User:Egm6321.f10.team03.sigillo/Hwk4

=Problem 1: Solution of a L2-ODE-VC=

Given
The following Linear second order ODE with Varying Coefficients is given:
 * {| style="width:100%" border="0"

$$\,F=2xy+\left( {{x}^{2}}-\sin (x) \right){y}'+\cos (x){{y}'}'=0$$
 * $$\,(Eq. 1.1) $$
 * style= |
 * }

Find
It has to be shown that Eq. 1.1 is exact (or either can be made exact), the function $$\Phi $$ has to be found and the solution for y needs to be calculated.

Exactness condition 1
The first exactness conditon is the one where the form of the given function F in Eq. 1.1 has to match the pattern $$g(x,y,p)+f(x,y,p)y''=0$$. Looking at Eq. 1.1 it can be seen that:
 * {| style="width:100%" border="0"

$$\underbrace{2xy+\left( {{x}^{2}}-\sin (x) \right){y}'}_{g(x,y,p)}+\underbrace{\cos \left( x \right)}_{f(x,y,p)}{{y}'}'=0$$ and thus the first condition of exactness is met.
 * $$\,(Eq. 1.2) $$
 * style= |
 * }

Exactness condition 2.1
As known, the second exactness condition for an L2-ODE has to sub-conditions, the first one says that:
 * {| style="width:100%" border="0"

$${{f}_{xx}}+2p{{f}_{xy}}+{{p}^{2}}{{f}_{yy}}\overset{!}{\mathop{=}}\,{{g}_{xp}}+p{{g}_{yp}}+{{g}_{y}}$$ where in this particular problem the single values are:
 * $$\,(Eq. 1.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"



$$\begin{align} & {{f}_{x}}=-\sin \left( x \right) \\ & {{f}_{xx}}=-\cos \left( x \right) \\ & {{f}_{xy}}={{f}_{xp}}={{f}_{yy}}={{f}_{y}}={{f}_{yp}}=0 \\ \end{align}$$

and
 * $$\,(Eq.set 1.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"



$$\begin{align} & {{g}_{x}}=2y+p\left( 2x-\cos \left( x \right) \right) \\ & {{g}_{y}}=2x \\ & {{g}_{xp}}=2x-\cos \left( x \right) \\ & {{g}_{yp}}={{g}_{pp}}=0 \\ \end{align}$$

Inserting the values of Eq.set 1.4 and 1.5 into Eq. 1.3 the first sub-condition of the second exactness condition becomes:
 * $$\,(Eq.set 1.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=2x-\cos \left( x \right)-2x$$ which is equal to
 * $$\,(Eq. 1.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=-\cos \left( x \right)$$ and thus the first sub-condition is also met.
 * $$\,(Eq. 1.7) $$
 * style= |
 * }

Exactness condition 2.2
The second sub-conditino of the second exactness condition for L2-ODE's, says that:
 * {| style="width:100%" border="0"

$${{f}_{xp}}+p{{f}_{yp}}+2{{f}_{y}}\overset{!}{\mathop{=}}\,{{g}_{pp}}$$ inserting the appropriate values from Eq.set 1.4 and 1.5 in to Eq. 1.8 yields to:
 * $$\,(Eq. 1.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,0=0$$ and which clearly states that the left and right side of Eq. 1.8 are equal. '''Thus also the second sub-condition is met and therefore the given function Equation in Eq. 1.1 is exact!'''
 * $$\,(Eq. 1.9) $$
 * style= |
 * }

Find $$\Phi $$
From Eq. (3) of the lecture notes in pp. 15-3, we know that function $$\Phi $$ is composed of the following sum:
 * {| style="width:100%" border="0"

$$\Phi =h(x,y)+\int{f\left( x,y,p \right)}\,dp$$ The second term of Eq. 1.0 can be easily calculated to be:
 * $$\,(Eq. 1.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{f\left( x,y,p \right)}\,dp=\cos \left( x \right)p$$ And thus the function $$\Phi$$ can be expressed as:
 * $$\,(Eq. 1.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Phi =h\left( x,y \right)+\cos \left( x \right)p$$ Now, from Eq. (3) of the lecture notes in pp. 15-2, it's know that
 * $$\,(Eq. 1.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}$$ which in our specific case is equal to
 * $$\,(Eq. 1.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}=2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p$$ If we take the derivatives of $$\Phi$$ in Eq. 1.12 with respect to x and with respect to y, we get:
 * $$\,(Eq. 1.14) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{\Phi }_{x}}={{h}_{x}}-p\cdot \sin \left( x \right)$$
 * $$\,(Eq. 1.15) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{\Phi }_{y}}={{h}_{y}}$$ Inserting Eq. 1.15 and Eq. 1.16 into Eq. 1.13 enables to write Eq. 1.14 as
 * $$\,(Eq. 1.16) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p={{h}_{x}}-p\cdot \sin \left( x \right)+{{h}_{y}}p={{h}_{x}}+\left( {{h}_{y}}-\sin \left( x \right) \right)p$$ and by comparison of the left and right side of the Eq. 1.17, it can be seen that:
 * <p style="text-align:right;">$$\,(Eq. 1.17) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{x}}=2xy$$
 * <p style="text-align:right;">$$\,(Eq. 1.18) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{y}}={{x}^{2}}$$ The function h(x,y) can also be obtained by integrating $${{h}_{x}}$$ with respect to x, that for our particular case yields to:
 * <p style="text-align:right;">$$\,(Eq. 1.19) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{{{h}_{x}}\,dx}=h\left( x,y \right)={{x}^{2}}y+{{k}_{1}}(y)$$ If we differentiate Eq. 1.20 with respect to y and compare the result with Eq. 1.19, then the value for $${{k}_{1}}(y)$$ can be found:
 * <p style="text-align:right;">$$\,(Eq. 1.20) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{dh\left( x,y \right)}{dy}={{x}^{2}}+{{k}_{1}}^{\prime }(y)$$
 * <p style="text-align:right;">$$\,(Eq. 1.21) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{h}_{y}}={{x}^{2}}+{{k}_{1}}^{\prime }(y)\overset{!}{\mathop{=}}\,{{x}^{2}}$$ thus,
 * <p style="text-align:right;">$$\,(Eq. 1.22) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{k}_{1}}^{\prime }\left( y \right)=0\Rightarrow {{k}_{1}}\left( y \right)=C$$ With the knowledge of Eq. 1.23 we can rewrite Eq. 1.20 as follows for the functino h(x,y):
 * <p style="text-align:right;">$$\,(Eq. 1.23) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,h(x,y)={{x}^{2}}y$$ and thus the function $$\Phi$$ can be written as being:
 * <p style="text-align:right;">$$\,(Eq. 1.24) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Phi \left( x,y,p \right)={{x}^{2}}y+\cos \left( x \right)p+C={{x}^{2}}y+\cos \left( x \right){y}'+C=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\,(Eq. 1.25) $$
 * style= |
 * }

Sove for y
At this point solving for y is not very difficult anymore, since we only need to integrate p which is the derivative of y. For that, Eq. 1.25 is rewritten as:
 * {| style="width:100%" border="0"

$$p\left( x \right)=\frac{-({{x}^{2}}y+C)}{\cos \left( x \right)}=\frac{-{{x}^{2}}y}{\cos \left( x \right)}-\frac{C}{\cos \left( x \right)}={y}'$$ Integrating Eq. 1.26 with respect to x gives the solution for y:
 * <p style="text-align:right;">$$\,(Eq. 1.26) $$
 * style= |
 * }

Contributing Members
Solved and posted by Egm6321.f10.team03.sigillo 01:01, 19 October 2010 (UTC)