User:Egm6321.f10.team2.patterson/Homework 5

= Problem 6: Show Agreement of Particular Integrals =

Given
Show that the following equation:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}

$$ (eq 6.1)
 * 
 * }

is equivalent to the following particular integral found in the Differential Equations text book by A.C.King:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)=\oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$ (eq 6.2)
 * 
 * }

given that the following equations holds true:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

W={{u}_{1}} {{u}_{2}}'-{{u}_{1}}' {{u}_{2}}=\left| \begin{matrix} {{u}_{1}} & {{u}_{2}} \\ {{u}_{1}}' & {{u}_{2}}' \\ \end{matrix} \right|

$$ (eq 6.3)
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

(\frac{u_{2}}{u_{1}})'=\frac{1}{h}

$$ (eq 6.4)
 * 
 * }

And considering the following conversion between the notation used in the lectures and A.C. King's notation:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\oint^x{f\left( s \right)}\,ds=\int^x{f\left(s\right)}\,ds =\int{f\left(x\right)}\,dx

$$  w/o constant (eq 6.5)
 * 
 * }

Solution
The first step to tackling this problem requires the application of the Integration by Parts formula to equation 6.1


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

$$\displaystyle \int w\,dv=wv-\int v\,dw $$
 * style="width:95%" |
 * style="width:95%" |

(eq 6.6)
 * 
 * }


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

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{w}\underbrace{\frac{1}{h\left( x \right)}\,dx}_{dv}}$$

(eq 6.7) The remaining unknowns in the right hand side of equation 6.6 are shown below:
 * 
 * }


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

$$\displaystyle dw=h\left( x \right) f\left( x \right)dx
 * style="width:95%" |
 * style="width:95%" |

$$

(eq 6.8)
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle v= \int {\frac{1}{h\left( x \right)}}\,dx
 * style="width:95%" |
 * style="width:95%" |

$$

(eq 6.9) Equations 6.7, 6.8 and 6.9 can be combined and rewritten into the form shown in equation 6.5 to yield the following:
 * 
 * }
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)=u_{1}\left( x \right)\left[\left( \oint^x{h\left(s \right)f\left(s \right)}\,ds \right)\int{\frac{1}{h\left(x \right)}}\,dx- \left[\oint^x{\left(\oint^x{\frac{1}{h\left(s\right)}}\,ds\right) h\left(s\right)f\left(s\right)}\,ds\right] \right]

$$

(eq 6.10)
 * 
 * }

The relation shown in equation 6.4 can be used to simplify equation 6.10 as follows:
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)=u_{1}\left( x \right)\left[\left( \oint^x{\frac{u_{1}^{2}\left(s\right)}{u_{2}'\left(s\right)u_{1}\left(s\right)-u_{2}\left(s\right)u_{1}'\left(s\right)}f\left(s \right)}\,ds \right)\frac{u_{2}\left(x\right)}{u_{1}\left(x\right)}- \left[\oint^x{\frac{u_{2}\left(s\right)}{\cancel{u_{1}\left(s\right)}} \frac{u_{1}^\cancel\left(s\right)}{u_{2}'\left(s\right)u_{1}\left(s\right)-u_{2}\left(s\right)u_{1}'\left(s\right)}f\left(s\right)}\,ds\right] \right]

$$

(eq 6.11) Now, multiplying through by u1(x) and using the relation shown in equation 6.3 yields the following:
 * 
 * }
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)=\left( \oint^x{\frac{u_{1}\left(s\right)u_{2}\left(x\right)}{W\left(s\right)}f\left(s \right)}\,ds \right)- \left(\oint^x{\frac{u_{2}\left(s\right)u_{1}\left(x\right)}{W\left(s\right)}f\left(s\right)}\,ds\right) = \oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$

(eq 6.12)
 * 
 * }

= Problem 7: Find Two Homogeneous Solutions using a Trial Solution =

Given
Find u1 and u2 of Equation 7.1 using the trial solution shown in equation 7.2:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

(x +1) y'' - (2x+3)y' + 2y = 0

$$ (eq 7.1)
 * 
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y(x)=e^{rx}

$$ (eq 7.2)
 * 
 * }

Solution
The first step is to differentiate the trial solution as follows:
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = e^{rx} $$ (eq 7.3)
 * 
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'(x) = r e^{rx} $$ (eq 7.4)
 * 
 * }


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

$$ \displaystyle y''(x) = r^2 e^{rx} $$ (eq 7.5) Next, plug the values into equation 7.1 to yield:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$ (eq 7.6) Since erx will not equal zero for any value of x, divide equation 7.6 by this value and factor out the x to yield:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0 $$ (eq 7.7) In order to satisfy equation 7.7 for any x, the following equation is solved:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle r(r-2)=(r-2)(r-1)=0 $$ (eq 7.8)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle r=2 $$ (eq 7.9) By plugging equation 7.9 into the trial solution in equation 7.3 we have obtained the first homogeneous solution:
 * <p style="text-align:right">
 * }
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle u_1(x)=e^{2x} $$ (eq 7.10) The second homogeneous solution is obtained by the following:
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$$ (eq 7.11) Equation 7.l is rearranged into the following form:
 * <p style="text-align:right">
 * }


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

$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0 $$ (eq 7.12) The value for a1 shown in equation 7.12 is then plugged into equation 7.11 along with the first homogeneous solution to solve for the second solution as shown:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$$ (eq 7.13)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$$ (eq 7.14)
 * <p style="text-align:right">
 * }
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle u_2(x)=\frac{-1}{4}(2x+3)$$ (eq 7.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }