User:Egm6321.f11.team1/report2

= R*2.1 Verify the solution of the specific Legendre differential equation=

Given
The Legendre differential equation is
 * $$\displaystyle L_2(y):=(1-x^2)y''-2xy'+n(n+1)y=0$$ .

For the specific case $$\displaystyle n=1$$ , the Legendre differential equation becomes
 * {| style="width:100%" border="0" align="left"

$$ The two homogeneous solutions are given:
 * $$\displaystyle (1-x^2)y''-2xy'+2y=0$$
 * $$\displaystyle (Eq.2.1.1)
 * $$\displaystyle (Eq.2.1.1)
 * }
 * }
 * $$\displaystyle y_H^1(x)=x$$ ;
 * $$\displaystyle y_H^2(x)=\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1$$ .

== Find == Verify that 
 * $$L_2(y_H^1)=L_2(y_H^2)=0$$ .

Solution
For $$y_H^1$$ ,  its first derivative is
 * $$\displaystyle \frac{dy_H^!}{dx}=1$$ ;

and its second derivative is 
 * $$\displaystyle \frac{d^2y_H^1}{dx^2}=0$$ .<br\>

Take them into $$\displaystyle (Eq.2.1.1)$$ ,<br\>
 * $$\displaystyle L_2(y_H^1)=(1-x^2)\cdot 0-2x\cdot 1+2\cdot x=0$$ .<br\>

For $$y_H^2$$ ,<br\> <br\> its first derivative is<br\>
 * $$\displaystyle \frac{dy_H^2}{dx}=\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{2}\cdot\frac{1-x}{1+x}\cdot\left(\frac{1+x}{1-x}\right)'$$ <br\>


 * $$\displaystyle =\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}$$ ;<br\>

<br\> and its second derivative is<br\>
 * $$\displaystyle \frac{d^2y_H^2}{dx^2}=\frac{1}{2}\cdot\frac{1-x}{1+x}\cdot\frac{2}{(1-x)^2}+\frac{1-x^2+2x^2}{(1-x^2)^2}$$ <br\>
 * $$\displaystyle =\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}$$ .<br\>

Take them into $$\displaystyle (Eq.2.1.1)$$ ,<br\>
 * $$\displaystyle L_2(y_H^2)=(1-x^2)\cdot \left[\frac{1}{1-x^2}+\frac{1+x^2}{(1-x^2)^2}\right]-2x\cdot \left[\frac{1}{2}log\left(\frac{1+x}{1-x}\right)+\frac{x}{1-x^2}\right]+2\cdot \left[\frac{x}{2}log\left(\frac{1+x}{1-x}\right)-1\right]$$ <br\>
 * $$\displaystyle =1+\frac{1+x^2}{1-x^2}-x\cdot log\left(\frac{1+x}{1-x}\right)-\frac{2x^2}{1-x^2}+x\cdot log\left(\frac{1+x}{1-x}\right)-2$$<br\>
 * $$\displaystyle =0$$ .<br\>

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by EGM6321.f11.team1.pan

= R*2.2 Verify the solution of an equation= == Find == Verify that<br\>
 * $$\displaystyle p(x)=k_1e^{-x}+x-1$$ <br\>

is indeed the solution for <br\>
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$$
 * $$\displaystyle p'+p=x$$.
 * <p style="text-align:right;">$$\displaystyle (Eq.2.2.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.2.1)
 * }
 * }

Solution
Derivate of $$\displaystyle p(x)=k_1e^{-x}+x-1$$ is <br\>
 * $$\displaystyle p'(x)=-k_1e^{-x}+1$$ .<br\>

Substituting its into $$\displaystyle (Eq.2.2.1)$$ ,<br\>
 * $$\displaystyle p'(x)+p(x)=-k_1e^{-x}+1+k_1e^{-x}+x-1=x$$ .<br\>

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by EGM6321.f11.team1.pan

= R*2.3 Show that the equation is linear in $${y}'$$ but an N1-ODE in general = Show that (2)p.7-6 is linear in y', and that (2) is in general an N1-ODE. But (2） is not the most general N1-ODE as represented by (1)p.7-6. Give an example of a more general N1-ODE.

Given
General N1-ODEs:
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$$  \displaystyle G({y}',y,x) =0 $$     (3-1) (1)p.7-6 Particular class of N1-ODEs: Linear in y'
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$$  \displaystyle M(x,y)+N(x,y){y}' =0 $$     (3-2) (2)p.7-6
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== Find == 1)Show that (2)p.7-6 is linear in y' 2)that (2) is in general an N1-ODE. 3)Give an example of a more general N1-ODE.

1) Show that (2)p.7-6 is linear in y'
For a operator to be linear, it has to satisfy the following:
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$$  \displaystyle F(\alpha u+\beta v)=\alpha F(u)+\beta F(v) ,\forall \alpha ,\beta \in \mathbb{R} $$     (3-3) Then we begin to check, Let Y = y',thus
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$$  \displaystyle G(x,y,Y)=\underbrace{M(x,y)}_{(1)}+\underbrace{N(x,y)Y}_{(2)} $$     (3-4) for (1)  We can regard this as a constant M; for (2)
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$$  \displaystyle F(x,y,Y)=N(x,y)Y $$     (3-5)
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$$  \displaystyle F(x,y,\alpha u+\beta v)=N(x,y)(\alpha u+\beta v)=\alpha N(x,y)u+\beta N(x,y)v $$     (3-6)
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$$  \displaystyle \alpha F(x,y,u)+\beta F(x,y,v)=\alpha N(x,y)u+\beta N(x,y)v $$     (3-7) So, Formula 3-6 is equal to 3-7, i.e.:
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$$  \displaystyle F(x,y,\alpha u+\beta v)=\alpha F(x,y,u)+\beta F(x,y,v) $$     (3-8) Then we can say that 3-2 i.e. (2)p.7-6 is linear in Y = y'
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2) Show that (2) is in general an N1-ODE
For a operator to be linear, it has to satisfy the following:
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$$  \displaystyle f(\alpha y_1+\beta y_2)=\alpha f(y_1)+\beta f(y_2) ,\forall \alpha ,\beta \in \mathbb{R} $$ Then we begin to check, So,
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$$  \displaystyle f(x,\alpha y_1+\beta y_2)\neq \alpha f(x,y_1)+\beta f(x,y_2) $$     (3-12) CONCLUSION: (2) is in general an N1-ODE
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3) Give an example of a more general N1-ODE
(2） is not the most general N1-ODE as represented by (1)p.7-6.
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$$  \displaystyle G(y',y,x)=0 $$     (3-13) A more general N1-ODE example can be:
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$$  \displaystyle M(x,y)+N(x,y)(y')^n=0 $$     (3-14) n is the power of y' This problem is solved by ourselves Contributed by Zexi
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= R*2.4 Verify that $$(4x^7+\sin y)+(x^2y^3)y'=0$$ is a N1-ODE =

Given

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$$  \displaystyle (4x^7+\sin y)+(x^2y^3)y'=0 $$     (4-1) == Find == Formula 4-1 is a N1-ODE
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Solution
1) The highest order of derivative is 1 and there is only one dependent variable in equation so it's a X1-ODE. (X can be L or N) 2) Verify Non-linearity For a operator to be linear, it has to satisfy the following:
 * {| style="width:100%" border="0"

$$  \displaystyle F(\alpha u+\beta v)=\alpha F(u)+\beta F(v) ,\forall \alpha ,\beta \in \mathbb{R} $$     (4-2) Then we begin to check,
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$$  \displaystyle F(x,y)=(4x^7+\sin y)+(x^2y^3)y' $$     (4-3)
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$$  \displaystyle F(x,{\alpha}{y_1}+{\beta}{y_2})=4x^7 + sin({\alpha}{y_1}+{\beta}{y_2}) +{x^2}({\alpha}{y_1}+{\beta}{y_2})^3({\alpha}{y_1}+{\beta}{y_2})' $$     (4-4)
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$$  \displaystyle {\alpha}F(x,{y_1})+{\beta}F(x,{y_2})={\alpha}(4x^7+sin{y_1}+x^2{y_1}^3{y_1}')+{\beta}(4x^7+sin{y_2}+x^2{y_2}^3{y_2}') $$     (4-5)
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$$  \displaystyle F(x,{\alpha}{y_1}+{\beta}{y_2})\neq {\alpha}F(x,{y_1})+{\beta}F(x,{y_2}) $$     (4-6) So, it's nonlinear. CONCLUSION: The equation is a N1-ODE This problem is solved by ourselves Contributed by Zexi
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= R*2.5 Conversion to Particular N1-ODE =

== Find ==

Explain the following equations:


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

$$
 * $$\bar M(x,y)+\bar N(x,y)(y')^3=0$$
 * <p style="text-align:right;">$$\displaystyle (Equation\;2.5.1)
 * <p style="text-align:right;">$$\displaystyle (Equation\;2.5.1)
 * }
 * }


 * $$M(x,y):=\left [\bar M(x,y)\right]^{1/3}$$


 * $$N(x,y):=\left [\bar N(x,y)\right]^{1/3}$$

Basic Equations
General N1-ODE form:


 * $$G\left(y',y,x\right)=0$$

Particular N1-ODE form:


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

$$M\left(x,y\right)+N(x,y)y'=0$$
 * <p style="text-align:right;">$$\,(Equation\;2.5.2) $$
 * style= |
 * }

Solution
The relation among the three equations we are to explain is illustrated by the ability to convert the first from a GENERAL form N1-ODE to a PARTICULAR form N1-ODE. The following steps show how redefining M and N as given will convert one type of equation into another.

Use definition of M and N and take cube power of each side.

$$ \left[M(x,y)\right]^{1/3}=\bar M(x,y) $$

$$ \left[N(x,y)\right]^{1/3}=\bar N(x,y) $$

Enter this result into Equation 2.5.1.

$$\left[M(x,y)\right]^3+\left[N(x,y)\right]^3(y')^3=0$$

Subtract one term to the right side.

$$\left[M(x,y)\right]^3=-\left[N(x,y)\right]^3(y')^3$$

$$\left[M(x,y)\right]^3=\left[-\left[N(x,y)\right](y')\right]^3$$

Take the cube root and add to the left side.

$$M(x,y)=-\left[N(x,y)\right](y')$$

$$M\left(x,y\right)+N(x,y)(y')=0$$

This now matches Equation 2.5.2. We can see that we were able to convert an apparent general form N1-ODE to a particular form N1-ODE.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by Egm6321.f11.team1.colsonfe 02:31, 17 September 2011 (UTC)

= R*2.6 Linear Independence of Homogeneous Solutions =

Given
Two homogeneous solutions:


 * $$y^1_H(x)=x$$


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

== Find ==

Show that $$y^1_H(x)$$ and $$y^2_H(x)$$ are linearly independent.

Plot $$y^1_H(x)$$ and $$y^2_H(x)$$.

Basic Equations
Linear Independence:


 * $$\forall\alpha\in\mathbb R, y^1_H(\cdot)\ne\alpha y^2_H(\cdot)$$

i.e, for any given $$\alpha$$, show that


 * $$\exists\hat x$$ such that $$y^1_H(\hat x) \ne\alpha y^2_H(\hat x)$$

Solution
By definition of linear independance, we merely need to show there is a value $$\hat x$$ where $$y^1_H(\hat x) \ne\alpha y^2_H(\hat x)$$ for some $$\displaystyle\alpha$$.

