User:Egm6321.f10.team6.Kim.MK/hw2

= Problem 3 Show linear independency =

Given
Given the following homogeneous function L2_ODE_VC (Legendre equation with n=1)


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


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

$$\left(1-{{x}^{2}}\right){{y}^{''}}-2x{{y}^{'}}+2y=0$$

(3.1) and its two homogeneous solutions
 * 
 * }


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


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

$$y_{H}^{1} ( x ) = x $$

(3.2)
 * 
 * }


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


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

$$y_{H}^{2} ( x ) = {\frac{x}{2}} log ({\frac{1+x}{1-x}} - 1)$$ (3.3)
 * 
 * }

Required
Plot $$ y_{H}^{1} $$ and $$ y_{H}^{2} $$

and show $$ y_{H}^{1}{~} {\neq} {~}{\alpha} y_{H}^{2} $$ such that $$ x {\in} {\mathbb{R}} $$ $$ {\alpha} {\in} {\mathbb{R}} $$

(Hint : Find $$ {\hat{x}} $$ such that $$ y_{H}^{1} - {\alpha} y_{H}^{2}{~} {\neq} {~} 0 $$)

Solution
Using $$ {\mathbf{MATLAB^{TM}}}$$ plot two homogeneous solutions,



Matlab code: problem2_3.m

Given

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


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

$$ {\phi} \left(x,y\right) = x^{2} y^{3/2} + log (x^{3} y^{2}) = k $$

(4.1)
 * 
 * }

Required
Find $$ F(x,y,y') = {\frac { d{\phi}(x,y)}{d x}}$$ and verify that F is exact N1_ODE and invent 3 more

Solution

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


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

$$ F(x,y,y') = {\frac {d{\phi}(x,y)}{d x}} = {\frac} + {\frac} {\frac{dy}{dx}}$$

(4.1)
 * 
 * }


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


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

$$\begin{align} & M(x,y):={\phi}_{x}(x,y): =\frac{\partial \phi(x,y)}{\partial x} \\ & N(x,y):={\phi}_{y}(x,y): =\frac{\partial \phi(x,y)}{\partial y} \\ \end{align}$$

(4.2)
 * 
 * }

So, that is
 * {| style="width:100%" border="0"


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

$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' $$

(4.3)
 * 
 * }

From the given $${{\phi}\left(x,y\right)}$$, we can get M(x,y) and N(x,y)


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


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

$$\begin{align} &M\left(x,y\right) :={\phi}_{x}(x,y)= 2 x y^{\frac{3}{2}} + {\frac{3}{x}}\\ &N(x,y) := {\phi}_{y}(x,y)= {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}}\\ \end{align} $$      (4.4)
 * 
 * }

Follow the $$ 1^{st} $$ condition of exactness,


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


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

$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' = \left( 2 x y^{\frac{3}{2}} + {\frac{3}{x}} \right) + \left( {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}} \right) y' = 0 $$      (4.5)
 * 
 * }


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


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

$$ y' = - \frac{M\left(x,y\right)}{N(x,y)} $$      (4.6)
 * 
 * }

And then, apply $$ 2^{nd} $$ condition of exactness,


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

$$ \underbrace{M_{y}\left(x,y\right)}_{\frac{\partial M(x,y)}{\partial y}}= \underbrace{N_{x}(x,y)}_{\frac{\partial N(x,y)}{\partial x}} $$      (4.7)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$ {\frac {{\partial}M(x,y)} {\partial y} } = {\frac {{\partial} \left( 2 x y^{\frac{3}{2}}+{\frac{3}{x}}\right)} {\partial y}} = 3 x y^{\frac{1}{2}} = {\frac {{\partial}\left( {\frac{3}{2}} y^{\frac{1}{2}} x^{2} + {\frac{2}{y}}\right)} {\partial x}} = {\frac {{\partial} N(x,y)}{\partial x}} $$      (4.8)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

So, F is exact N1_ODE. And 3 more equations,


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

$$ {\phi}(x, y) = {\frac{1}{2}} x^{4} sin(y^{2}) = k $$ (4.9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$ {\phi}(x, y) = {\frac{1}{4}}( x^{4} + {6} x^{2} y^{2} + y^{4})= k $$ (4.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$ {\phi}(x, y) = e^ = k $$ (4.11)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }