User:Egm6321.f12.team1.cla/report2

=Problem 2.1* - Two linearly-independent homogeneous solutions:=

Link to lecture notes

Given:
The Legendre differential operator equation
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 * $$L_{2}(y)=(1-x^2)y''-2xy'+n(n+1)y=0$$
 * $$L_{2}(y)=(1-x^2)y''-2xy'+n(n+1)y=0$$

 (1.1)
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Two linearly-independent homogeneous solutions
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 * $$y^1_{H}(x)=x$$
 * $$y^1_{H}(x)=x$$

 (1.2)
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 * $$y^2_{H}(x)=\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1$$
 * $$y^2_{H}(x)=\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1$$



 (1.3)
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Find:
Show that
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$$L_{2}(y^1_{H}(x))=L_{2}(y^2_{H}(x))=0$$  (1.4)
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Solution:
The solution is found by taking the first and second derivatives of equations 1.2 and 1.3 and then plugging them into equation 1.1
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$$\frac{d}{dx}y^1_{H}(x)=1$$  (1.5)
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$$\frac{d^2}{dx^2}y^1_{H}(x)=0$$  (1.6)
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$$\frac{d}{dx}y^2_{H}(x)=\frac{d}{dx}\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-\frac{d}{dx}(1)$$  (1.7)
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$$=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{2}\left(\frac{1-x}{1+x}\right)\left(\frac{2}{(1-x)^2}\right)$$  (1.8)
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$$=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\left(\frac{x}{1-x^2}\right)$$  (1.8)
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 * $$\frac{d^2}{dx^2}y^1_{H}(x)=\frac{d}{dx}\left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]$$
 * $$\frac{d^2}{dx^2}y^1_{H}(x)=\frac{d}{dx}\left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]$$

 (1.9)
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 * $$=\frac{1}{2}\left(\frac{1-x}{1+x}\right)\left(\frac{2}{(1-x)^2}\right)+\frac{1(1-x^2)-(-2x)(x)}{(1-x^2)^2}$$
 * $$=\frac{1}{2}\left(\frac{1-x}{1+x}\right)\left(\frac{2}{(1-x)^2}\right)+\frac{1(1-x^2)-(-2x)(x)}{(1-x^2)^2}$$

 (1.10)
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 * $$=\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}$$
 * $$=\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}$$

 (1.11)
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Now plug in derivatives into the Legrenge operator and see if you get the solution zero


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 * $$L_{2}(y^1_{H}(x))=(1-x^2)(0))-2x(1)+2x$$
 * $$L_{2}(y^1_{H}(x))=(1-x^2)(0))-2x(1)+2x$$

 (1.12)
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 * $$L_{2}(y^1_{H}(x))=0$$
 * $$L_{2}(y^1_{H}(x))=0$$

 (1.13)
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 * $$L_{2}(y^2_{H}(x))=(1-x^2)\left[\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}\right]-2x\left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\left(\frac{x}{1-x^2}\right)\right]+2\left[\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1\right]$$
 * $$L_{2}(y^2_{H}(x))=(1-x^2)\left[\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}\right]-2x\left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\left(\frac{x}{1-x^2}\right)\right]+2\left[\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1\right]$$

 (1.14)
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 * $$L_{2}(y^2_{H}(x))=1+\frac{1+x^2}{1-x^2}-xlog\left(\frac{1+x}{1-x}\right)-\frac{2x^2}{1-x^2}+xlog\left(\frac{1+x}{1-x}\right)-2$$
 * $$L_{2}(y^2_{H}(x))=1+\frac{1+x^2}{1-x^2}-xlog\left(\frac{1+x}{1-x}\right)-\frac{2x^2}{1-x^2}+xlog\left(\frac{1+x}{1-x}\right)-2$$

 (1.15)
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 * $$L_{2}(y^2_{H}(x))=\frac{2x^2}{1-x^2}-xlog\left(\frac{1+x}{1-x}\right)-\frac{2x^2}{1-x^2}+xlog\left(\frac{1+x}{1-x}\right)$$
 * $$L_{2}(y^2_{H}(x))=\frac{2x^2}{1-x^2}-xlog\left(\frac{1+x}{1-x}\right)-\frac{2x^2}{1-x^2}+xlog\left(\frac{1+x}{1-x}\right)$$

<p style="text-align:right"> (1.16)
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 * $$L_{2}(y^2_{H}(x))=0$$
 * $$L_{2}(y^2_{H}(x))=0$$

<p style="text-align:right"> (1.17)
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Therefore equation 1.4 is verified
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$$L_{2}(y^1_{H}(x))=L_{2}(y^2_{H}(x))=0$$ <br\>
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=Problem 2.2* Verify a solution = <br\> Link to lecture notes

Given:
<br\>

Original Equation:

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 * $$p'+p=x$$
 * $$p'+p=x$$

<p style="text-align:right"> (2.1) <br\>
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Solution:

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 * $$p(x)=k_{1}e^{-x}+x-1$$
 * $$p(x)=k_{1}e^{-x}+x-1$$

<p style="text-align:right"> (2.2) <br\>
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Find:
<br\> Verify that $$p(x)=k_{1}e^{-x}+x-1$$ is the solution of equation (2.1) <br\>

Solution:
<br\> First plug equation (2.2) into (2.1) <br\>
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 * $$x=(k_{1}e^{-x}+x-1)'+(k_{1}e^{-x}+x-1)$$
 * $$x=(k_{1}e^{-x}+x-1)'+(k_{1}e^{-x}+x-1)$$

<p style="text-align:right"> (2.3) <br\> Now take the derivative of the terms in p' <br\>
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 * $$x=(-k_{1}e^{-x}+1-0)+(k_{1}e^{-x}+x-1)$$
 * $$x=(-k_{1}e^{-x}+1-0)+(k_{1}e^{-x}+x-1)$$

<p style="text-align:right"> (2.4) <br\> Group common terms <br\>
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 * $$x=(-k_{1}e^{-x}+k_{1}e^{-x})+(-1+1)+x$$
 * $$x=(-k_{1}e^{-x}+k_{1}e^{-x})+(-1+1)+x$$

<p style="text-align:right"> (2.5) <br\> Canceling terms results in the following <br\>
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 * $$x=x$$
 * $$x=x$$

<p style="text-align:right"> (2.6) <br\> This verifies that equation (2.2) is a solution to (2.1) <br\>
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=Problem 2.3*Show linearity & general N1-ODE= <br\>

Given:

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$$M(x,y) + N(x,y)y' = 0$$ <p style="text-align:right"> (3.1) <br\>
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To Show:
The equation 3.1 is linear in y' and that it is a N1-ODE <br\>

Solution:
<br\> For a function to be linear it needs to satisfy two conditions. For a function f(.)
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$$f(x+y) = f(x) + f(y)$$ <p style="text-align:right"> (3.2)
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$$f(a*x) = a f(x)$$ <p style="text-align:right"> (3.3)
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From equation 3.1 we have


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$$y' = -\frac{M(x,y)}{N(x,y)}$$
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To show that the function is a first order ODE(N1-ODE)
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$$f(\alpha y'_1 + \beta y'_2) = -\alpha\frac{M(x_1,y_1)}{N(x_1,y_1)} -\beta\frac{M(x_2,y_2)}{N(x_2,y_2)} = -\alpha f(y'_1)-\beta f(y'_2)$$ <p style="text-align:right"> (3.4)
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The given function is thus shown to be linear in y' and hence is an N1-ODE

<br\>

=Problem 2.4*: Orthogonality of Legendre Functions=

Refer to Case Functions and linearly-independent Homogeneous Solutions from the lecture notes for reference.

Given:

 * Given the following equations (3) and (4) from lecture notes linearly-independent Homogeneous Solutions


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$$y_{H}^{1}(x)=x=:P_{1}(x)$$ <p style="text-align:right"> (4.1)
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 * and
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$$y_{H}^{2}(x)=\frac{x}{2}\log \left (\frac{1+x}{1-x} \right )-1=:Q_{1}(x)$$ <p style="text-align:right"> (4.2)
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Find:

 * Show that
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$$\exists x$$ such that $$y_{H}^{1}(\hat{x})\neq \alpha y_{H}^{2}(\hat{x})$$ <p style="text-align:right"> (4.3)
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 * for any given $$\alpha$$

Solution:

 * Let us prove the relation by contradiction.
 * Let us assume $$\exists \alpha$$ such that

<br\>
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 * $$y_{H}^{1}(\hat{x})= \alpha y_{H}^{2}(\hat{x})\qquad \qquad \forall x \in \mathbb R$$
 * $$y_{H}^{1}(\hat{x})= \alpha y_{H}^{2}(\hat{x})\qquad \qquad \forall x \in \mathbb R$$

<p style="text-align:right"> (4.4)
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 * $$\Rightarrow x = \alpha \frac{x}{2}\log \left (\frac{1+x}{1-x} \right )-\alpha \qquad \qquad \forall x \in \mathbb R$$
 * $$\Rightarrow x = \alpha \frac{x}{2}\log \left (\frac{1+x}{1-x} \right )-\alpha \qquad \qquad \forall x \in \mathbb R$$

<p style="text-align:right"> (4.5)
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 * Taking derivative with respect to x
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$$1=\alpha \frac{d}{dx}\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-\frac{d}{dx}(\alpha)\qquad \qquad \forall x \in \mathbb R$$ <p style="text-align:right">
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 * $$\Rightarrow 1 = \frac{\alpha}{2}log\left(\frac{1+x}{1-x}\right)+\frac{\alpha x}{2}\left(\frac{1-x}{1+x}\right)\left(\frac{2}{(1-x)^2}\right) \qquad \qquad \forall x \in \mathbb R$$
 * $$\Rightarrow 1 = \frac{\alpha}{2}log\left(\frac{1+x}{1-x}\right)+\frac{\alpha x}{2}\left(\frac{1-x}{1+x}\right)\left(\frac{2}{(1-x)^2}\right) \qquad \qquad \forall x \in \mathbb R$$

<p style="text-align:right"> (4.6)
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 * The Left Hand Side of the equation is a constant, but the Right Hand Side can never be reduced to a constant.

<br\><br\>
 * So the assumption that we made was incorrect.

<br\>
 * therefore
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$$\exists x$$ such that $$y_{H}^{1}(\hat{x})\neq \alpha y_{H}^{2}(\hat{x})$$ for any given $$\alpha\quad \blacksquare$$ <p style="text-align:right"> (4.7)
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=Problem 2.5*-Show N1-ODE=

Given:

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 * $$\phi (x,y)=x^2y^\frac{3}{2}+log(x^3y^2)=k$$
 * $$\phi (x,y)=x^2y^\frac{3}{2}+log(x^3y^2)=k$$

<p style="text-align:right"> (5.1)
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Find:

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 * $$G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$
 * $$G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$

<p style="text-align:right"> (5.2)
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 * and show that (5.1)is an N1-ODE

Solution:

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 * $$\frac{d}{dx}\phi(x,y)=2xy^\frac{3}{2}+x^2\left(\frac{3}{2}y^\frac{1}{2}y'\right)+\frac{1}{x^3y^2}\left(3x^2y^2+2x^3yy' \right )=0$$
 * $$\frac{d}{dx}\phi(x,y)=2xy^\frac{3}{2}+x^2\left(\frac{3}{2}y^\frac{1}{2}y'\right)+\frac{1}{x^3y^2}\left(3x^2y^2+2x^3yy' \right )=0$$

<p style="text-align:right"> (5.3)
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 * $$\frac{d}{dx}\phi(x,y)=2xy^\frac{3}{2}+\frac{3}{2}x^2y^\frac{1}{2}y'+\frac{3}{x}+\frac{2}{y}y'=0$$
 * $$\frac{d}{dx}\phi(x,y)=2xy^\frac{3}{2}+\frac{3}{2}x^2y^\frac{1}{2}y'+\frac{3}{x}+\frac{2}{y}y'=0$$

<p style="text-align:right"> (5.4)
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 * $$\frac{d}{dx}\phi(x,y)=\left(2xy^\frac{3}{2}+\frac{3}{x}\right)+\left(\frac{3}{2}x^2y^\frac{1}{2}+\frac{2}{y}\right)y'=0$$
 * $$\frac{d}{dx}\phi(x,y)=\left(2xy^\frac{3}{2}+\frac{3}{x}\right)+\left(\frac{3}{2}x^2y^\frac{1}{2}+\frac{2}{y}\right)y'=0$$

<p style="text-align:right"> (5.5)
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Observing Equation(5.5)$$\frac{d}{dx}\phi(x,y)$$has three variables: y',y and x. Letting
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 * $$G(y',y,x):=\frac{d}{dx}\phi(x,y)$$
 * $$G(y',y,x):=\frac{d}{dx}\phi(x,y)$$

Therefore
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 * $$G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$<br\>
 * $$G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$<br\>

<p style="text-align:right"> (5.6)
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 * $$G(y',y,x)=\underbrace{\left(2xy^\frac{3}{2}+\frac{3}{x}\right)}_{M(x,y)}+\underbrace{\left(\frac{3}{2}x^2y^\frac{1}{2}+\frac{2}{y}\right)y'}_{N(x,y)}=0$$
 * $$G(y',y,x)=\underbrace{\left(2xy^\frac{3}{2}+\frac{3}{x}\right)}_{M(x,y)}+\underbrace{\left(\frac{3}{2}x^2y^\frac{1}{2}+\frac{2}{y}\right)y'}_{N(x,y)}=0$$

<p style="text-align:right"> (5.7) Define that
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 * $$\left\{\begin{matrix}M(x,y):=2xy^\frac{3}{2}+\frac{3}{x}
 * $$\left\{\begin{matrix}M(x,y):=2xy^\frac{3}{2}+\frac{3}{x}

\\ N(x,y):=\frac{3}{2}x^2y^\frac{1}{2}+\frac{2}{y} \end{matrix}\right.$$ We can find
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 * $$G(y',y,x)=M(x,y)+N(x,y)y'=0$$
 * $$G(y',y,x)=M(x,y)+N(x,y)y'=0$$

<p style="text-align:right"> (5.8) Equation (5.8) is satisfied with the first exactness condition of N1=ODE Find
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 * $$M_y(x,y)=\frac{\partial M(x,y)}{\partial y}=3xy^{\frac{1}{2}}$$
 * $$M_y(x,y)=\frac{\partial M(x,y)}{\partial y}=3xy^{\frac{1}{2}}$$

<p style="text-align:right"> (5.9) and
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 * $$N_x(x,y)=\frac{\partial N(x,y)}{\partial x}=3xy^{\frac{1}{2}}$$
 * $$N_x(x,y)=\frac{\partial N(x,y)}{\partial x}=3xy^{\frac{1}{2}}$$

<p style="text-align:right"> (5.10) We find
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 * $$M_y(x,y)=N_x(x,y)$$
 * $$M_y(x,y)=N_x(x,y)$$

<p style="text-align:right"> (5.11) Equation (5.11) is satisfied with the Second exactness condition of N1-ODE. Therefore, Equation(5.1)is a N1-ODE <br\>
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=Problem 2.6*: Symmetry of second derivatives=

Find:
Find the minimum degree of differentiability of the function $$\phi(x,y)$$ such that the following equation
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 * $$\frac {\partial^2 \phi (x,y)} {\partial x \partial y} = \frac {\partial^2 \phi (x,y)} {\partial y \partial x}$$
 * $$\frac {\partial^2 \phi (x,y)} {\partial x \partial y} = \frac {\partial^2 \phi (x,y)} {\partial y \partial x}$$

<br\> is satisfied. State the full theorem and provide a proof. Link to the Lecture Note
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Solution:
According to Mean Value Theorem, if $$f(x)$$ is continuous in $$[x,x+h]$$ and differentiable in $$(x,x+h)$$, then there exists a point $$x_1 \in (x,x+h)$$ such that
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 * $$f(x+h)-f(x) = h \frac{\partial f(x)} {\partial x} \vert _{x=x_1}$$
 * $$f(x+h)-f(x) = h \frac{\partial f(x)} {\partial x} \vert _{x=x_1}$$

Let us define
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 * $$f(h,k):=[\phi(x+h,y+k)-\phi(x+h,y)]-[\phi(x,y+k)+\phi(x,y)]$$
 * $$f(h,k):=[\phi(x+h,y+k)-\phi(x+h,y)]-[\phi(x,y+k)+\phi(x,y)]$$

<br\> Applying Mean Value Theorem on the two terms in square braces
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 * $$f(h,k) = k \frac{\partial \phi(x+h,y_1)} {\partial y} - k \frac{\partial \phi(x,y_2)} {\partial y}$$
 * $$f(h,k) = k \frac{\partial \phi(x+h,y_1)} {\partial y} - k \frac{\partial \phi(x,y_2)} {\partial y}$$


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 * $$\Rightarrow\frac{f(h,k)}{k} = \frac{\partial \phi(x+h,y_1)} {\partial y} - \frac{\partial \phi(x,y_2)} {\partial y}$$
 * $$\Rightarrow\frac{f(h,k)}{k} = \frac{\partial \phi(x+h,y_1)} {\partial y} - \frac{\partial \phi(x,y_2)} {\partial y}$$

<br\> As $$k \rightarrow 0, y_1 \rightarrow y$$ and $$y_2 \rightarrow y $$
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 * $$\Rightarrow \lim_{k \to 0}\frac{f(h,k)}{k} = \frac{\partial \phi(x+h,y)} {\partial y} - \frac{\partial \phi(x,y)} {\partial y} $$
 * $$\Rightarrow \lim_{k \to 0}\frac{f(h,k)}{k} = \frac{\partial \phi(x+h,y)} {\partial y} - \frac{\partial \phi(x,y)} {\partial y} $$

<br\> Applying Mean Value Theorem
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 * $$\lim_{k \to 0}\frac{f(h,k)}{k} = h \frac{\partial \phi(x_1,y)} {\partial y}$$, where $$x_1 \in (x,x+h)$$
 * $$\lim_{k \to 0}\frac{f(h,k)}{k} = h \frac{\partial \phi(x_1,y)} {\partial y}$$, where $$x_1 \in (x,x+h)$$


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 * $$\Rightarrow \lim_{k \to 0}\frac{f(h,k)}{hk} = \frac{\partial \phi(x_1,y)} {\partial y}$$
 * $$\Rightarrow \lim_{k \to 0}\frac{f(h,k)}{hk} = \frac{\partial \phi(x_1,y)} {\partial y}$$

<br\>
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As $$h \rightarrow 0, x_1 \rightarrow x$$ <br\>
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 * $$\lim_{h \to 0} \lim_{k \to 0}\frac{f(h,k)}{hk} = \frac{\partial^2 \phi(x,y)}{\partial x \partial y})$$
 * $$\lim_{h \to 0} \lim_{k \to 0}\frac{f(h,k)}{hk} = \frac{\partial^2 \phi(x,y)}{\partial x \partial y})$$

<p style="text-align:right"> (6.1) <br\> Rearranging the terms of $$f(h,k)$$
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 * $$f(h,k):=[\phi(x+h,y+k)-\phi(x,y+k)]-[\phi(x+h,y)+\phi(x,y)]$$
 * $$f(h,k):=[\phi(x+h,y+k)-\phi(x,y+k)]-[\phi(x+h,y)+\phi(x,y)]$$

<br\> Following similar arguments, we can show that <br\>
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 * $$\lim_{k \to 0} \lim_{h \to 0}\frac{f(h,k)}{hk} = \frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$
 * $$\lim_{k \to 0} \lim_{h \to 0}\frac{f(h,k)}{hk} = \frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$

<p style="text-align:right"> (6.2) <br\> Equation 6.1 and 6.2 implies that
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y} = \frac{\partial^2 \phi(x,y)}{\partial y \partial x}$$
 * $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y} = \frac{\partial^2 \phi(x,y)}{\partial y \partial x}$$

<p style="text-align:right"> (6.3) <br\> To prove the relation, the only assumption we made was in the application of Mean Value Theorem, where we assumed that <br\><br\>
 * }
 * {| style="width:100%" border="0"

<br\><br\> These are the three conditions necessary for equation 6.3 to hold. Since $$h \rightarrow 0, k \rightarrow 0$$, they can together be stated as <br\><br\>
 * style="width:100%" |
 * 1) $$f(h,k)$$ is differentiable in $$(x,x+h)$$ and $$(y,y+k)$$
 * 2) $$\frac{\partial^2 \phi(x,y)}{\partial y}$$ is differntiable in $$(x,x+h)$$
 * 3) $$\frac{\partial^2 \phi(x,y)}{\partial x}$$ is differntiable in $$(y,y+k)$$
 * }
 * }


 * $$\phi(x,y), \frac{\partial^2 \phi(x,y)}{\partial x}, \frac{\partial^2 \phi(x,y)}{\partial y}$$ are required to be differentiable at $$(x,y)$$ for equation 6.3 to hold. That is $$\phi(x,y)$$ needs to be Double Differentiable.

Pavel Bhowmik (talk) 01:41, 16 October 2012 (UTC)

=Problem 2.7*: Verify a proposed solution=

Given:
Given the following equation
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$M(x,y)+N(y,x)y' = 0$$
 * $$M(x,y)+N(y,x)y' = 0$$

<p style="text-align:right"> (7.1) <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * with $$M(x,y) = 75x^4, N(x,y) = cos(y)$$
 * with $$M(x,y) = 75x^4, N(x,y) = cos(y)$$

<p style="text-align:right"> (7.2) and the proposed solution $$y(x) = sin^{-1}(k-15x^5)$$ Link to the Lecture Note
 * }

Verify:
Verify that the proposed solution is indeed the solution of the given equation

Solution:
The proposed solution
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$y(x) = sin^{-1}(k-15x^5)$$
 * $$y(x) = sin^{-1}(k-15x^5)$$

<p style="text-align:right"> (7.3) <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\Rightarrow sin(y) = k-15x^5$$
 * $$\Rightarrow sin(y) = k-15x^5$$

<p style="text-align:right"> (7.4) <br\> Differentiating with respect to $$x$$
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$ cos(y)y' = -75x^4$$
 * $$ cos(y)y' = -75x^4$$

<p style="text-align:right"> (7.5) <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$ 75x^4+cos(y)y' = 0$$
 * $$ 75x^4+cos(y)y' = 0$$

<p style="text-align:right"> (7.6) <br\> Which is the same as the given differential equation. <br\> So $$y(x) = sin^{-1}(k-15x^5)$$ is indeed the solution of the given equation. Pavel Bhowmik (talk) 01:41, 16 October 2012 (UTC)
 * }

=Problem 2.8* Explain why solving for the integrating factor h(x,y) is usually not easy.= <br\> Link to lecture notes and

Given:
<br\>
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$f(x,y)=M(x,y)+N(x,y)y'$$
 * $$f(x,y)=M(x,y)+N(x,y)y'$$

<p style="text-align:right"> (8.1) <br\> Using Euler's Integration Method there is a function h(x,y) so that, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$h(x,y)[M(x,y)+N(x,y)y']=0$$
 * $$h(x,y)[M(x,y)+N(x,y)y']=0$$

<p style="text-align:right"> (8.2) <br\> where h(x,y) is the integrating factor. <br\>
 * }

Find:
<br\> Why finding h(x,y) is difficult <br\>

Solution:
<br\> Rewriting equation we get, <br\>
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$
 * $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$

<p style="text-align:right"> (8.3) <br\> Solving the differential equation assuming h is only a function of x or y (not both) is easy because the either $$h_{y}$$ or $$h_{x}$$ becomes 0 allowing you to use separation of variables to find the integrating factor. This is demonstrated below assuming h(x,y) is only a function of x. <br\> Consider the equation, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$
 * $$h_{x}N-h_{y}M+h(N_{x}-M_{y})=0$$

<p style="text-align:right"> (8.4) <br\> Because we are assuming the integration factor is only a function of x $$h_{y}$$ becomes 0, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$h_{x}N-0+h(N_{x}-M_{y})=0$$
 * $$h_{x}N-0+h(N_{x}-M_{y})=0$$

<p style="text-align:right"> (8.5) <br\> Divide terms by h and N to separate variables, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\frac{h_{x}}{h}+\frac{(N_{x}-M_{y})}{N}=0$$
 * $$\frac{h_{x}}{h}+\frac{(N_{x}-M_{y})}{N}=0$$

<p style="text-align:right"> (8.6) <br\> Further simplification results in, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\frac{h_{x}}{h}=-\frac{(N_{x}-M_{y})}{N}$$
 * $$\frac{h_{x}}{h}=-\frac{(N_{x}-M_{y})}{N}$$

<p style="text-align:right"> (8.7) <br\> From this point integrating both sides yields the integration factor assuming it is only a function of x. Assuming the integration factor is only a function of y also allows for a simple solution using separation of variables. However trying to find a function h(x,y)=0 means not being able to remove M(x,y) and N(x,y) from equation and making it very difficult to solve, <br\> Again start by dividing by h, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\frac{h_{x}}{h}N-\frac{h_{y}}{h}M+(N_{x}-M_{y})=0$$
 * $$\frac{h_{x}}{h}N-\frac{h_{y}}{h}M+(N_{x}-M_{y})=0$$

<p style="text-align:right"> (8.8) <br\> Then subtract $$(N_{x}-M_{y})$$ to the left side of the equation, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\frac{h_{x}}{h}N-\frac{h_{y}}{h}M=-(N_{x}-M_{y})$$
 * $$\frac{h_{x}}{h}N-\frac{h_{y}}{h}M=-(N_{x}-M_{y})$$

<p style="text-align:right"> (8.9) <br\> Without the assumption of the integrating factor being only a function of one variable makes it difficult to simplify the equation further because separation of variables is not possible at this point. If we were to know the functions of M and N and both were homogeneous of the same degree the equation below could be used to find the integrating factor, <br\>
 * }
 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\frac{1}{xN+yM}$$
 * $$\frac{1}{xN+yM}$$

<p style="text-align:right"> (8.10) <br\> Equation from
 * }

=Problem* 2.9: Find IMF for Equation= <br\>

Given:

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

$$ \frac{h_y}{h} = \frac{1}{N}(N_x + M_y)$$ <p style="text-align:right"> (9.1)
 * style="width:100%" |
 * style="width:100%" |
 * }

<br\>

Find:
h from the equation 9.1 <br\>

Solution:
<br\> Under the assumption $$h_x = 0$$ Equation (4) 11-1
 * {| style="width:100%" border="0"

$$ h_x N - h_y M + h(N_x - M_y) = 0 $$ <p style="text-align:right"> (9.2) Becomes
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h_y M + h(N_x - M_y) = 0$$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * }

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

$$ \frac{h_y}{h} = -\frac{1}{M}(N_x - M_y) $$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * }


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

$$ \int\frac{h_y}{h} = -\int \frac{1}{M}(N_x - M_y) $$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * }


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

$$ \log h(y) = -\int\limits_{}^{y}n(y) + k $$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * }

Where $$ n(y)= \frac{1}{M}(N_x - M_y) $$

The solution to h this is
 * {| style="width:100%" border="0"

$$h = e^{-\int\limits_{}^{y}n(x) + k} $$ <p style="text-align:right"> (9.3) Referenced from the lecture pea1.sec11.wmv and Pea1.f12.sec11.djvu <br\>
 * style="width:100%" |
 * style="width:100%" |
 * }

=Problem 2.10* Euler Integration factor method= </Br>
 * Refer to lecture notes Euler Integration factor method section 11 for reference.

Given:

 * Given the following


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

$${y}'+\frac{1}{x}y=x^{2}$$ <p style="text-align:right"> (10.1)
 * style="width:100%" |
 * style="width:100%" |
 * }

Show:

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

$$h(x)=x$$ <p style="text-align:right"> (10.2)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * and
 * {| style="width:100%" border="0"

$$y(x)=\frac{x^{3}}{4}+\frac{k}{x}$$ <p style="text-align:right"> (10.3)
 * style="width:100%" |
 * style="width:100%" |
 * }

Solution:

 * From Equation $${y}'+\frac{1}{x}y=x^{2}$$ we can deduce that


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

$$a_{0}(x):=\frac{1}{x}$$ <p style="text-align:right"> (10.4)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * and
 * {| style="width:100%" border="0"

$$b(x):=x^{2}$$ <p style="text-align:right"> (10.5)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * From lecture notes Euler Integration factor method section 11 page 4, the value of h(x) is defined by
 * {| style="width:100%" border="0"

$$h(x)=exp\left [ \int_{ }^{x}a_{0}(s)ds+k_{1} \right ]$$ <p style="text-align:right"> (10.6)
 * style="width:100%" |
 * style="width:100%" |
 * }


 * and From lecture notes Euler Integration factor method section 11 page 5 the value of y(x) is defined by


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

$$y(x)=\frac{1}{h(x)}\left [\int_{ }^{x} h(s)b(s)ds+k_{2}\right ]$$ <p style="text-align:right"> (10.7)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * We can therefore find a value for $$h(x)$$ by applying $$a_{0}$$ into the equation.


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

$$h(x)=exp\left [ \int_{ }^{x}\frac{1}{x}dx\right ]$$ <p style="text-align:right"> (10.8)
 * style="width:100%" |
 * style="width:100%" |
 * }


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

$$h(x)=exp\left [ \ln x \right ]$$ <p style="text-align:right"> (10.9)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * which yields to


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

$$h(x)=x$$ <p style="text-align:right"> (10.10)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * Applying $$h(x)=x$$ into $$y(x)=\frac{1}{h(x)}\left [\int_{ }^{x} h(s)b(s)ds+k_{2}\right ]$$ yields to
 * {| style="width:100%" border="0"

$$y(x)=\frac{1}{x}\left [\int_{ }^{x} s\cdot s^{2}ds+k_{2}\right ]$$ <p style="text-align:right"> (10.11)
 * style="width:100%" |
 * style="width:100%" |
 * }


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

$$y(x)=\frac{1}{x}\left [\int_{ }^{x} s^{3}ds+k_{2}\right ]$$ <p style="text-align:right"> (10.12)
 * style="width:100%" |
 * style="width:100%" |
 * }


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

$$y(x)=\frac{1}{x}\left [ \frac{x^{4}}{4}+k\right ]$$ <p style="text-align:right"> (10.13)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * Therefore
 * {| style="width:100%" border="0"

$$y(x)=\frac{x^{3}}{4}+\frac{k}{x}$$ <p style="text-align:right"> (10.14)
 * style="width:100%" |
 * style="width:100%" |
 * }

=Problem 2.11*-Solve the general L1-ODE-VC=
 * Refer to lecture notes Euler Integration factor method section 11 <br\>

Given:

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

$$ a_1(x)y'+a_0(x)y=b(x)$$ <p style="text-align:right"> (11.1) Part 1 $$ a_1(x)=1, a_0(x)=x+1, b(x)=x^2+4 $$ Part 2 N/A Part 3 $$ a_1(x)=x^2+1, a_0(x)=x^3, b(x)=x^4 $$
 * style="width:100%" |
 * style="width:100%" |
 * }

Find:
Part 1 $$y(x)$$ Part 2 $$y(x)$$ in terms of $$a_1(x),a_0(x),b(x)$$ Part 3 $$y(x)$$