Let us then choose $$\hat x=0$$.Then

$$y^1_H(\hat x)=\hat x=0$$

$$y^2_H(\hat x)=\frac{\hat x}{2}\log\left(\frac{1+\hat x}{1-\hat x}\right)-1=-1$$

Now lets consider any given $$\displaystyle\alpha$$ for our chosen $$\hat x=0$$,

$$y^1_H(\hat x) \overset{?}{=}\alpha y^2_H(\hat x)$$

$$0 \overset{?}{=}-\alpha$$

It is clear that the above equality IS NOT true for all $$\displaystyle\alpha$$.

Therefore, for $$\hat x=0$$ and $$\alpha=1$$, $$y^1_H(\hat x) \ne\alpha y^2_H(\hat x)$$ for some $$\displaystyle\alpha$$, which means that $$y^1_H(x)$$ and $$y^2_H(x)$$ are linearly independent.


 * HW2_6plots.gif

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by Egm6321.f11.team1.colsonfe 02:31, 17 September 2011 (UTC)

= R*2.7 Find G(y',y,x)=0, and verify it is an N1-ODE=

Given
Function $$\displaystyle \phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3t^2)=k $$ .<br\>

== Find == Find $$\displaystyle G(y',y,x)=\frac{d}{dx}\phi(x,y)=0$$ .<br\> And show that this is an N1-ODE.

Solution
According to the Chain Rule, <br\>
 * $$\displaystyle \frac{d\phi(x,y)}{dx}=\frac{\partial\phi(x,y)}{\partial x}+\frac{\partial\phi(x,y)}{\partial y}\frac{dy}{dx}$$ <br\>
 * $$\displaystyle =2xy^{\frac{3}{2}}+3x^2y^2\cdot\frac{1}{x^3y^2}+y'\left(\frac{3}{2}x^2y^{\frac{1}{2}}+2x^3y\frac{1}{x^3y^2}\right)$$<br\>
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle =2xy^{\frac{3}{2}}+\frac{3}{x}+\left(\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y}\right)y'$$
 * <p style="text-align:right;">$$\displaystyle (Eq.2.7.1)
 * <p style="text-align:right;">$$\displaystyle (Eq.2.7.1)
 * }
 * }

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

Let <br\>
 * $$\displaystyle M(x,y)=2xy^{\frac{3}{2}}+\frac{3}{x}$$ ;<br\>
 * $$\displaystyle N(x,y)=\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{y} $$ .<br\>

Therefore, $$\displaystyle (Eq.2.7.1)$$ can be shown as <br\>
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$ ,

which is the particular class of N1-ODE (linear in $$\displaystyle y'$$). <div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by EGM6321.f11.team1.pan

= R*2.8 First Exactness Condition =

Given

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$$M\left(x,y\right)\cos y'+N(x,y)\log y'=0$$
 * <p style="text-align:right;">$$\,(Equation\;2.8.1) $$
 * style= |
 * }

== Find ==

Does Equation 2.8.1 satisfy the first exactness condition?

Basic Equations
First exactness condition for N1-ODE


 * $$M\left(x,y\right)+N(x,y)y'=0$$

Solutions
Let us define

$$\bar M\left(x,y\right):=M(x,y)$$

$$\bar N\left(x,y\right):=N(x,y)$$

Then Equation 2.8.1 becomes

$$\bar M\left(x,y\right)\cos y'+\bar N\left(x,y\right)\log y'=0$$

$$\bar M\left(x,y\right)+\bar N\left(x,y\right)\frac{\log y'}{\cos y'}=0$$

$$\bar M\left(x,y\right)+\bar N\left(x,y\right)F(y')=0$$

According to notes, if the function $$F$$ has no explicit inverse, then it CANNOT be put in the form which satisfies the First Exactness Condition.

The function $$F(y')=\frac{\log y'}{\cos y'}$$ has no inverse $$F^{-1}\left(y'\right)$$, and so therefore this N1-ODE fails to satisfy the First Exactness Condition.

<div style="width: 80%; margin-left: auto; margin-right: auto; padding: 4px; border: 2px solid #FF0000; background-color: #FFDDDD; text-align: left;"> Solution by Egm6321.f11.team1.colsonfe 04:30, 17 September 2011 (UTC)

= R*2.9 State and proof the full theorem =

Given
Review calculus, and find the minimum degree of differentiability of the function such that (2) p.9-3 is satisfied. State the full theorem and provide a proof. p.9-3 (2): $$ \frac{\partial^2 \phi(x,y)}{\partial x \partial y}=\frac{\partial^2 \phi(x,y)}{\partial y \partial x} $$

== Find ==

See reference for full problem statement.

Clairaut's theorem
In mathematical analysis, Clairaut's theorem or Schwarz's theorem,[1] named after Alexis Clairaut and Hermann Schwarz, states that if $$f:\mathbb R \rightarrow \mathbb R$$ has continuous second partial derivatives at any given point in $$ \mathbb R^n $$,say $$(a_1,...,a_n)$$, then for $$ 1<i,$$  $$ j<n,$$

$$\frac{\partial^2 f}{\partial x_i \partial x_j}(a_1,...,a_n)=\frac{\partial^2 f}{\partial x_j \partial x_i}(a_1,...,a_n)$$

In words, the partial derivations of this function are commutative at that point. One easy way to establish this theorem (in the case where n = 2, i = 1, and j = 2, which readily entails the result in general) is by applying Green's theorem to the gradient of f.

http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives#Clairaut.27s_theorem

Solution
The minimum degree of differentiability of the function such that (2) p.9-3 is satisfied is 3.

As Clairaut's theorem mentions, functions that satisfy (2) p.9-3 must have continuous second derivative.

For instance,

$$ z=f(x,y)=x^2+y^2 $$

$$\frac{\partial^2 f(x,y)}{\partial x\partial y}=\frac{\partial^2 f(x,y)}{\partial y\partial x}=2$$

The function

$$ z=f(x,y)=x^2+y^2 $$

can only be differentialed by 3 degrees

$$ \frac{\partial^2 f(x,y)}{\partial^2 x}=0 $$

And functions with degree of differentiability less than 3 can not satisfy (2) p. 9-3

For example f(x,y)=x+y

So The minimum degree of differentiability of the function such that (2) p.9-3 is satisfied is 3.

= R*2.10 Verify the solution for the N1-ODE =

Given

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$$  \displaystyle y(x)=\sin ^{-1}(k-15x^5) $$     (10-1) (1) p.10-4
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$$  \displaystyle \frac{d\phi (x,y(x))}{dx}=75x^4+(\cos y)y'=0 $$     (10-2) (2) p.8-6
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== Find == Verify that 10-1 is the solution for the N1-ODE 10-2

Solution
We just need to put 10-1 into 10-2 and see if the equation is still in effect.
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$$  \displaystyle \frac{d\phi (x,y(x))}{dx}=75x^4+(\cos (\sin ^{-1}(k-15x^5)))(\sin^{-1}(k-15x^{5}))' $$     (10-3) We can easily verify that the equation 10-3 is equal to 0 with the help of Matlab. i.e.
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$$  \displaystyle 75x^4+(\cos (\sin ^{-1}(k-15x^5)))(\sin^{-1}(k-15x^{5}))'=0 $$     (10-4) Here are the codes:
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syms x syms k syms y y=asin(k-15*x^5); dy=diff(y,x); 75*x^4+cos(y)*dy

This problem is solved by ourselves

= R2.11 Explain Difficulty In Solving For Integrating Factor = Question from 11-1: Why is it usually difficult to solve for the integrating factor h(x,y) in the following equation? $$h_xN - h_yM + h(N_x - M_y) = 0$$ The solution requires an understanding of how the Euler Integrating Factor Method (IFM) is derived with the reduction of order method. The solution expands upon concepts introduced in meetings 7-11.

First, the general nonlinear ODE of order n without the missing function y(x) is:

$$G(y^{(n)},y^{(n-1)},...,y'',y',y,x) = 0 $$

If we remove y, then the general nonlinear ODE without the function is:

$$G(y^{(n)},y^{(n-1)},...,y'',y',y,x) = 0 $$

Reduce the order by substituting p(x) := y'(x) below

$$G(p^{(n-1)},p^{(n-2)},...,p',p,x) = 0$$

The substitution gives the following solution for y(x).

$$y(x) = \int p(x)dx + k$$

where k is the integration constant. The key is in the dummy integration variable derivation below.

$$\int_a^x p(z)dz = \int_a^x y'(z)dz = y(x) - y(a)$$

Rearrangment gives

$$y(x) = \int_a^x p(z)dz + y(a) = P(x) - P(a) + y(a) = P(x) + k$$

The derivation requires the astute reader to note the distinction beween capital "P" and small "p." In addition, k = -P(a) + y(a)

Now, the Euler IFM applies to G(y',y,x)=0, and for a particular class of N1 ordinary differential equations, a solution linear in y' is

M(x,y) + N(x,y)y' = 0 where

$$ y' = \frac{dy}{dx}$$

So a particular solution is M(x,y) + N(x,y)dy = 0

<h3 style="color:red">The derivations to this point are from meeting 7 A more general solution for f(y') of the particular case is

$$\bar{M}(x,y) + \bar{N}(x,y)F(y') = 0$$

for the nonlinear case. But the assumption is that F(*) has an inverse. This leads to the first exactness conditions of N1 ordinary differential equations.

First Exactness Condition of N1-ODEs

If an N1-ODE is exact then it has a particular form of

M(x,y) + N(x,y)y' = 0 where

$$ y' = \frac{dy}{dx}$$

<h3 style="color:red">The appended derivations to this point are from meeting 8 Second Exactness Condition of N1-ODEs

Now, we must derive the 2nd exactness condition. A general ND-ODE is exact if for G the following relations exist:

$$\frac{d\phi}{dx} = G$$

$$\phi(x,y) = k$$

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

Let $$\phi(x,y)$$ be a smooth function, then for the 2nd exactness condition we have

$$M_y(x,y) = \frac {\partial^2\phi(x,y)}{\partial x\partial y} = N_x(x,y) = \frac{\partial^2\phi(x,y)}{\partial y\partial x}$$ <h3 style="color:red">The appended derivations to this point are from meeting 9 If an N1-ODE satisfies the 1st exactness condition but not the 2nd, then the N1-ODE can be transformed by using the Euler Integrating Factor Method (IFM). Assume the conditions below.

$$M(x,y) + N(x,y)y' = 0 \And M_y(x,y) \ne N_x(x,y)$$

The transformation of the given N1-ODE to a form that fulfills both exactness conditions begins by finding an integrating factor h(x,y).

For condition 1, the substitution of h gives

$$(hM) + (hN)y' = \bar{M} + \bar{N}y' = 0$$

The key is to derive a solution using

$$\bar{M} = hM \And \bar{N} = hN$$

The derivations are as follows:

$$\bar{M}_y = h_yM + hM_y = \bar{N}_x = h_xN + hN_x$$

So the Euler IFM gives the following solution using h(x,y) as the transformation.

$$h_xN - h_yM + h(N_x - M_y) = 0 $$

<h3 style="color:red">The cumulative derivations to this point are from meetings 10-11 The solution h(x,y) is difficult to find since a separation of variables may not be possible for x and y. It is difficult to solve for h(x,y) in the equation when N(x,y) and M(x,y) are multivariate.

In addition, the following variables may have non-zero values, which complicates integrations to solve for h. The non-zero conditions complicate integration for h

$$h_x \ne 0$$ and $$h_y \ne 0$$

$$N_x \ne 0$$ and $$M_y \ne 0$$

So the solution of h is often cumbersome to solve because of the multivariate dependencies of h(x,y), M(x,y), and N(x,y) in the following equation.

$$\frac {h_x}{h} = -\frac{1}{N}(N_x - M_y)$$

The integration of this equation above is difficult with dependencies on both x and y and the non-zero conditions listed.

References for R2.11 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 7) 5 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 8) 6 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 9) 8 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 10) 8 Sep 2011.

Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) 13 Sep 2011.

= R*2.12 Find Integrating Factor = Question from 11-3 : Find h for the following equation where h is the Euler Integration Factor Method. $$\frac{h_y}{h} = \frac{1}{M}(N_x - M_y) =: m(y)$$

The solution provides a condition in which simplification allows easier integration for one variable y

$$h_x(x,y) = 0$$

The condition above simplifies the integral for the Euler Integration Factor Method and the transformation involving h(x,y).

$$h_xN - h_yM + h(N_x - M_y) = -h_y + h(N_x - M_y) = 0$$

Rearrangement gives

$$h_yM = h(N_x - M_y) = 0$$

$$\frac{h_y}{h} = \frac{1}{M}(N_x - M_y) =: m(y)$$

$$\int \frac{h_y}{h} = - \int \frac {1}{N}(N_x - M_y)$$

Substitution of m(s) for the right side of the equation gives

$$ \log h(y) = \int\limits^{y} m(s)ds + k$$

So the solution is

$$h(y) = \exp {[\int\limits^{y} m(s)ds + k]}$$

<h3 style="color:red">The derivations used notes from meeting 11 References for R2.12 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) 13 Sep 2011.

= R*2.13 Prove Given Equation = Question from 11-5 : If the following equation is given $$y' + \frac{1}{x} = x^2$$

then show that

$$h(x) = x \And y(x) = \frac{x^3}{4} + \frac{k}{x}$$

The equation does not fulfill the 2nd exactness condition, so an integration factor h can be derived by rearranging the equation.

$$ a_0(x)y + y' = M(x,y) + N(x,y)y' = b(x)$$

Recall that h is defined as follows:

$$If h_y(x,y) = 0 then$$

$$\frac{h_x}{h} = \frac{1}{M}(N_x - M_y) =: n(x) = +a_0(x)$$

This gives h(x) as

$$h(x) = \exp[\int\limits^{x} a_0(s)ds + k_1]$$

Recall that

$$a_0(x) = \frac{1}{x}$$

So that $$h(x) = \exp[\int\limits^{x} \frac{1}{s}] = \exp(\ln {x}) = x $$

Now that we have solved for h(x), the next step is to solve for y(x). The transformation with h(x) is done here.

$$h(y' + a_0y) = hb$$

$$hy' + h'y = hb = (hy)'$$

Integration (hy)' to get

$$\int\limits^{x}(hy)' = \int\limits^{x}h(s)b(s)ds + k_2$$

Divide both sides by h(x) to get y(x).

$$y(x) = \frac{1}{h(x)}[\int\limits^{x}h(s)b(s)ds + k_2]$$

Recall that the given equation gives b as shown below.

$$h(x) = x \And b(x) = x^2$$

So substitute the functions back into the integral for y(x).

$$y(x) = \frac{1}{x}[\int\limits^{x} x^2 ] = \frac{1}{x}[\frac{x^4}{4} + k]$$ Simplification gives the solution for y(x) below.

$$y(x) = \frac{x^3}{4} + \frac{k}{x}$$

References for R2.13 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) 13 Sep 2011.

= R*2.14 Solve the general L1-ODE-VC=

== Part 1 == Solve the general L1-ODE-VC

Given
$$ a_1(x)\,y'+a_0(x)\,y=b(x)$$ <p style="text-align:right;">$$\displaystyle (Equation\;14.1) $$

where

$$ a_1(x)=1,

a_0(x)=x,

b(x)=2x+3 $$

Solution
Substituting the conditions into 14.1

$$ y'+xy=2x+3$$ <p style="text-align:right;">$$\displaystyle (Equation\;14.2) $$

Equation 14.2 is of the form

$$ y'+P(x)y=Q(x)$$

The integrating factor for this form is

$$ I(x)=\exp[P(x)dx]$$

For equation 14.2, the integrating factor is

$$ I(x)=\exp[\int xdx]=\exp[\frac{x^2}{2}] $$

Applying the integrating factor for equation 14.2:

$$ \exp[\frac{x^2}{2}]\frac{dy}{dx}+\exp[\frac{x^2}{2}]xy=2x\exp[\frac{x^2}{2}]+3\exp[\frac{x^2}{2}] $$ <p style="text-align:right;">$$\displaystyle (Equation\;14.3) $$

(Eq.14.3) reduces to

$$ \frac{d}{dx}(y\exp(\frac{x^2}{2}))=2x\exp[\frac{x^2}{2}]+3\exp[\frac{x^2}{2}] $$

Integrating

$$ y\exp(\frac{x^2}{2})=\int 2x\exp[\frac{x^2}{2}]dx+\int3\exp[\frac{x^2}{2}]dx $$

Results in the following equation:

$$ y\exp(\frac{x^2}{2})=2\exp[\frac{x^2}{2}]+\int 3\exp [\frac{x^2}{2}]dx $$

There is no known anti-derivative of the last term, so using mathematical analysis tools the last therm results in the error-function. So that rearranging the equation we get for y(x):

$$ y=2+3\exp[\frac{x^2}{2}][\sqrt{\frac{\pi}{2}}erfi(\frac{x}{\sqrt2})] $$

==Part 2 == A general solution from the integrating factor method

Given
Assume $$ a(x)\ne0 $$ for all x, rearranging equationg 14.1 gives $$ y'+\frac{a_0(x)}{a_1(x)}y=\frac{b(x)}{a_1(x)} $$ <p style="text-align:right;">$$\displaystyle (Equation\;14.4) $$

Find
Find an expression for $$ y(x) $$ in terms of $$ a_0, a_1 $$ and b

Solution
Define $$P(x)=\frac{a_0(x)}{a_1(x)}$$ and $$Q(x)=\frac{b(x)}{a_1(x)}$$

Equation 14.4 is then in the form

$$y'+P(x)y=Q(x)$$

The integrating factor for this form is

$$I(x)=\exp[\int{P(x)}dx]$$

Introducing a dummy factor t and substituting in for P(x)

$$\exp[\int^x{\frac{a_0(t)}{a_1(t)}}dt]\frac{dy}{dx}+\exp[\int^x{\frac{a_0(t)}{a_1(t)}}dt]\frac{a_0(x)}{a_1(x)}y=\frac{b(x)}{a_1(x)}\exp[\int^x{\frac{a_0(t)}{a_1(t)}}dt] $$

Reducing

$$\frac{d}{dx}[\exp[\int^x\frac{a_0(t)}{a_1(t)}]y]=\frac{b(x)}{a_1(x)}\exp[\int^x\frac{a_0(t)}{a_1(t)}dt]$$

Integrating

$$y=A\exp[-\int\frac{a_0(t)}{a_1(t)}dt]+\exp[-\int^x\frac{a_0(t)}{a_1(t)}dt]\int^x\frac{b(s)}{a_1(s)}ds\exp[-\int^s\frac{a_0(t)}{a_1(t)}dt]ds$$

==Part 3 == Solution to linear, first order, ordinary differential equation with varying coefficients using Euler integrating factor method.

Solution
Given an equation in the form of 14.1, where

$$a_1(x)=x^2+1, a_0(x)=x,$$ and $$b(x)=2x,$$

Equation 14.1 becomes

$$(x^2+1)y'+xy=2x$$

Rearranging

$$y'+\frac{x}{x^2+1}y=\frac{2x}{x^2+1}$$<p style="text-align:right;">$$\displaystyle (Equation\;14.5) $$

Equation 14.5 is in the form

$$y'+P(x)y=Q(x)$$|| <p style="text-align:right;">$$\displaystyle (Equation\;14.6) $$

The integrating for this form is

$$I(x)=\exp[\int{\frac{x}{x^2+1}}dx]=\exp[\frac{1}{2}\ln(x^2+1)]=\sqrt {x^2+1} $$ <p style="text-align:right;">$$\displaystyle (Equation\;14.7) $$

Multiply 14.6 by 14.7 and simplify

$$\frac{d}{dx}(y\sqrt {x^2+1})=\frac{2y}{\sqrt {x^2+1}}$$ <p style="text-align:right;">$$\displaystyle (Equation\;14.8) $$

Integrating

$$\int^x\frac{d}{dx}(y\sqrt {x^2+1})=\int^x \frac{2y}{\sqrt {x^2+1}}dy$$<p style="text-align:right;">$$\displaystyle (Equation\;14.9) $$

Gives

$$Y=2$$

=R*2.15 Show the necessity of constant k1 and k2=

Given
Since (2)-(3) p.11-3 is an L1-ODE-VC, there should be only one integration constant, not two.

k1 is contained in (3) p11.4

$$ h(x)= \exp[ \int a_0(s)ds +k_1]$$

k2 is contained in (1) p11.5

$$y(x)=\frac{1}{h(x)}[\int^xh(x)b(s)ds+k_2]$$

== Find ==

Show that the integration constant k1 in (3 ) p.11-4 is not necessary, i.e, only k2 in (1) p.11-5 is necessary.

Solution
To show this we will use (6)p10.3 and show that k1 cancels out of the equation, but we must find the contants k1 and k2 from equations(eq8.1) and (eq8.2)

$$y(x)=\frac{1}{h(x)}\int^xh(s)b(s)ds$$

When we perform the integration in the RHS of (eq8.1) it becomes

$$\exp[\int^x a_0(s)ds]=\exp[\int a_0(x)dx+k]$$

Using the properties for exponential functions

$$\exp[a+b]=\exp[a]exp[b]$$

Now (eq8.3) becomes

$$\exp[\int a_0(x)dx]\underbrace{\exp[k]}_{=:k_1}$$

Plugging this back into (eq8.1) gives us

$$h(x)= k_1\exp[\int a_0(x)dx]$$

To find k2 we will evaluate the integral in (eq8.2)

$$\int^x h(s)b(s)ds=\int h(x)b(x)dx+k_2$$

Finally plug equations (eq8.5) and (eq8.6) into (6) p.10-3

$$y(x)=\frac{1}{\cancelto{}{k_1}\exp[\int a_0(x)dx]}[\int \cancelto{}{k_1}\exp[\int a_0(x)dx]b(x)dx+k_2]$$

This shows k1 is not necessary

$$y(x)=\frac{1}{\exp[\int a_0(x)dx]}[\int \exp[\int a_0(x)dx]b(x)dx+k_2]$$

=R*2.16 LVQ Solution compared to King =

Given
The solution of

$$\displaystyle a_1(x) \cdot y^{(1)}+a_0(x)y^{(0)}=b(x)$$  <p style="text-align:right;">$$\displaystyle (3) p. 11-3 $$

is

$$\displaystyle y(x) = \frac{1}{h(x)} \left[\int^x h(s)b(s)ds + k_2 \right] $$ <p style="text-align:right;">$$\displaystyle (1) p. 11-5 $$

From lecture notes [[media:pea1.f11.mtg11.djvu|Mtg 11 (b)]]

== Find ==

Show that $$\displaystyle (1) P. 11-5 $$ agrees with $$\displaystyle King p. 512 $$ shown below


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

y = A{y}_{H}\left(x \right)+{y}_{P}\left(x \right) $$ $$
 * <p style="text-align:right;">$$\displaystyle (2.16.1)
 * style= |
 * }

Solution
Start by using the integration factor defined by $$\displaystyle (3) p. 11-4$$

$$ h(x) = \exp \left[ \int^{x} {a}_{0} \left( s\right)ds + k_1 \right]$$ <p style="text-align:right;">$$\displaystyle (3) p. 11-4 $$

Using the properties for exponential functions $$ \exp \left[ a+b\right] = \exp \left[ a\right]\exp \left[b\right]$$

Now $$\displaystyle (3) p. 11-4$$ becomes


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

\exp \left[ \int {a}_{0} \left( x\right)dx \right]\underbrace{\exp \left[k\right]}_{=: k_1} $$ $$
 * <p style="text-align:right;">$$\displaystyle (2.16.2)
 * style= |
 * }

Plugging this back into $$\displaystyle (3) p. 11-4 $$ gives us

$$ h \left( x\right)= k_{1}\exp \left[ \int {a}_{0} \left( x\right)dx\right] $$ <p style="text-align:right;">$$\displaystyle (Eq 2.16.3) $$

To find $$ {k}_{2}$$ we will evaluate the integral in $$\displaystyle (1) p. 11-5$$

$$\int^{x} h\left( s\right)b\left( s\right)ds= \int h\left( x\right)b\left( x\right)dx + k_{2}$$ <p style="text-align:right;">$$\displaystyle (Eq 2.16.4) $$

Next plug equations $$\displaystyle Eq 2.16.3 $$ and $$\displaystyle Eq 2.16.4 $$ into $$\displaystyle (1) p. 11-5 $$ and canceling $$\displaystyle k_1 $$

$$y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]}\left[ \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\right]$$ <p style="text-align:right;">$$\displaystyle (Eq 2.16.5) $$

From $$\displaystyle K p. 512 $$ the equation for the solution of the homogeneous differential equation and the particular integral solution of the inhomogeneous differential equation are given below, respectively


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

A{y}_{H}\left(x \right)= A\exp \left\{ - \int ^{x}{P} \left( t\right)dt\right\} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 2.16.6)
 * style= |
 * }


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

{y}_{P}\left(x \right)= \exp \left\{ - \int ^{x}{P} \left( t\right)dt\right\}\int ^{x}{Q} \left( s\right)\exp \left\{  \int ^{x}{P} \left(t\right)dt\right\}ds $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 2.16.7)
 * style= |
 * }

Now, using $$\displaystyle (Eq 2.16.5) $$ which is the expanded form of $$\displaystyle (3) p. 11-4 $$ we can further expand and rearrange to get


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

y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]} \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +\frac{k_2}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 2.16.8)
 * style= |
 * }


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

y \left( x\right)= \exp \left[  -\int {a}_{0} \left( x\right)dx\right] \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\exp \left[  -\int {a}_{0} \left( x\right)dx\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 2.16.9)
 * style= |
 * }

Now we have $$\displaystyle (3) p. 11-4 $$ in the form of $$\displaystyle (Eq 2.16.1)$$ where


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

A{y}_{H}\left(x \right)= k_2\exp \left[ -\int {a}_{0} \left( x\right)dx\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 2.16.10)
 * style= |
 * }


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

{y}_{P}\left(x \right)= \exp \left[ -\int {a}_{0} \left( x\right)dx\right] \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 2.16.11)
 * style= |
 * }

By comparing $$\displaystyle (Eq 2.16.6) $$ with $$\displaystyle (Eq 2.16.10)$$ and $$\displaystyle (Eq 2.16.7) $$ with $$\displaystyle (Eq 2.16.11) $$ we can see that these equations are in agreement

= R*2.17 Redo the equation for the solution of the homogeneous differential equation done R*2.16 using (2) p. 12-2 =

Given

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

$$ From lecture notes [[media:pea1.f11.mtg11.djvu|Mtg 11 (b)]]
 * $$\displaystyle {{y}^{'}}+{{a}_{0}(x)}y=0$$
 * <p style="text-align:right;">$$\displaystyle (2) p. 12-2
 * <p style="text-align:right;">$$\displaystyle (2) p. 12-2
 * }
 * }

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


 * Find $$\displaystyle y_H$$
 * style= |
 * }
 * }

Solution
Since $$\displaystyle (2) p. 12-2 $$ is the homogeneous portion of

$$\displaystyle a_1(x) \cdot y^{(1)}+a_0(x)y^{(0)}=b(x)$$ with $$\displaystyle  a_1(x)=1$$ <p style="text-align:right;">$$\displaystyle (3) p. 11-3 $$

Equation $$\displaystyle (2) p. 12-2 $$  is seperable, So we can solve directly for $$ y_H$$

Equation $$\displaystyle (2) p. 12-2 $$ can be rearranged as follows

<p style="text-align:left;"> $$\displaystyle \frac{y'}{y} = -{a}_{0} $$ <p style="text-align:right;">$$\displaystyle (Eq 2.17.1) $$

