+\beta v)=\alpha\,N(x,y)\,u+\beta\,N(x,y)\,v$$
 * While


 * $$\displaystyle \alpha\,G_N(u,y,x)+\beta\,G_N(v,y,x)=\alpha\,N(x,y)\,u+\beta\,N(x,y)\,v$$
 * Thus,


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


 * style="width:95%" |
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$$\displaystyle G_N(\alpha u+\beta v,y,x)=\alpha\,G_N(u,y,x)+\beta\,G_N(v,y,x)$$
 * (3.4)
 * }
 * In conclusion, we have proved that the second part of $$\displaystyle M(x,y)+N(x,y)\,y'$$ is linear in $$\displaystyle y'$$, while the first part is an intersection constant that is nonlinear in $$\displaystyle y'$$


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

Therefore, Equation 3.1 is affine in $$\displaystyle y'$$
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 * }

Step 1: To prove Equation 3.1 is nonlinear

 * Denote that


 * In general, to prove $$\displaystyle G(y,x)$$ is nonlinear in $$\displaystyle y$$
 * One just needs to show that


 * Substitute $$\displaystyle y$$ with $$\displaystyle \alpha u+\beta v$$, then Equation 3.2 becomes
 * $$ \displaystyle G(y',\alpha u+\beta v,x)=M(x,\alpha u+\beta v)+N(x,\alpha u+\beta v)\,\frac{d(\alpha u+\beta v)}{dx} $$
 * {| style="width:100%" border="0"


 * style="width:95%" |
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$$ \displaystyle =M(x,\alpha u+\beta v)+\alpha N(x,\alpha u+\beta v)\,\frac{du}{dx}+\beta N(x,\alpha u+\beta v)\,\frac{dv}{dx} $$
 * (3.7)
 * }
 * $$ \displaystyle \alpha G(y',u,x)+\beta G(y',v,x)= \alpha M(x,u)+\alpha N(x,u)\,\frac{du}{dx}+\beta M(x,v)+\beta N(x,v)\,\frac{dv}{dx} $$
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$ \displaystyle =\alpha[M(x,u)+M(x,v)]+\alpha N(x,u)\,\frac{du}{dx}+\beta N(x,v)\,\frac{dv}{dx} $$
 * (3.8)
 * }


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


 * style="width:95%" |
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$$\displaystyle G(y',\alpha u+\beta v,x) \neq \alpha G(y',u,x)+\beta G(y',v,x)$$
 * (3.9)
 * }
 * Therefore, Equation 3.1 is nonlinear

Step 2: To prove Equation 3.1 is First Order ODE

 * From (3.1), we can see that the highest order of derivative is y', which is the first order derivative of y with respect to x, thus it is a "First Order" equation;
 * Also, since $$\displaystyle y'=\frac{dy}{dx}$$, we know that "x" is the only independent variable for the derivative y', therefore this is an "Ordinary Differential Equation"


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In general, we can conclude that Equation 3.1 is in general a N1-ODE
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 * }

Part c): Example of More General N1-ODEs

 * General N1-ODE is expressed as
 * $$\displaystyle G(y',y,x)=0$$, where $$\displaystyle G$$ is a nonlinear function
 * Equation (3.1) is just one particular case of $$\displaystyle G(y',y,x)=0$$ that is affine in $$\displaystyle y'$$
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An example of a more general N1-ODE could be $$\displaystyle e^x+tan(y')+cos(x^3)+\frac{x^{17}}{14}+log\,y=0$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * The homogeneous solutions, $$ \displaystyle y^1_H(x) $$, and, $$ \displaystyle y^2_H(x) $$, are linearly independent
 * Part 1)
 * Show that
 * $$ \displaystyle \forall \alpha \in \mathbb R, $$


 * Part 2)
 * Graphically display
 * $$ \displaystyle y^1_H(x) $$ and $$ \displaystyle y^2_H(x) $$.

Given

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$$ \displaystyle y^1_H(x)=x $$
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 * style="width:95%" |
 * (4.2)
 * }
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$$ \displaystyle y^2_H(x)= \frac{x}{2}\log\left(\frac{1+x}{1-x} \right)-1 $$
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 * style="width:95%" |
 * (4.3)
 * }

Nomenclature

 * $$ \displaystyle \forall $$ - For All
 * $$ \displaystyle \in $$ - Within
 * $$ \displaystyle \mathbb R $$ - Set of Real Numbers
 * $$ \displaystyle (\cdot) $$ - Linear Operator

Part 1: Verification of Nonlinearity

 * Step 1 - We restate linear dependence as; When two functions are linearly dependent, there exists a value, $$ \displaystyle \alpha $$, such that, $$ \displaystyle \forall \hat x \in \mathbb R $$,
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$$ \displaystyle y^1_H(\hat x)=\alpha y^2_H(\hat x) $$
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 * style="width:95%" |
 * (4.4)
 * }


 * Step 2 - We will assume linear dependence, and for any arbitrary value of, $$ \displaystyle \hat x $$, we can solve for, $$ \displaystyle \alpha $$, Therefore;
 * for $$ \displaystyle \hat x=0 $$
 * $$ \displaystyle y^1_H(0)=\alpha y^2_H(0) $$
 * $$ \displaystyle 0=\alpha (0-1) $$


 * Step 3 - Given our assumption of of linear dependence, we can choose any arbitrary value for, $$ \displaystyle \hat x $$, other than what was chosen in step 2. Then with Equation 4.4, the functions will be equal if they are linearly dependent, and not equal if they are linearly independent.
 * for $$ \displaystyle \hat x= 0.5, \alpha=0$$


 * $$ \displaystyle y^1_H(0.5)=0 \times (y^2_H(0.5)) $$


 * $$ \displaystyle 0.5=0 \times (y^2_H(0.5)) $$


 * and in fact


 * Step 4 - Therefore we conclude Equation 4.1, that is;
 * $$ \displaystyle \forall \alpha \in \mathbb R, $$
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$$ \displaystyle y^1_H(\cdot) \ne \alpha y^2_H(\cdot) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * (4.7)
 * }

Part 2: Graph of yH1(x), and yH2(x)

 * The plot below of, $$ \displaystyle y^1_H(x) $$ and $$ \displaystyle y^2_H(x) $$, demonstrate that the functions are linearly independent.
 * $$ \displaystyle f(x):=y^1_H(x)=x $$
 * $$ \displaystyle g(x):=y^2_H(x)= \frac{x}{2}\log\left(\frac{1+x}{1-x} \right)-1 $$



Problem Statement

 * Consider the following function $$ \displaystyle \phi \left ( x, y \right ) = x^2y^\frac{3}{2} + \log(x^3y^2) = k $$
 * a) Find $$ \displaystyle G(y',y,x) = \frac{d}{dx}\phi(x,y)=0 $$
 * b) Show that G(y',y,x) is an N1-ODE

Given

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

$$ \displaystyle \phi \left ( x, y \right ) = x^2y^\frac{3}{2} + \log(x^3y^2) = k $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.1)
 * }

Nomenclature

 * N1-ODE stands for Nonlinear First Order Differential Equation
 * $$ \displaystyle \frac{dy}{dx} = y' $$


 * $$ \displaystyle \frac{1}{x} = x^{-1} ; \displaystyle \frac{1}{y} = y^{-1} $$


 * $$ \displaystyle M_y(x,y) = \frac {\partial M(x,y)}{\partial y}; N_x(x,y) = \frac{\partial N(x,y)}{\partial x} $$

Part a): Derivation of G(y',y,x) Form

 * Using Chain Rule, we can take $$ \displaystyle \frac{d}{dx} \phi(x,y(x)) $$, which will become following.
 * $$ \displaystyle \frac{d\phi}{dx} = \frac{\partial \phi} {\partial x} \cancelto{1}\frac{dx}{dx} + \frac{\partial \phi} {\partial y} \frac{dy}{dx} = \frac{\partial \phi} {\partial x} + \frac{\partial \phi} {\partial y} \frac{dy}{dx} $$
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{d\phi}{dx} = \frac{\partial \phi} {\partial x} + \frac{\partial \phi} {\partial y} y' $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.2)
 * }
 * In order to substitute Equation 5.1 to Equation 5.2, take the derivative of $$ \displaystyle \phi \left ( x, y \right ) = x^2y^\frac{3}{2} + \log(x^3y^2) $$ respect to x and y.
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial \phi} {\partial x} = 2xy^\frac{3}{2} + \frac {3x^2y^2} {x^3y^2} = 2xy^\frac{3}{2} + 3x^{-1} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.3)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \frac{\partial \phi} {\partial y} = \frac{3}{2}x^2y^\frac{1}{2} + \frac {2x^3y} {x^3y^2} = \frac{3}{2}x^2y^\frac{1}{2} + 2y^{-1} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.4)
 * }
 * Also k will become 0 since k is a constant value. Therefore the expression of $$ \displaystyle \frac{d\phi}{dx} $$ becomes following.
 * {| style="width:100%" border="0"|-

$$ \displaystyle G(y',y,x) = \frac{d}{dx} \phi(x,y)= 2xy^\frac{3}{2} + 3x^{-1} + \left (\frac{3}{2}x^2y^\frac{1}{2} + 2y^{-1}\right )y' = 0$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">(5.5)
 * }

Part b): Proof of N1-ODE

 * Exact N1-ODEs are subset of all sets of N1-ODEs. Also, in order to satisfy exactness condition for N1-ODEs, two conditions have to meet.
 * {| style="width:100%" border="0"

1. $$ \displaystyle G(x,y,z) = M(x,y) + N(x,y)y' = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.6)
 * }
 * {| style="width:100%" border="0"

2. $$ \displaystyle M_y(x,y) = N_x(x,y) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.7)
 * }
 * This means that if we prove that given equation is exact, then we can conclude that the given equation is N1-ODE. Using Equation 5.5 can prove the first condition that is written as Equation 5.6. This is demonstrated below.
 * $$ \displaystyle G(y',y,x) = M(x,y) + N(x,y)y' = \left (2xy^\frac{3}{2} + 3x^{-1}\right ) + \left (\frac{3}{2}x^2y^\frac{1}{2} + 2y^{-1}\right )y' = 0$$
 * In addition, we can identify both M and N.
 * {| style="width:100%" border="0"

$$ \displaystyle M(x,y) = 2xy^\frac{3}{2} + 3x^{-1}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.8)
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle N(x,y) = \frac{3}{2}x^2y^\frac{1}{2} + 2y^{-1}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.9)
 * }
 * Then, we can test the second condition.
 * $$ \displaystyle \frac {\partial M(x,y)}{\partial y} = \frac{3}{2} \times 2xy^\frac{1}{2} = 3xy^\frac{1}{2} $$
 * $$ \displaystyle \frac {\partial N(x,y)}{\partial x} = 2 \times \frac{3}{2}x^2y^\frac{1}{2} = 3xy^\frac{1}{2} $$
 * {| style="width:100%" border="0"

$$ \displaystyle M_y(x,y) = N_x(x,y) = 3xy^\frac{1}{2} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(5.10)
 * }
 * As a result, since Equation 5.10 satisfies the condition given in Equation 5.7, we can say that this equation is an Exact N1-ODE.
 * Furthermore, we can observe the linearity briefly using an answer from Part a). There are two criteria that a differential equation must meet in order to be linear.


 * 1) The dependent variable y and all its derivatives are of the first degree; that is, the power of each term involving y is 1.
 * 2) Each coefficient depends on only the independent variable x.
 * The Equation 5.5 has coefficient depends on y, and also a y term does not have a power of 1. Therefore the equation is a nonlinear differential equation.
 * {| style="width:100%" border="0"|-

Since we have tested that the given equation is nonlinear and exact, we can conclude that the given equation is Exact N1-ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * The minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$, and state the full theorem and provide a proof.

Given

 * Review calculus such that


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

$$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(6.1)
 * }

Nomenclature

 * $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial\phi(x,y)}{\partial x}\frac{\partial\phi(x,y)}{\partial y} $$
 * $$ \frac{\partial^2 \phi(x,y)}{\partial y \partial x}=\frac{\partial\phi(x,y)}{\partial y}\frac{\partial\phi(x,y)}{\partial x} $$

Solution

 * We have to know Clairaut's Theorem first.
 * If $$\displaystyle f:R^{n}\rightarrow R $$
 * {| style="width:100%" border="0"

$$\displaystyle f_{x_{i},y_{j}}(a_1, \dots, a_n) $$ = $$\displaystyle f_{y_{j},x_{i}}(a_1, \dots, a_n) $$ ,suppose that $$\displaystyle 1\leq i,j\leq n $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(6.2)
 * }
 * {| style="width:100%" border="0"|-

According to Clairaut's Theorem 6.2, the minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$ must be three.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }


 * Proof of Clairaut's Theorem:
 * According to Mean Value Theorem, if $$ \displaystyle f:[a,b]\rightarrow\mathbb{R} $$ is continuous on $$ \displaystyle [a,b]$$ and differentiable on $$ \displaystyle (a,b)$$, and there exist c which is $$\displaystyle a<c<b $$, such that


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

$$f ' (c) = \frac{f(b) - f(a)}{b - a}.$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(6.3)
 * }
 * Let K is the near the point which exist $$\displaystyle x,y\in K $$
 * {| style="width:100%" border="0"

$$f_{x,y}' (a+n,b)-f_{x,y} ' (a,b)=\lim_{h\rightarrow 0}\frac{f(a+n,b+h)-{f(a+n,b)}}{h}-\lim_{h\rightarrow 0}\frac{f(a,b+h)-{f(a,b)}}{h}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(6.4)
 * }
 * We can define the function after fixed h
 * $$\displaystyle z_{h}(t)=f(a+t,b+h)-f(a+t,b)$$
 * and after derivate $$z_{h}(t)$$
 * $$\displaystyle z_{h}'(t)=f'(a+t,b+h)-f'(a+t,b)$$
 * According to Mean Value theorem,
 * $$f_{x,y}' (a+n,b)-f_{x,y} ' (a,b)=\lim_{h\rightarrow 0}\frac{z_{h}(t)-{z_{h}(0)}}{h}=\lim_{h\rightarrow 0}\frac{nz_{h}'(\bar{n})}{h}$$
 * $$\frac{f_{x,y}'(a+n,b)-f_{x,y}'(a,b)}{n}=\lim_{h\rightarrow 0}\displaystyle f_{x,y}'(a+\bar{n},\bar{y})$$
 * $$\displaystyle f_{x,y}''(a+\bar{n},\bar{y})=\displaystyle f_{x,y}'(a,b)$$
 * so we let
 * $$D=\frac{\partial^{2}}{\partial y\partial x}f(a,b)$$
 * then given $$\displaystyle\delta > 0$$
 * {| style="width:100%" border="0"
 * then given $$\displaystyle\delta > 0$$
 * {| style="width:100%" border="0"

$$|\frac{ f_{x,y}'(x,y)-f_{x,y}'(a,b)}{r}-D|<\displaystyle\delta $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(6.5)
 * }
 * because $$\displaystyle\delta $$ is chosen by ourselves and $$\displaystyle r\neq0 $$
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\frac{\partial^{2}}{\partial y\partial x}f(x,y)=\frac{\partial^{2}}{\partial x\partial y}f(x,y)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(6.6)
 * }
 * so we can get the Equation 6.1
 * {| style="width:100%" border="0"|-

$$\displaystyle \frac{\partial^{2}\phi(x,y)}{\partial x\partial y}=\frac{\partial^{2}\phi(x,y)}{\partial y\partial x}$$.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">(6.7)
 * }

Problem *2.7
===Problem Statement ===
 * Verify that
 * $$y(x)=\sin^{-1}(k-15x^5)$$
 * is indeed the solution for the N1-ODE
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle \frac {d \phi(x,y(x))}{dx}=75x^4+(\cos y)y' = 0$$
 * <p style="text-align:right">(7.1)
 * }

Given

 * For
 * $$y=\arcsin(x)$$
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$${y}'=\frac{1}{\sqrt{1-x^2}}$$
 * <p style="text-align:right">(7.2)
 * }

Nomenclature

 * $$y=y(x)$$
 * $$\arcsin(x)=\sin^{-1}(x)$$
 * $${y}'=\frac{dy}{dx}$$

Solution

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


 * style="width:95%" |
 * style="width:95%" |

$$y(x)=\arcsin(k-15x^5)$$
 * <p style="text-align:right">(7.3)
 * }
 * So that
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\sin(y)=k-15x^5$$
 * <p style="text-align:right">(7.4.1)
 * }
 * According to Equation 7.2 and 7.4, the following equation can be acquired
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$y'=\frac {-75x^4}{\sqrt{1-(k-15x^5)^2}}=\frac {-75x^4}{\sqrt{1-\sin^2 y}}$$
 * <p style="text-align:right">(7.4.2)
 * }
 * Since the range of Equation 7.3 is $$[-\frac \pi 2,\frac \pi 2]$$. And in this specific situation, y can't equal to $$\pm \frac \pi 2$$
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\sqrt{1-\sin^2y}=\cos y$$
 * <p style="text-align:right">(7.5)
 * }
 * According to Equation 7.5 and 7.6
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$y'=\frac {-75x^4} {\cos y}$$
 * <p style="text-align:right">(7.6)
 * }
 * From Equation 7.1 and Equation 7.6
 * $$75x^4+(\cos y)y'=75x^4+(\cos y)\frac {-75x^4}{\cos y}=0$$
 * {| style="width:100%" border="0"|-

We have found that $$y(x)=\sin^{-1}(k-15x^5)$$ is indeed the solution for the N1-ODE $$\displaystyle \frac {d \phi(x,y(x))}{dx}=75x^4+(\cos y)y' = 0$$.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * Explain why solving the integrating factor $$ \displaystyle h(x,y) $$ is not easy.

Given

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

$$ \displaystyle h_{x}N-h_{y}M+h\left ( N_{x}-M_{y} \right )=0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(8.1)
 * }

Solution

 * The given Equation 8.1 is a non-linear partial differential equation with a function $$ \displaystyle h $$ which is not given.
 * And also $$ \displaystyle h_x $$ in general non-linear, and in this PDE we have partial derivations with respect to $$ \displaystyle x $$ and $$ \displaystyle y $$ which are two independent variables. Solving is should be in two dimensional domine. It is also the same situation with $$ \displaystyle N_x $$ and $$ \displaystyle M_y $$.
 * And it has terms $$ \displaystyle h_{x} $$,$$ \displaystyle h_{y} $$,$$ \displaystyle N_{x} $$ and $$ \displaystyle M_{y} $$.
 * {| style="width:100%" border="0"|-

With reasons listed above, Equation 8.1 is usually not easy to solve.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Problem Statement

 * Find h using $$\frac {h_{y}}{h} =\frac{1}{M}(N_x-M_y)=:m(y)$$

Given

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


 * style="width:95%" |
 * style="width:95%" |

$$\int \frac {y'} {y}=log(y)$$
 * <p style="text-align:right">(9.1)
 * }
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

if $$\log(x)=y$$ then $$e^y=x$$
 * <p style="text-align:right">(9.2)
 * }

Nomenclature

 * $$\exp(x)=e^x$$

Solution

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


 * style="width:95%" |
 * style="width:95%" |

$$\frac {h_{y}}{h} =\frac{1}{M}(N_x-M_y)=:m(y)$$


 * <p style="text-align:right">(9.3)
 * }
 * Integrate both sides of the Equation 9.3 and according to Equation 9.1
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\log[h(y)]=\int^y m(y)ds +k$$
 * <p style="text-align:right">(9.4)
 * }
 * {| style="width:100%" border="0"|-

Then according to Equation 9.2, the final answer is $$h(y)=\exp[\int^y m(y)ds +k]$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">(9.5)
 * }

Problem Statement

 * Show that
 * $$\displaystyle h(x)=x$$
 * and
 * $$ \displaystyle y(x)=\frac{x^{3}}{4}+\frac{k}{x}$$

Given

 * Specific non-homogenous L1-ODE-VC
 * {| style="width:100%" border="0"

$$\displaystyle y'+\frac{1}{x}y=x^2$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.1)
 * }

Nomenclature

 * L1:First Order Linear Equation
 * ODE: Ordinary Differential Equation
 * VC: Variable Coefficients
 * $$\displaystyle y'=\frac{dy}{dx}$$

Solution

 * We can get an equation from (1)p11-2, which is written in below
 * {| style="width:100%" border="0"

$$\displaystyle h_xN+h(N_x -M_y)=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.2)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle \frac{h_x}{h}=-\frac{1}{N}(N_x-M_y)$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.3)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle\int\frac{h_{x}}{h}=-\displaystyle\int\frac{1}{N}(N_{x}-M_{y})$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.4)
 * }
 * from (3)p7-4
 * $$\displaystyle\int\frac{h_{x}}{h}=\log h(x)$$
 * and
 * $$-\displaystyle\int\frac{1}{N}(N_{x}-M_{y})=\displaystyle\int n(s)ds+k $$
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$\displaystyle h(x)=\exp[\int^{x} n(s)ds+k]$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(10.5)
 * }
 * $$\displaystyle h(x)=\exp[ln(x)+k]$$
 * When k=0, then h(x)=x
 * the given Equation 10.1
 * $$\displaystyle y'+\frac{1}{x}y=x^2$$
 * $$\displaystyle h(y'+\frac{y}{x})=hx^{2}$$
 * According to (5)p11-4, which is $$\displaystyle ha_{0}=h_{x}=h'$$, we can get
 * $$\displaystyle hy'+h'y=hx^{2}$$
 * Then according to (6)p11-4, which is $$\displaystyle hy'+h'y=(hy)'$$
 * $$\displaystyle (hy)'=hx^{2}$$
 * Because h(x)=x, so
 * $$\displaystyle (xy)'=x^{3}$$
 * Then we integrate the righ and left sides, the equation becomes
 * $$\displaystyle xy=\frac{x^{4}}{4}+k$$
 * {| style="width:100%" border="0"|-
 * Then we integrate the righ and left sides, the equation becomes
 * $$\displaystyle xy=\frac{x^{4}}{4}+k$$
 * {| style="width:100%" border="0"|-
 * {| style="width:100%" border="0"|-

