User:Egm6321.f11.team3/Hwk2

Given
The Legendre equation is given as

where n is the order of the equation. When n=1,

Two homogeneous solutions are given as:

Find
Verify that:

Solution
We solved this problem on our own.

Substitute equation 2.1.3 into 2.1.2 as the independant variable since the order of polynomial is 1.

Therfore, the terms cancel and $$\displaystyle L_2(y^1_H)=0.$$

Now to prove the second equation, substitute equation 2.1.4 into 2.1.2 as the independant variable since the order of polynomial is 1.

Therefore,

By rearranging terms we get following and it is clear the equation does equal zero.

Robert Anderson 23:38, 18 September 2011 (UTC)

Find
Verify that:

is the solution for:

Solution
We solved this problem on our own.

Thus (2.2.1) is indeed the solution for (2.2.2).

Find
Show that:

is linear in $$\displaystyle y'$$, and that (2.3.1) is in general an N1-ODE.

But (2.3.1) is not the most general N1-ODE

Give an example of a more general N1-ODE.

Solution
We solved this problem on our own.

If f(.) is a linear function, f(.) satisfies with

Let

Thus,$$\displaystyle y^{'} $$ is linear. To show that (2.3.1) is in general an N1-ODE,Let

Thus (2.3.1) is in general an N1-ODE

is a more general N1-ODE.

Given
Mtg8.8-1

Find
Verify equation (2.4.1) is a N1-ODE

Solution
Because the highest differential coefficient in this equation is y',so it is first-ODE. Then this equation includes some components,such as siny and y^3,are nonlinear,it is a nonliner equation. We also can proof it by using the linearity equation below:

You can easliy find that:

So it is nonlinear.

Find
Explain how



can be converted into the form

using



Solution
We solved this problem on our own



Taking cube root on both sides





We know that,



Thus the given equation is converted into the form,



Prashant Gopichandran 19:02, 21 September 2011 (UTC)

Given
The homogeneous solutions $$\displaystyle y_{H}^{1}=x,\, \, y_{H}^{2}=\frac{x}{2}log(\frac{1+x}{1-x})-1$$ are linearly independent. Mtg9,9-1

Find
Show that $$\displaystyle \forall \alpha\in \mathbb{R},\, y_{H}^{1}(\cdot )\neq \alpha y_{H}^{2}(\cdot )$$

Solution
The method how to find linear indenpendence: In order to find:
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$$\forall \alpha\in \mathbb{R},\, y_{H}^{1}(\cdot )\neq \alpha y_{H}^{2}(\cdot )$$ We can proof in this way:
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$$show\, \, that\, \exists \hat{x}\, such\, \, that\, \, y_{H}^{1}(\hat{x})\neq \alpha y_{H}^{2}(\hat{x})$$ So,in our question,suppose that there exists such $$\alpha _{1}$$ for $$\hat{x}_{1}$$
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$$y_{H}^{1}(\hat{x}_{1})=\alpha _{1}y_{H}^{2}(\hat{x}_{1})$$ and there exists such $$\alpha _{2}$$ for $$\hat{x}_{2}$$
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$$y_{H}^{1}(\hat{x}_{2})=\alpha _{2}y_{H}^{2}(\hat{x}_{2})$$ If $$\alpha _{2}\neq \alpha _{1}$$,They are linearly-independent. Let $$\hat{x}_{1}=1$$
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$$0.5=\alpha _{1}\frac{0.5}{2}log(\frac{1+0.5}{1-0,5})-1\Rightarrow \alpha _{1}=-0.57$$ Then,let:
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$$0.2=\alpha _{1}\frac{0.2}{2}log(\frac{1+0.2}{1-0,2})-1\Rightarrow \alpha _{1}=-0.2$$
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So $$\alpha _{1}\neq \alpha ^{2}$$


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Given
Considering the following equation,

Find
and show that 2.7.2 is a non-linear first order ordinary differential equation.

Solution
We solved this problem on our own.

The general form of a N1-ODE is given in lecture 7-6 as: and the particuluar form is: Now determine the partial differential terms using the differentiation rules

The general form equation will therefore be And is therefore a N1-ODE.

Robert Anderson 02:51, 21 September 2011 (UTC)

Given
The necessary criteria for a 1st order ordinary differential equation to satisfy the first exactness condition is that it has the particular form:

Find
Does equation (2.8.2) satisfy the first exactness condition?

Solution
We solved this problem on our own.

Rearranging equation (2.8.2) results in:

Because $$\displaystyle f(\cdot) $$  has no explicit inverse, we cannot find particular form for (2.8.2). Thus, (2.8.2) cannot meet the first exact condition.