Integration of both sides of this equation gives

<p style="text-align:left;"> $$ \ln \left( y\right)= -\int^{x} {a}_{0} \left( s\right)ds + C$$   <p style="text-align:right;">$$\displaystyle (Eq 2.17.2)$$

Solving for $$y$$ gives

$$ y= \exp\left[ -\int^{x} {a}_{0} \left( s\right)ds + C \right]= A\exp\left[ -\int^{x} {a}_{0} \left( s\right)ds\right]= y_H $$<p style="text-align:right;">$$\displaystyle (Eq 2.17.3) $$

= R*2.18 - Find the integrating factors to make L1-ODE-VC exact =

Given
$$\displaystyle \left[ x^4 y + 10 \right] + \frac{1}{2} x^2 y' = 0 $$ <p style="text-align:right;">$$\displaystyle (4) p. 12.4 $$

From lecture notes [[media:pea1.f11.mtg12.djvu|Mtg 12]]

== Find ==

Determine if $$\displaystyle (4) p. 12-4 $$ is exact, if it is not find the integrating factor $$\displaystyle h $$ to make it exact

Solution
The first exactness condition for an N1-ODE is

$$\displaystyle M(x,y) + N(x,y) y'= 0$$ <p style="text-align:right;">$$\displaystyle (2) p. 7-6 $$

Since the equation is already in this form the first exactness condition is fulfilled

The second exactness condition requires that

$$\displaystyle M_y(x,y)=N_y(x,y)$$ <p style="text-align:right;">$$\displaystyle (3) p. 9-3 $$

Where:

$$\displaystyle M(x,y) = \left[x^4y+10\right]$$ <p style="text-align:right;">$$\displaystyle (2.18.1) $$

And:

$$\displaystyle N(x,y) = \left (\frac{1}{2}x^2\right)$$ <p style="text-align:right;">$$\displaystyle (2.18.2) $$

Taking the derivatives:

$$\displaystyle M_y(x,y)=x^4$$ <p style="text-align:right;">$$\displaystyle (2.18.3) $$

$$\displaystyle N_x(x,y)= x$$ <p style="text-align:right;">$$\displaystyle (2.18.4) $$

Since $$\displaystyle M_y(x,y) \ \ne N_y(x,y) $$ <p style="text-align:right;">$$\displaystyle (2.18.5) $$

Therefore equation $$\displaystyle (4) p. 12-4 $$ is not exact

So we re-write the ODE


 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at the first exactness condition for the integrating factor we obtain:

$$\displaystyle \frac{1}{N}\left(N_x - M_y\right) = -f(x) $$<p style="text-align:right;">$$\displaystyle (2.18.6) $$ replacing values found above $$\displaystyle \frac{2}{x^2}\left(x - x^4\right) = -f(x) $$<p style="text-align:right;">$$\displaystyle (2.18.7) $$ applying knowledge from integrating factor we know $$\displaystyle h(x)=\exp \int^{x}f(s)ds $$<p style="text-align:right;">$$\displaystyle (2.18.8) $$ replacing f(x) into the integral and evaluating it, obtain $$\displaystyle h(x) = \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right)$$<p style="text-align:right;">$$\displaystyle (2.18.9) $$

Lets verify that the integrating factor has made the ODE exact by checking the second condition of exactness $$\displaystyle \bar{M}_y = \bar{N}_x $$<p style="text-align:right;">$$\displaystyle (2.18.10) $$ $$\displaystyle\underbrace{(hM)}_{\bar{M}}dx = \underbrace{(hN)}_{\bar{N}}dy$$ <p style="text-align:right;">$$\displaystyle (2.18.11) $$

Calculating the respective values we obtain

$$\displaystyle \bar{N}_x = x^2 \exp\left(\frac{2x^3}{3}\right)$$<p style="text-align:right;">$$\displaystyle (2.18.12) $$ $$\displaystyle \bar{M}_y =x^2 \exp\left(\frac{2x^3}{3}\right)$$<p style="text-align:right;">$$\displaystyle (2.18.13) $$

Since $$\displaystyle (2.18.10)$$ is satisfied(i.e. $$\displaystyle ((2.18.12) = (2.19.13))$$ the non-homogenous L1_ODE_VC has been made exact by using the appropriate integrating factor $$\displaystyle h $$.

= References =

= Table of Assignments =

{| class="prettytable" style="margin: 1em auto 1em auto;"
 * colspan="2" | Problem Assignments


 * Problem #
 * Solved by
 * Solved by


 * 1
 * Jing Pan
 * Jing Pan


 * 2
 * Jing Pan
 * Jing Pan


 * 3
 * Zexi Zheng
 * Zexi Zheng


 * 4
 * Zexi Zheng
 * Zexi Zheng


 * 5
 * Fenner Colson
 * Fenner Colson


 * 6
 * Fenner Colson
 * Fenner Colson


 * 7
 * Jing Pan
 * Jing Pan


 * 8
 * Fenner Colson
 * Fenner Colson


 * 9
 * Yi Zhao
 * Yi Zhao


 * 10
 * Zexi Zheng
 * Zexi Zheng


 * 11
 * Manuel Steele
 * Manuel Steele


 * 12
 * Manuel Steele
 * Manuel Steele


 * 13
 * Manuel Steele
 * Manuel Steele


 * 14
 * Yi Zhao
 * Yi Zhao


 * 15
 * Yi Zhao
 * Yi Zhao


 * 16
 * Ben Neri
 * Ben Neri


 * 17
 * Ben Neri
 * Ben Neri


 * 18
 * Ben Neri
 * Ben Neri


 * -}