User:Egm6321.f12.team07.liu/report5

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Problem 5: First Integral for a class of exact L2-ODE-VC
Problem 5 from lecture notes section 21-5.

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

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

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Problem 6: Solve certain L2-ODE-VC
Report problem 6 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

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This solution was produced without referring to previous solutions. 1, Show exactness:
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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:

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Problem 12: analyze the hypergeometric differential equation and verify the hypergeometric fuction to be a solution of it
Report problem 12 from.

Given: the hypergeometric differential equation and the hypergeometric function
The hypergeometric differential equation:

The hypergeometric function:

where: $$k=1,2,\cdots $$, $$(a)_k=a(a+1)(a+2)\cdots (a+k+1)$$, and $$(b)_k, (c)_k$$ have the same expressions as $$(a)_k$$.

Find: Show exactness and verify solution
1, Is ($$) exact?

2, Is ($$) in the power form

3, Verify that ($$) is indeed a solution of ($$)

Solution

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

1, Show exactness:

First write ($$) into:

Where $$p=y'$$.

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:

As for ($$), we have: $$\left\{\begin{matrix}LHS1=1\\ RHS1=ab-a-b-1\end{matrix}\right. \Rightarrow LHS1\neq RHS1$$ and $$LHS2=RHS2=0$$

Thus, ($$)satisfies the 1st but not satisfies the 2nd exactness conditions, i.e. it is not exact.

2, Power form?

We could see the power form as following:

Comparing ($$) with ($$), we know:

$$\alpha $$ and $$r$$ can not be found to satisfy $$ax^r=x(1-x)$$.

Therefore, ($$) is not in the power form.

3, Verify that ($$) is indeed a solution of ($$)

If the statement that "($$) is indeed a solution of ($$)" is true, we could obtain:

where: $$k=1,2,\cdots $$, $$(a)_k=a(a+1)(a+2)\cdots (a+k+1)$$, and $$(b)_k, (c)_k$$ have the same expressions as $$(a)_k$$;

if we institute ($$), ($$) and ($$) into the left hand side of ($$)(which we call $$LHS_{hde}$$), we could have:

Next we will verify if ($$) is true.

Unify each expression to the form of "$$\sum_{k=0}^{\infty} [\cdots \cdots ] \frac{x^{k+1}}{k!}$$":

Sum all the five expressions above, we obtain:

considering the following expression of $$(a)_k$$(Similar expressions could be derived as for $$(b)_{k}$$ and $$(c)_{k}$$.):

We have:

thus, ($$) is true.

Therefore, ($$) is one solution of ($$).