User:EGM6321.f12.team7.Zhou/R2

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Problem 1: Verify the Homogeous Solutions
Report problem 2.1 from

Given: Two homogeous solutions
The Legendre differential equation is given by,

When the $$n=1$$, we have,

We also have two linearly-independent solutions, which are given by,

Find
Verify that,$$ L_0(Y^2_H(x))= L_0(Y^1_H(x))=0$$.

Solution

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This solution was prepared without referring to previous solutions.
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For $$ L_0(Y^1_H(x))$$, we have,

Then the $$\displaystyle L_0(Y^1_H(x))$$ comes to,

For $$\displaystyle L_0(Y^2_H(x))$$, we have,

So the $$\displaystyle L_0(Y^2_H(x))$$ changes into,

So we have,

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Problem 6: Explain the Condition of Mixed Partial Derivitive Equality
Report problem 2.6 from lecture notes [ section 9]

Given:The second exactness condition
THe second exactness condition of $$N2-ODE$$ is given by,

Where $$M(x,y), N(x,y)$$ is nonlinear founctions.

FInd: Find the Condition
Find the minimun degree of differentiability and proof the theorem.

Solution
The theorem is: Suppose there is a founction(x_1, y_1). If the partial derivatives$$\frac{\partial^2 f}{\partial x_1 \partial y_1}$$ and $$\frac{\partial^2 f}{\partial y_1 \partial x_1}$$ exist and are continuous at (x_1, y_1), then we have,

*The minimun Degree of Differentiability
The mixed partial derivitives$$\frac{\partial^2 f}{\partial x_1 \partial y_1}$$ and $$\frac{\partial^2 f}{\partial y_1 \partial x_1}$$ exists and is continuous, which implies that the mimimum degree of differentiability is two, since from lecture note we know that a founction is continuous at a point doesn't means that it is differentiable at the same point.

*Proof of the Thoerem
According to the definition of derivative,which is given by,

So, for the partial derivatives, suppose we have two real number $$a$$ $$b$$, then we have,

Change the order of derivatives, we have,

Obviously we have,

The theorem has been proved.

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Problem 8: Solving equations for integrating factor
Report problem 2.8 from lecture notes

Given:Euler Integrating Factor Method
When we find a N1-ODE can not fit the first exactness condition but fail on the second one, we can use Euler Intergrating Factor to make the equation exact, which is given by,

Apply the second exactness condition, we have,

Where,

$$h_{x}:=\frac{\partial h}{\partial x},$$ $$h_{y}:=\frac{\partial h}{\partial y},$$ $$N_{x}:=\frac{\partial N}{\partial x},$$ $$M_{y}:=\frac{\partial M}{\partial y}$$

Solution

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This solution was prepared without referring to previous solutions. In ($$), since $$h$$ is nonlinear and we treat x and y as two independent variables, so the equation turns into a $$N1-PDE$$, so we have to solve this problem in two dimensional domain. What's more, the coefficients of $$h$$, $$h_{x}$$ and $$h_{y}$$ have varying coefficients ($$N_{x}, $$$$M_{y}, $$$$M, $$$$N$$). Hence it is not easy to solve this equation. And that's also the reason to make assumpetion that $$h$$ is only a founction of $$x$$ or $$y$$.
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