User:EGM6321.f10.team6.cook/hw2

= General Form of First Order Differential Equations =

Given
The general form for a first order nonlinear ordinary differential equation is,

Required
Verify the given equation is nonlinear, and give an example of a linear first order ordinary differential equation.

Derivation
Consider a general two-dimensional first order differential equation $$\begin{align}du(x,y)\end{align}$$ which has continuous partial derivatives so that its total differential is

If $$\begin{align} {u=\text{const}} \end{align}$$ then $$\begin{align} du=0 \end{align}$$ (exact equation)

Defining $$M,N$$

So that,

Taking the partial derivatives of $$M,N$$

By assumption of continuity through Schwarz's theorem the mixed derivatives are equal

Which is a necessary and sufficient requirement for eq 1.1.

Demonstrating Nonlinearity
In the derivation only $$\begin{align}du=0 \end{align}$$ and continuous functions were required. Thus, to demonstrate the nonlinearity of the general form only one example must be shown where the general form is nonlinear. It can also be concluded immediately because linearity was not required of $$\begin{align}M\end{align}$$ and $$\begin{align}N\end{align}$$ that nonlinearity is included in the general form. By example, let

Then checking that $$\begin{align}f(cy)=cf(y)\end{align}$$

and thus this function is nonlinear, making the general equation nonlinear in general as one such case has been found.

== Bernoulli Equation ==

Bernoulli's equation is a nonlinear ordinary differential equation which can be transformed into a linear ordinary differential equation.

Equation (1.8) is known as the Bernoulli equation and is nonlinear for $$a$$ greater than or equal to 2 ($$n$$ is an integer). The nonlinearity may be seen clearly by letting $$ \begin{align} {y}={c}{y} \end{align} $$.

It may be written in the form of equation 1.1, like so

Note that the requirement of equation 1.7 is not met and the equation is neither exact nor linear.

Linearization
The equation can be linearized by letting

Differentiating

Which arrives at a new first order linear ordinary differential equation

Note that the partial of $$M$$ with respect to $$y$$ is zero, as is the partial of $$N$$ with respect to $$x$$; thus, equation 1.12 is both exact as well as linear.

Given

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$$ {\phi} \left(x,y\right) = x^{2} y^{3/2} + log (x^{3} y^{2}) = k $$ (4.1)
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Required
Find $$ F(x,y,y') = {\frac {d {\phi}}{d x}} (x,y) $$ and verify that F is exact N1_ODE and invent 3 more

Solution

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$$ F(x,y,y') = {\frac {d{\phi}(x,y)}{d x}} = {\frac} + {\frac} {\frac{dy}{dx}}$$ (4.1)
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$$\begin{align} & M(x,y):={\phi}_{x}(x,y): =\frac{\partial \phi(x,y)}{\partial x} \\ & N(x,y):={\phi}_{y}(x,y): =\frac{\partial \phi(x,y)}{\partial y} \\ \end{align}$$ (4.2)
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So, that is
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$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' $$      (4.3)
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From the given $${{\phi}\left(x,y\right)}$$, we can get M(x,y) and N(x,y)


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$$\begin{align} &M\left(x,y\right) :={\phi}_{x}(x,y)= 2 x y^{\frac{3}{2}} + {\frac{3}{x}}\\ &N(x,y) := {\phi}_{y}(x,y)= {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}}\\ \end{align} $$      (4.4)
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Follow the $$ 1^{st} $$ condition of exactness,


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$$F(x,y,y') = M\left(x,y\right)+N(x,y)y' = \left( 2 x y^{\frac{3}{2}} + {\frac{3}{x}} \right) + \left( {\frac{3}{2}} y^{\frac{1}{2}}x^{2} + {\frac{2}{y}} \right) y' = 0 $$      (4.5)
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$$ y' = - \frac{M\left(x,y\right)}{N(x,y)} $$      (4.6)
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And then, apply $$ 2^{nd} $$ condition of exactness,


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$$ \underbrace{M_{y}\left(x,y\right)}_{\frac{\partial M(x,y)}{\partial y}}= \underbrace{N_{x}(x,y)}_{\frac{\partial N(x,y)}{\partial x}} $$      (4.7)
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$$ {\frac {{\partial}M(x,y)} {\partial y} } = {\frac {{\partial} \left( 2 x y^{\frac{3}{2}}+{\frac{3}{x}}\right)} {\partial y}} = 3 x y^{\frac{1}{2}} = {\frac {{\partial}\left( {\frac{3}{2}} y^{\frac{1}{2}} x^{2} + {\frac{2}{y}}\right)} {\partial x}} = {\frac {{\partial} N(x,y)}{\partial x}} $$      (4.8)
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So, F is exact N1_ODE.

And 3 more equations,


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$$ {\phi}(x, y) = {\frac{1}{2}} x^{4} sin(y^{2}) = k $$ (4.9)
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$$ {\phi}(x, y) = {\frac{1}{4}}( x^{4} + {6} x^{2} y^{2} + y^{4})= k $$ (4.10)
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$$ {\phi}(x, y) = e^ = k $$ (4.11)
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Required
Find the function of $$f\left( y' \right)$$ such that there is no analytical solution to $$f\left( y' \right)=-\frac{M\left( x,y \right)}{N\left( x,y \right)}$$.

== Solution == The solution of the problem is equivalent to find a N1_ODE that does not satisfy the 1st condition of exactness. In order to a N1_ODE to be exact, it must be in the form of:

The function chosen to test the first condition of exactness is:

From Equation 5.3 we can identify the functions, $$M\left( x,y \right)$$, $$N\left( x,y \right)$$ and $$f\left( y' \right)$$ as:

If we can rearrange Equation 5.3 to fit in the form of the Equation 5.2, than the 1st condition of exactness, if not we found Non-Exact N1_ODE.

