User:Egm6321.f12.team6.Hu.M/report2

=Report 2=

==Problem *2.8 Explain the Difficulty in Solving the Integrating Factor $$ \displaystyle h(x,y) $$ ==

Problem Statement

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

Given

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$$ \displaystyle h_{x}N-h_{y}M+h\left ( N_{x}-M_{y} \right )=0 $$
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 * (8.1)
 * }

Nomenclature

 * $$ \displaystyle h_{x} = \frac {\partial h}{\partial x} $$
 * $$ \displaystyle N_{x} = \frac {\partial N}{\partial x} $$
 * $$ \displaystyle M_{y} = \frac {\partial N}{\partial y} $$

Solution

 * 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 $$ \displaystyle M_{y} $$.
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With reasons listed above, Equation 8.1 is usually not easy to solve.
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 * }

Problem Statement

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Solve $$ \displaystyle a_1(x)\,y'+a_0(x)\,y=b(x) $$.
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 * (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:
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$$ \displaystyle h(x)=\exp \left[ \int^x n(s)ds+k_1\right] $$
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 * (11.2)
 * }
 * To obtain $$ \displaystyle y(x) $$,with Equation 11.2 we have:
 * $$ \displaystyle y'+a_0y=b \Rightarrow h(y'+a_0y)=hb $$
 * $$ \displaystyle \frac{h_x}{h}=a_0 \Rightarrow ha_0=h_x=h' $$
 * $$ \displaystyle hy'+h'y=hb $$
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$$ \displaystyle \Rightarrow y(x)=\frac{1}{h(x)} \left[ \int^x h(s)b(s)ds+k_2\right] $$
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 * (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:
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$$ \displaystyle y(x)=\frac{1}{h(x)} \left[ \int^x h(s)b(s)ds+k_2\right]=\frac{1}{\exp \left[ \frac{(x+1)^2}{2}\right]} \left[ \int^x \exp \left[\frac{(s+1)^2}{2} \right](s^2+4)ds+k_2\right] $$
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 * (11.4)
 * }
 * So, we have final solution:
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$$ \displaystyle y(x)=\frac{1}{\exp \left[ \frac{(x+1)^2}{2}\right]}\left[\int\exp\left[\frac{(x+1)^2}{2}\right](x^2+4)dx \right] $$
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 * }

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:
 * $$ \displaystyle h(x)=\exp \left[ \int^x \frac{a_0(s)}{a_1(s)}ds\right] $$
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$$ \displaystyle y(x)=\frac{1}{\exp \left[ \int^x \frac{a_0(s)}{a_1(s)}ds\right]} \left[ \int^x \exp \left[ \int^x \frac{a_0(s)}{a_1(s)}ds\right]\frac{b(s)}{a_1(s)}ds+k_2\right] $$
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 * }

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:
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$$ \displaystyle h(x)=\exp\left[\int^x\frac{s^3}{s^2+1}ds \right]=\exp\left[\frac{1}{2}\int\frac{x^2}{x^2+1}dx^2\right]$$
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 * (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:
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$$ \displaystyle y(x)=\frac{1}{\exp\left\{\frac{1}{2}\left[x^2-ln(x^2+1)\right]\right\}}\left[\int\exp\left\{\frac{1}{2}\left[x^2-ln(x^2+1)\right]\right\}\frac{x^4}{x^2+1}dx+k_2\right] $$
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 * }

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.