User:Egm6321.f12.team2/Report4

=Introduction=

Summary
In this report, the topics of exactness in linear and non-linear second order differential equations are explored, both in direct and general examples. If an equation is not exact, it is solved using an exponent integration factor.

Team Members
Zhe Wang Yitien Yan Ranjit Alexander Karl Van Wyk Anubhav Mohan

=R*4.1 Exactness of L2-ODE-VC=

Verify that
is an exact L2-ODE-VC.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

First Exactness Condition
The first exactness condition is as follows, where $$p=y'$$

To verify for exactness, ($$) can be written as follows, so ($$) meets the first exactness condition for N2-ODE's.

Second Exactness Condition
The second exactness condition is as follows, and From the above two equations we can see that, {| style="width:100%" border="0" $$f_{xx} = \frac {\partial (\sqrt x)}{\partial x \partial x} = -\frac{1}{4 x^{3/2}}$$ $$f_{xy} = \frac {\partial (\sqrt x)}{\partial x \partial y} = 0$$ $$f_{yy} = \frac {\partial (\sqrt x)}{\partial y \partial y} = 0$$ $$f_{xp} = \frac {\partial (\sqrt x)}{\partial x \partial p} = 0$$ $$f_{yp} = \frac {\partial (\sqrt x)}{\partial y \partial p} = 0$$ $$f_{y} = \frac {\partial (\sqrt x)}{\partial y} = 0$$ $$g_{xp} = \frac {\partial (2xp+3y)}{\partial x \partial p} = 2$$ $$g_{yp} = \frac {\partial (2xp+3y)}{\partial y \partial p} = 0$$ $$g_{y} = \frac {\partial (2xp+3y)}{\partial y} = 3$$ $$g_{pp} = \frac {\partial (2xp+3y)}{\partial p \partial p} = 0$$
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Substituting the above values into equations ($$) and($$), we get, The L2-ODE-VC does not meet the first rule of the second exactness condition, so the equation is not exact.

Given
The following N2-ODE,

Find m,n
Find $$m,n \, \in \mathbb{R}$$ such that equation ($$) is exact.

Find L1-ODE-VC
Additionally, show that the first integral is a L1-ODE-VC of the form, with $$p(x):= y'(x)$$.

Find y(x)
Solve ($$) for $$y(x)$$.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Find m,n
In order to find $$m,n$$, we recall the second exactness condition for N2-ODEs:

where,

Thus, we have two equations ($$) and ($$), and two unknowns $$m,n$$. Start by finding every element of the 2nd exactness condition equations. Derivation of $$f_{xx}$$ Derivation of $$2pf_{xy}$$ Derivation of $$p^2f_{yy}$$ Derivation of $$g_{xp}$$ Derivation of $$pg_{yp}$$ Derivation of $$-g_y$$ Derivation of $$f_{xp}$$ Derivation of $$pf_{yp}$$ Derivation of $$2f_y$$ Derivation of $$g_{pp}$$ Inject the terms in equations ($$)-($$) into ($$) in order to solve for n. Next, combine the terms in equations ($$)-($$) and ($$), and inject into equation($$) to solve for m. For simplicity, choose $$x=1$$. Therefore, ($$) can be solved for m, Check which value of $$m$$ is still valid when substituting into ($$), When $$m=\frac{3}{2}$$, Therefore, $$m=\frac{3}{2}$$ is not valid for all $$x$$. Now check if ($$) is valid for all $$x$$ when $$m=\frac{1}{2}$$, Therefore, $$m=\frac{1}{2}$$ is valid for all $$x$$.

Find L1-ODE-VC
Substituting $$m=\frac{1}{2},n=0$$ into equation ($$), the following is obtained, Recalling the form of a second order ODE, From equations ($$) and ($$), it is seen that, We start by solving for $$\phi_p$$ where $$p=y'$$, Next,taking the total derivative with respect to x, Set ($$) and ($$) equal to each other yields, In order to find h(x,y), the most obvious case is chosen first where, From ($$), Taking the partial derivative with respect to x of ($$) and setting it equal to ($$) yields, Therefore, combining ($$),($$) and ($$) yields $$\phi$$,

Find y(x)
In order to solve for y(x), we determine whether or not equation ($$) is exact. First exactness condition, Thus, equation ($$) satisfies the first exactness condition. Second exactness condition, Thus, the second exactness condition is not satisfied and equation ($$) is not exact. First, we rewrite equation ($$) as, We find an integrating factor, $$h(x,y)$$ that will make the equation ($$) exact. Multiplying equation ($$) through by the integrating factor, and using the fact that $$h'=h\frac{2x^\frac{3}{2}-1}{x}$$, the following is obtained,

Given
The form of general N2-ODE is The ($$) is considered as follows:

A generated class of exact L2-ODE-VC is

Find
Show that ($$) and ($$) leads to ($$).

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

From ($$) and ($$), we can obtain

Integrating the both sides of ($$), the following is obtained.

Using ($$),

where

Integrating the term of R(x)y in ($$),

With ($$), ($$) can be rewritten as follows:

To find k(y), $$\displaystyle \phi_y $$ is found and compared to Q(x) because $$\displaystyle \phi_y = Q(x) $$ in ($$).

Frrom ($$), we can find the following:

Therefore, k(y) is verified with constant k and ($$) is resulted in

Given
The following N2-ODE,

Prove exactness
Prove that equation ($$) is exact.

Find L1-ODE-VC
Additionally, Find the first integral, $$\phi(x,y,p)$$ which is a L1-ODE-VC with $$p(x):= y'(x)$$.

Find y(x)
Solve ($$) for $$y(x)$$.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.

Proof of exactness
The second exactness condition for N2-ODEs:

where,

Start by finding every element of the 2nd exactness condition equations. Derivation of $$f_{xx}$$ Note here that For $$g_{y}$$ For $$g_{xp}$$

Inject the terms in equations ($$)-($$) into ($$) and ($$) to test for exactness gives Where both the statements are true. Hence the equation is exact.

Find L1-ODE-VC
Recalling the form of a second order ODE, From equations ($$) and ($$), it is seen that, We start by solving for $$\phi_p$$ where $$p=y'$$, Next,taking the total derivative with respect to x, Set ($$) and ($$) equal to each other yields, In order to find h(x,y), the most obvious case is chosen first where, From ($$), Taking the partial derivative with respect to x of ($$) and setting it equal to ($$) yields, Therefore, combining ($$),($$) and ($$) yields $$\phi$$,

Find y(x)
In order to solve for y(x), we determine whether or not equation ($$) is exact. First exactness condition, Thus, equation ($$) satisfies the first exactness condition. Second exactness condition, Thus, the second exactness condition is not satisfied and equation ($$) is not exact. First, we rewrite equation ($$) as, We find an integrating factor, $$h(x,y)$$ that will make the equation ($$) exact. Multiplying equation ($$) through by the integrating factor, and using the fact that $$h'=hx^2sec\;x$$, the following is obtained,

Given
{| style="width:100%" border="0" $$ \begin{align} \\& \displaystyle g_{0}=\frac{\partial}{\partial y}\left(\frac{d\phi}{dx}\right)=\phi_{xy} \\& \frac{d}{dx}g_{1}=\frac{d}{dx}\left(\phi_{xy'}+\phi_{yy'}y'+\phi_{y}+\phi_{y'y'}y''\right) \\& \frac{d}{dx}g_{2}=\frac{d}{dx}\left(\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y''\right) \end{align} $$
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Find
Show the equivalence of 2nd exactness condition.

Solution
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. From the lecture notes (1)-(2) p.21-8, the second exactness condition is: After Substituting the equations given in this problem into ($$): Collecting each term according to its order of derivatives of $$x$$ and $$y$$:

Since the coefficients $$ C_{0}, C_{1}, C_{2}, C_{3} $$ are linearly independent, we can conclude that $$ C_{0}=C_{1}=C_{2}=C_{3}=0 $$. This yields the symmetry of the mixed 2nd partial derivatives of the first integral $$ \phi $$: {| style="width:100%" border="0" $$ \begin{align} \\& \phi_{xy}=\frac{d}{dx}\left(\phi_{y}\right)=\phi_{yx} \\& \phi_{xy'}=\phi_{y'x} \\& \phi_{yy'}=\phi_{y'y} \end{align} $$
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