Find
Review Calculus, and fine the minimum degree of differentiability of the function $$\displaystyle \phi(x,y) $$ such that (2.9.1) is satisfied. State the full theorem and provide a proof.

Solution
We solved this problem on our own.

Clairaut's Theorem: Given $$\displaystyle f:D^{n}\rightarrow R $$, suppose that for all $$\displaystyle 1\leq i,j\leq n $$, the mixed partial derivative

$$\displaystyle f_{x_{i},x_{j}}(x) $$ exists in a neighborhood of a, If $$\displaystyle f_{x_{i},x_{j}}(x) $$ is continuous at a for all$$\displaystyle 1\leq i,j\leq n $$,

then $$\displaystyle f_{x_{i},x_{j}}(a) $$ = $$\displaystyle f_{x_{j},x_{i}}(a) $$

Thus, according to Clairaut's Theorem, if $$\displaystyle f_{x_{i},x_{j}}(a) $$ is continuous at a for all $$\displaystyle 1\leq i,j\leq n $$, then minimum degree of differentiability is three.

Proof:

First, we have to know Mean Value Theorem.

If $$\displaystyle g(x):[a,b]\rightarrow R $$ is continuous on $$\displaystyle [a,b] $$ and differentiable on $$\displaystyle (a,b) $$, then there exists some $$\displaystyle a<c<b $$, such that

Let $$\displaystyle N $$ be a neighborhood of the point a with the property that $$\displaystyle f_{x_{i},x_{j}} $$

exists for all $$\displaystyle  x\in N $$ and all  $$\displaystyle  i,j\in \left \{ 1,2,...,n \right \} $$

Define a new function $$\displaystyle g:N\rightarrow R $$ by

Consider

0< Note that $$\displaystyle g(a+te_{i}) $$ is differentiable for $$\displaystyle 0<t<h $$. We can therefore apply the Mean Value Theorem to $$\displaystyle g(a+te_{i}) $$

On the other hand,

Thus,

We get

Now we look at

The Mean Value Theorem implies that there exists some $$\displaystyle c_{j}\in(0,h) $$ such that

Thus, we can get

Now,we look at the limit

Since $$\displaystyle 0<c_{i}<h,0<c_{j}<h,\lim_{h\rightarrow 0}(a+c_{i}e_{i}+c_{j}e_{j})=a $$

By hypothesis, the mixed partial derivative is continuous at a. This implies

On the other hand, the entire computation can be rewritten with the roles of i and j exchanged. Thus we can get

Thus, we complete the proof.

Given
The equation: Mtg10,10-5

Find
Verify that (2.10.1) is indeed the solution for N1-ODE (2.10.2)

Solution
We solved this problem on our own

We can take the derivative of $$\phi (x,y)$$:
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$$\displaystyle \frac{\mathrm{d} \phi (x,y)}{\mathrm{d} x}=75x^{4}+cosy\cdot {y}'$$
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Find
Explain why solving for the integrating operator h(x,y) is not easy.

Solution
The given equation is a non-linear partial differential equation with terms like hx,hy,Nx,My.

So solving for the integrating factor h(x,y) is not easy.

Prashant Gopichandran 19:08, 21 September 2011 (UTC)

Given
Suppose $$\displaystyle h_{x}(x,y)=0,$$thus h is a function of y only.then $$\displaystyle \frac{hy}{h}=\frac{1}{M}(N_{x}-M_{y})=:m(y)$$

Find
Find h(y)

Solution
$$\displaystyle \int ^{y}\frac{hy}{h}=\int^{y} \frac{1}{M}(N_{x}-M_{y})$$

$$\displaystyle \Rightarrow logh(y)=\int ^{y}m(s)ds+k$$

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

Find
Show that

Solution
We solved this problem on our own.

First we put the equation in the form of (1)p11-2

$$\displaystyle h_xN+h(N_x -M_y)=0$$

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

$$\displaystyle dy+(\frac{y}{x}-x^2)=0$$

With using the first term for N and the second for M, we have

$$\displaystyle h(x)=exp[-\int\frac{1}{N}(N_x-M_y)]=exp[\int\frac{1}{1}(\frac{1}{x}-0)dx+k]$$

$$\displaystyle h(x)=exp[ln(x)+k]$$

For k=0, h(x)=x

Now we find y(x),

$$\displaystyle y'+\frac{y}{x}=x^2$$

$$\displaystyle h(y'+\frac{y}{x})=hx^2$$

$$\displaystyle hy'+h'y=hx^2$$

$$\displaystyle (xy)'=x^3$$

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

After integration we have, $$\displaystyle xy=\frac{x^4}{4}+k$$

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

Robert Anderson 04:40, 21 September 2011 (UTC)

Find
1. Find $$\displaystyle y(x) $$ in terms of

2. Find $$\displaystyle y(x) $$ in terms of

3. Find $$\displaystyle y(x) $$ in terms of

Solution
We solved this problem on our own.