$$\displaystyle y=\frac{x^3}{4}+\frac{k}{x}$$, it is the result which we want to get.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">(10.6)
 * }

Problem Statement

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

Solve $$ \displaystyle a_1(x)\,y'+a_0(x)\,y=b(x) $$.
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(11.1)
 * }

Nomenclature

 * $$ \displaystyle \exp \left (x \right) = e^x $$
 * $$ \displaystyle h_{x} = \frac {\partial h}{\partial x} $$
 * $$ \displaystyle N_{x} = \frac {\partial N}{\partial x} $$
 * $$ \displaystyle M_{y} = \frac {\partial N}{\partial y} $$

Solution

 * From the lecture notes, we know that:
 * $$ \displaystyle \underbrace{(hM)}_{\displaystyle{\bar M}}+\underbrace{(hN)}_{\displaystyle{\bar N}}\, y'=0 $$
 * Apply 2nd exactness condition to find $$ \displaystyle h $$:
 * $$ \displaystyle {\displaystyle{\bar M_{y}}}={\displaystyle{\bar N_{x}}} $$
 * $$ \displaystyle \begin{cases} & \bar M_y=h_y M +h M_y\\& \bar N_x=h_x N+h N_x \end{cases} $$
 * Then we have $$ \displaystyle h_{x} N-h_{y} M+h(N_{x}-M_{y})=0 $$
 * Assume that $$ \displaystyle h_{y}(x,y)=0 $$, thus $$ \displaystyle h $$ is a function of $$ \displaystyle x $$ only, then we have:
 * $$ \displaystyle h_{x}N+h(N_{x}-M_{y})=0 $$
 * Dividing the equation by $$ \displaystyle h N $$ to obtain $$ \displaystyle h(x) $$
 * $$ \displaystyle \frac{h_x}{h}=-\frac{1}{N}(N_x-M_y) $$
 * Then we have:
 * {| style="width:100%" border="0"

$$ \displaystyle h(x)=\exp \left[ \int^x n(s)ds+k_1\right] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(11.2)
 * }
 * To obtain $$ \displaystyle y(x) $$,with Equation 11.2 we have:
 * $$ \displaystyle y'+a_0y=b \Rightarrow h(y'+a_0y)=hb $$
 * $$ \displaystyle \frac{h_x}{h}=a_0 \Rightarrow ha_0=h_x=h' $$
 * $$ \displaystyle hy'+h'y=hb $$
 * {| style="width:100%" border="0"

$$ \displaystyle \Rightarrow y(x)=\frac{1}{h(x)} \left[ \int^x h(s)b(s)ds+k_2\right] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(11.3)
 * }

Part 1

 * Use the given information in Q1 and substitute $$ \displaystyle a_1(x),\ a_0(x)$$ and $$ \displaystyle b(x) $$ in Equation 11.1, we have:
 * $$ \displaystyle y'+(x+1)y=x^2+4 $$
 * Then Equation 11.2 should be:
 * $$ \displaystyle h(x)=\exp \left[ \int^x (s+1)ds+k_1\right]=\exp \left[\int^x (s+1)d(s+1)+k_1\right]=\exp \left[ \frac{1}{2}(x+1)^2\right] $$
 * Equation 11.3 should be:
 * {| style="width:100%" border="0"

$$ \displaystyle y(x)=\frac{1}{h(x)} \left[ \int^x h(s)b(s)ds+k_2\right]=\frac{1}{\exp \left[ \frac{(x+1)^2}{2}\right]} \left[ \int^x \exp \left[\frac{(s+1)^2}{2} \right](s^2+4)ds+k_2\right] $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(11.4)
 * }
 * So, we have final solution:
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=\frac{1}{\exp \left[ \frac{(x+1)^2}{2}\right]}\left[\int\exp\left[\frac{(x+1)^2}{2}\right](x^2+4)dx \right] $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Part 2