Part 1:
Substitute $$ a_1(x)=1, a_0(x)=x+1, b(x)=x^2+4 $$ into Equation(11.1),we get
 * {| style="width:100%" border="0"

$$ y'+\underbrace{(x+1)}_{a_0(x)}y=\underbrace{x^2+4}_{b(x)}$$ <p style="text-align:right"> (11.2) Find Euler Integrating factor for Equation(11.2)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h(x)=exp[\int^{x}a_0(s)ds+k_1]=exp[\int^{x}(s+1)ds+k_1]$$ <p style="text-align:right"> (11.3)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h(x)=exp(\frac{x^2}{2}+x+k_1)$$ <p style="text-align:right"> (11.4) Apply h(x) to equation(11.2)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h(x)y'+h(x)(x+1)y=h(x)(x^2+4)$$ <p style="text-align:right"> (11.5)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ exp(\frac{x^2}{2}+x)y'+exp(\frac{x^2}{2}+x)(x+1)y=exp(\frac{x^2}{2}+x)(x^2+4)$$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dx}[exp(\frac{x^2}{2}+x)y]=exp(\frac{x^2}{2}+x)(x^2+4)$$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ exp(\frac{x^2}{2}+x)y=\int^xexp(\frac{s^2}{2}+s)(s^2+4)ds+k_2$$
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ y=\frac{1}{exp(\frac{x^2}{2}+x)}[\int^xexp(\frac{s^2}{2}+s)(s^2+4)ds+k_2]$$ Using Matlab to solve y
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ exp(\frac{x^2}{2}+x)y=exp(0.5*x^2 + x)*(x - 1.0) - erf(x*0.7071*i + 0.70715*i)*3.0407*i - 1.8991$$ <p style="text-align:right"> (11.6)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$y(x)=-exp(- x^2/2 - x)*(erf(x*0.7071*i + 0.7071*i)*3.0407*i - exp(0.5*x^2 + x)*(x - 1.0) + 1.8991)$$ <p style="text-align:right"> (11.7)
 * style="width:100%" |
 * style="width:100%" |
 * }

Part 2:
Equation(11.2)is equal to
 * {| style="width:100%" border="0"

$$ \underbrace{1}_{N(x,y)}*y'+\underbrace{\frac{a_0(x)}{a_1(x)}}_{M(x,y)}y=\frac{b(x)}{a_1(x)}$$ <p style="text-align:right"> (11.8) Euler Integrating factor for Equation(11.2) is
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h(x)=exp[\int^{x}\frac{a_0(s)}{a_1(s)}ds+k_1]$$ <p style="text-align:right"> (11.9) Apply h(x)to Equation (11.8) gives
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h(x)y'+h(x)\frac{a_0(x)}{a_1(x)}y=h(x)\frac{b(x)}{a_1(x)}$$ <p style="text-align:right"> (11.10) And refer to definition Lecture Section 14 Equation(2)P11-2
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ \frac{h_x}{h}=-\underbrace{\frac{1}{N}}_{1}(\underbrace{N_x}_{0}-\underbrace{M_y}_{\frac{a_0(x)}{a_1(x)}})$$ <p style="text-align:right"> (11.11)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ \frac{h_x}{h}=\frac{a_0(x)}{a_1(x)}$$ <p style="text-align:right"> (11.12)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h\frac{a_0(x)}{a_1(x)}=h'$$ <p style="text-align:right"> (11.13) Replace$$h(x)\frac{a_0(x)}{a_1(x)}$$ with $$h'(x)$$ in Equation(11.10)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ h(x)y'+h'(x)y=h(x)\frac{b(x)}{a_1(x)}$$ <p style="text-align:right"> (11.14)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ (hy)'=h(x)\frac{b(x)}{a_1(x)}$$ <p style="text-align:right"> (11.15)
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ y(x)=\frac{1}{h(x)}[\int^xh(s)\frac{b(s)}{a_1(s)}ds+k_2]$$ <p style="text-align:right"> (11.16) Apply Equation (11.9) into Equation (11.16), therefore
 * style="width:100%" |
 * style="width:100%" |
 * }
 * {| style="width:100%" border="0"

$$ y(x)=\frac{1}{exp[\int^{x}\frac{a_0(s)}{a_1(s)}ds+k_1]}[\int^x\frac{exp[\int^{x}\frac{a_0(s)}{a_1(s)}ds+k_1]b(s)}{a_1(s)}ds+k_2]$$ <p style="text-align:right"> (11.17)
 * style="width:100%" |
 * style="width:100%" |
 * }

Part 3:
Substitute $$ a_1(x)=x^2+1, a_0(x)=x^3, b(x)=x^4 $$ into Equation(11.17),we get
 * {| style="width:100%" border="0"

$$ y(x)=\frac{1}{exp[\int^{x}\frac{s^3}{s^2+1}ds+k_1]}[\int^x\frac{exp[\int^{x}\frac{s^3}{s^2+1}ds+k_1]x^4}{x^2+1}ds+k_2]$$ <p style="text-align:right"> (11.18)
 * style="width:100%" |
 * style="width:100%" |
 * }

=Contribution Table=

= References =