1.

2.

3.

Find
Show that the integration constant $$\displaystyle k1 $$ in (2.15.1) is not necessary. i.e., only $$\displaystyle k2 $$ in (2.15.2) is necessary

Solution
We solved this problem on our own.

According to above manipulation, we can see that $$\displaystyle k $$ is cancelled, but $$\displaystyle \frac{k2}{k} $$ remains. Thus, $$\displaystyle k1 $$ is not necessary in the equation (2.15.1) but $$\displaystyle k2 $$ is necessary in the equation (2.15.2)

Given
The form of L1-ODE in class: The form of L1-ODE in King's book The solution in class: The solution in King's book: Mtg12,12-2

Find
Show that the solution of (2.16.3) in (2.16.1) agrees with the result present in King's book (2.16.4).

Solution
We solved this problem on our own

Let:
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$$\displaystyle y_{H}(x)=\frac{1}{exp\int ^{x}a_{0}(s)ds},\, \, A=k,\, \, y_{P}(x)=\frac{\int ^{x}(exp\int ^{x}a_{0}(s)\cdot ds)\cdot b(s)ds}{exp\int ^{x}a_{0}(s)ds}$$ it is the same as King's equation below:
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$$\displaystyle y=\underbrace{Aexp(-\int ^{x}P(t)dt)}_{Ay_{H}(x)}+\underbrace{exp(-\int ^{x}P(t)dt)\cdot \int ^{x}Q(s)exp(\int ^{x}P(t)dt)\cdot ds}_{y_{P}(x)}$$
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Given




Find
Instead of identifying yH(x) from the given equations solve the homogeneous part of



Solution
We solved this problem on our own

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

Since the equation,



is homogeneous,b(s)=0.

$$\displaystyle \Rightarrow y_{H}=\frac{k}{h(x)}$$

=Problem R*2.18 Integrating Factor [h]=

Given

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$$\displaystyle \left [x^4 y + 10 \right ] + (\frac{1}{2} x^2)y'=0 $$  (2.18.1)
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which is not an exact L1-ODE-VC.

Find
Determine whether Eq (2.18.1) is exact or not. If it is not, make it exact by using integrating factor h (x, y)

Solution
1. First exactness condition,

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

with


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$$\displaystyle M(x,y) = x^4 y + 10$$

(2.18.2)
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and


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$$\displaystyle N(x) = \frac{1}{2} x^2$$

(2.18.3) so it satisfies the first condition.
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2. Second exactness condition


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$$\displaystyle M_{y} = x^4$$ (2.18.4)
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(2.18.5) so
 * $$\displaystyle N_{x} = x$$
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$$\displaystyle M_{y}(x,y) \ne N_{x}(x,y) $$

Therefore, Eq (2.18.1) is not exact.

3. Finding an integrating factor h(x,y) such that the following exact N1-ODE:

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

Assume $$\displaystyle h_{y}(x,y) = 0 $$, so $$\displaystyle h$$ becomes a function of x only,

then


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(2.18.6)
 * $$\displaystyle \frac{h_{x}}{h}= -\frac{1}{N}(N_{x}-M_{y})=: n(x)$$
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Substituting $$\displaystyle N$$ Eq (2.18.3), $$\displaystyle N_{x}$$  Eq (2.18.5),  and  $$\displaystyle M_{y}$$  Eq (2.18.4)  into  Eq (2.18.6)


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(2.18.7)
 * $$\displaystyle n(x)= -\frac {1}{\frac{1}{2} x^2}(x-x^4)$$
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(2.18.8) Changing the variable of function,
 * $$\displaystyle n(x)= 2(x^2 - \frac{1}{x}) $$
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(2.18.9)
 * $$\displaystyle n(s)= 2(s^2 - \frac{1}{s}) $$
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Definition from $$\displaystyle h(x)$$


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(2.18.10) Substituting $$\displaystyle n(s) $$ into Eq (2.18.10)
 * $$\displaystyle h(x) = exp \left[\int^x n(s)ds + k \right]$$
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(2.18.11)
 * $$\displaystyle h(x) = exp \left[2 \int^x (s^2 - \frac{1}{s})ds + k \right]$$
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(2.18.12)
 * $$\displaystyle h(x) = exp \left[\frac{2}{3} x^3 - 2\ln x + k \right]$$
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(2.18.13)
 * $$\displaystyle h(x) = \frac{1}{x^2}exp \left[\frac{2}{3} x^3 + k \right]$$
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