 * With the given condition, we have Equation 11.1 as:
 * $$ \displaystyle y'+\frac{a_0(x)}{a_1(x)}y=\frac{b(x)}{a_1(x)} $$
 * And with Equation 11.2 and Equation 11.3, we have:
 * $$ \displaystyle h(x)=\exp \left[ \int^x \frac{a_0(s)}{a_1(s)}ds\right] $$
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=\frac{1}{\exp \left[ \int^x \frac{a_0(s)}{a_1(s)}ds\right]} \left[ \int^x \exp \left[ \int^x \frac{a_0(s)}{a_1(s)}ds\right]\frac{b(s)}{a_1(s)}ds+k_2\right] $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Part 3

 * Use the given information in Q1 and substitute $$ \displaystyle a_1(x),\ a_0(x) $$ and $$ \displaystyle b(x) $$ in Equation 11.1, we have:
 * $$ \displaystyle (x^2+1)y'+x^3y=x^4 \Rightarrow y'+\frac{x^3}{x^2+1}y=\frac{x^4}{x^2+1} $$
 * Then Equation 11.2 should be:
 * {| style="width:100%" border="0"

$$ \displaystyle h(x)=\exp\left[\int^x\frac{s^3}{s^2+1}ds \right]=\exp\left[\frac{1}{2}\int\frac{x^2}{x^2+1}dx^2\right]$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">(11.5)
 * }
 * Let $$ \displaystyle t=x^2 $$
 * Then Equation 11.5 would be:
 * $$ \displaystyle h(x)=\exp\left[\int\frac{1}{2}\frac{t}{t+1}dt\right]=\exp\left[\frac{1}{2}\int(1-\frac{1}{t+1})dt\right]=\exp\left\{\frac{1}{2}\left[x^2-ln(x^2+1)\right]\right\} $$
 * Equation 11.3 should be:


 * So, we have final solution:
 * {| style="width:100%" border="0"|-

$$ \displaystyle y(x)=\frac{1}{\exp\left\{\frac{1}{2}\left[x^2-ln(x^2+1)\right]\right\}}\left[\int\exp\left\{\frac{1}{2}\left[x^2-ln(x^2+1)\right]\right\}\frac{x^4}{x^2+1}dx+k_2\right] $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Note

 * About the final answer of Part 1 and Part 3, I tried to integrate, but that's as far as I can go. So I have to leave the $$ \displaystyle \int $$ with it.

Team Contribution

 * {| class="prettytable"


 * colspan="4" | Team Contribution Table
 * colspan="4" | Compiled by: Kyung-Hoon Bang
 * colspan="4" | Compiled by: Kyung-Hoon Bang


 * Problems
 * Solved and Typed By
 * Solved and Inspected By
 * Skimmed By
 * Problem 2.1
 * Brandon Taylor
 * Kyung-Hoon Bang
 * All Team Members
 * Problem 2.2
 * Tianyu Gu
 * Brandon Taylor
 * All Team Members
 * Problem 2.3
 * Tianyu Gu
 * Brandon Taylor
 * All Team Members
 * Problem 2.4
 * Brandon Taylor
 * Tianyu Gu
 * All Team Members
 * Problem 2.5
 * Kyung-Hoon Bang
 * Tianyu Gu
 * All Team Members
 * Problem 2.6
 * Che-Ming Lee
 * Jiajun Han
 * All Team Members
 * Problem 2.7
 * Jiajun Han
 * Brandon Taylor
 * All Team Members
 * Problem 2.8
 * Mutian Hu
 * Jiajun Han
 * All Team Members
 * Problem 2.9
 * Jiajun Han
 * Mutian Hu
 * All Team Members
 * Problem 2.10
 * Che-Ming Lee
 * Mutian Hu
 * All Team Members
 * Problem 2.11
 * Mutian Hu
 * Kyung-Hoon Bang
 * All Team Members
 * }
 * Problem 2.10
 * Che-Ming Lee
 * Mutian Hu
 * All Team Members
 * Problem 2.11
 * Mutian Hu
 * Kyung-Hoon Bang
 * All Team Members
 * }
 * All Team Members
 * }
 * }