User:Egm6321.f12.team7/report4

=R*4=

Problem 1: Verify exactness of the equation
Report problem 4.1 from

Given: An L2-ODE-VC and two exactness conditions
An L2-ODE-VC is given by,

The two exactness conditions are given by,

Where,

Find
Use the exactness conditions to verify whether the L2-ODE-VC is exact.

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions. For the first condition, we have,
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

So the first exactness condition is satisfied. For the second exactness condition, we have,

So according to ($$), we have,

Obviously, $$RHS\neq LHS$$, so the equation is not exact.

Problem 2: Solve an N2-ODE
Report problem 4.2 from.

Find: solution for the N2-ODE
1. Find $$m,n \in \mathbb{R}$$ that ($$)is exact.

2. Show that the first integral of ($$) is a L1-ODE-VC

where

3. Solve for y(x)

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

1. The first exactness condition of N2-ODE is

Rearrange ($$) to get

Thus, it satisfies the first exactness condition, where

The second exactness condition is

where

thus, from ($$),

from ($$),

if ($$) is true for any x and y,

substitute ($$) into ($$),

thus, Thus, $$m=\frac{1}{2}, n=0$$ for ($$) to be exact.

2.

integral both sides of ($$)

then

compare with ($$),

Assume

so

thus, the first integral of ($$) is a L1-ODE-VC shown in ($$)

3. Rearrange ($$) to get

it is in the form of

Problem3: First Integral for a class of exact L2-ODE-VC
Problem R*4.3 from lecture notes section 21.

Given: A class of exact L2-ODE-VC
Given a class of linear, 2nd order, ordinary differential equations with varying coefficients of the form:

Find: A general form of the first integral
Show that the first integral of the form given below generates a class of exact L2-ODE-VC of the form given in ($$).

where

Solution: Derive the first integral

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

Integrating ($$) with respect to p yields:

The partial derivatives with respect to x and y are given by

Substituting back into the general form of L2-ODE-VC ($$)

where

Integrating ($$) results in

where

Substituting into ($$) produces

Taking the partial derivative with respect to y and comparing to Q(x) (since $$\phi_y = Q(x)$$ from ($$) and ($$))

Therefore, $$k_1 '$$ cannot be a function of y, and $$k_1$$ must be a constant. This means

Problem 4: Solve certain L2-ODE-VC
Report problem 4 from.

Given: A L2-ODE-VC
The linear 2nd-order ordinary differential equation with varying coefficients(L2-ODE-VC):

Find: Show exactness and solve through integration

 * 1) Show the equation is exact
 * 2) Find first integration $$\phi$$
 * 3) Solve for $$y(x)$$

Solution of the L2-ODE-VC

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

This solution was produced without referring to previous solutions. 1, Show exactness:
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

First write ($$) into:

Obiviously, ($$) represents that the L2-ODE-VC satisfies the 1st exactness conditin.

Then, as for the 2nd exactness condition, we need to check if ($$) satisfies the following expressions:

we have: $$LHS1=RHS1=-cosx$$, $$LHS2=RHS2=0$$

Thus, ($$)satisfies the 1st and 2nd exactness conditions i.e. it is exact.

2, Find $$\phi$$

As for ($$), we know that there exist a $$\phi$$ which satisfies:

According to ($$), we have:

Moreover, we obtain:

{{NumBlk|:|$$\left\{\begin{matrix} \phi _y=h_y\\ \phi _x=-psinx+h_x\end{matrix}\right.$$|$$}}

Institute ($$) into ($$) and make integration, we obtain:

Where $$k_1$$ is a constant.

Finally, institute ($$) into ($$), we have:

3, Solve for $$y(x)$$

Then our problem to solve $$y(x)$$ turns to solve equation:

i.e.

It is easy to find that: $$M_y\neq N_x$$

Thus ($$) is not exact, we consider using IFM:

Introduce $$h(x)$$ such that:

to render it exact, we have: $$\bar{M}_y=\bar{N}_x$$

More specificly:

Through integration, we obtain:

Moreover to get $$y(x)$$, we need to find the integration $$\bar{\phi}$$ of ($$):

we have:

According to ($$), we obtain:

Institute ($$) into ($$), we get:

Then we get:

i.e.

Plug ($$) back into ($$), we obtain the result:

Where $$h(x)$$ is decided by ($$)

Problem 5: Showing equivalence to symmetry of mixed 2nd partial derivatives of first integral
Based on lecture notes

section 22

Solution

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

On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * }

We know that:

Putting this in the form given in ($$),

Now we collect the like terms:

From here we get:

Now if we expand out both sides of ($$), we get:

Setting them equal to each other, we get:

Similarly, from ($$) and ($$), we see that:

= Contribution =