Doing some algebra with Equation 5.3 to arrange in the fashion of $$f\left( y' \right)=-\frac{M\left( x,y \right)}{N\left( x,y \right)}$$:

Equation 5.7 does not fit in the same arrangement of Equation 5.2, since we have $$\begin{align} y'+\cos y'\end{align}$$ on the left hand side of the equation instead of only $$\begin{align}y' \end{align} $$.

Therefore there is no analytical solution to the function chosen and also proves that Equation 5.3 is a Non Exact N1_ODE.

Required
Verify that equation 6.1 satisfies the first condition of exactness, equation 6.2.

The first condition of exactness:

where,

Solution
From the equation 6.1,

Total differential of function $$\phi\left(x,y\right)$$,

As$$\phi\left(x,y\right)=k=const$$, then $$ d\mathbf{\phi}=0 $$, We can rewrite the equation 6.5,

For the given problem,

So, we can verify that the equation 6.1 could be converted to the form of the first condition of exactness, equation 6.2.

The mixed derivatives of $$\begin{align}\phi\end{align}$$, $$\begin{align}\phi_{xy},\phi_{yx} \end{align}$$ are both zero, meeting the second condition of exactness. This is clearly seen from equation 6.7 as $$\begin{align}\phi_{x} \end{align}$$ is only a function of $$\begin{align}x\end{align}$$ while $$\begin{align} \phi_{y} \end{align} $$ is just a function of $$\begin{align}y \end{align}$$.

Given
The Euler Integrating Factor Method may be applied to homogeneous differential equations when the equations meet the first condition of exactness, but do not meet the second condition of exactness. That is to say, those equations that can me made to fit the form shown in equation 7.1, but do not meet the criteria shown in equation 7.2 can be made exact using the Euler method. One of the benefits of the Integrating Factor Method is that it can be applied to solve non-homogenous differential equations.

Required
(i) Solve the following non-homogeneous differential equation using the Integrating Factor Method:

(ii) Assume that $${{a}_{1\left( x \right)}}\ne 0$$ for any $$x$$ and find an expression for $${{y}_{\left( x \right)}}$$ in terms of $$\begin{align}{{a}_{0}}\end{align}$$, $$\begin{align}{{a}_{1}}\end{align}$$, $$\begin{align}{b}\end{align}$$ for the following form:

(iii) Apply the integrating factor method to the following equation:

== Solution ==

Explanation of Process
The Integrating Factor Method assumes that some function $${{h}_{\left( x,y \right)}}$$ can be found to make a differential equation exact when it meets the first criterion of exactness (equation 7.1) but not the second (equation 7.2). This factor is multiplied by the original equation to yield equation 7.6.

Applying the second condition of exactness and using the product rule to differentiate each term we obtain:

If we assume that the partial derivative of h with respect to y is zero, then h becomes a function of x only and we are left with the following:

Which can be rearranged to obtain:

Dividing by h and rearranging terms we obtain:

Integrating both sides with respect to x we obtain the following function we can use to obtain h:

If we return to our original non-exact equation given in equation 7.1 we can apply the integrating factor to obtain the following equation:

If we recall that $$h\left( {y}'+{{a}_{0}}y \right)=hb$$, we can rearrange the above to obtain the following:

The reader might recognize that the left side of equation 7.13 is really nothing more than $${{\left( hy \right)}^{\prime }}$$. Substituting into 7.13 we obtain:

After integrating we are left with the solution to y.

Solution to Part I
From equation 7.3 we obtain the following:

Substituting into equation 7.11 we obtain an expression for h:

Combing the result from equation 7.17 with equation 7.15 we obtain the solution to y.

Rearranging the terms:

Solving the first Integral

Using the substitution method

Substituting in the Euqation 7.21 we get

The integral in equation 7.24 cannot be solved using standard techniques. Numerical methods or approximations that employ series expansions must be used.

Solution to Part II
From equation 7.3 we obtain the following:

Substituting into equation 7.11 we obtain an expression for h:

Combing the result from equation 7.26 with equation 7.15 we obtain the solution to y.

Solution to Part III
First we need to rearrange Equation 7.5 to be in the form of:

Rearranging Equation 7.5 we get:

Calculating $${{M}_{y}}$$, $$N$$ and $${{N}_{x}}$$:

Substituting into equation 7.11 we obtain an expression for h:

From equation 7.27 we have a general solution in terms of $${{a}_{0}}$$, $${{a}_{1}}$$, $${b}$$. Substituting the terms given in the problem statement in equation 7.5 we obtain:

Using the substitution method to solve the integral

Substituting Equation 7.33 into 7.15:

Using the substitution method to solve the integral

From Equation 7.34 we get the final answer for $$y(x)$$

Required
1. Show that an integration constant is not necessary in

2. Show the formulation,

agrees with with that in King

which is in the form

3. Find $$y_h(x)$$ independently, i.e, solve $$y'+a_0y=0$$

Part 1
Suppose an integration constant was included

Then recognizing the constant is general, and using exponential rules

The function $$h(x)$$ appears inverted before the integral,

so that the constant can be moved outside of the integration and cancelled with the same constant in the inverted function outside the integration. Thus, the constant is superfluous.

Part 2
The derivation by King is as follows

Multiply by through by

To get

Noting that the left hand side is the product rule carried out for the total derivative

Integrating now that we have a total derivative (note that the integration variable goes from x to s on the right hand side)

Dividing through by the factor on the left hand side arrives at King's form,

Now let $$h(x)$$ be the integrating factor

Noting that integration generates a constant, we can drop $$A$$. Letting $$Q$$ become $$b$$

Thus it is shown that the two forms are equivalent.

Part 3
Finding a solution to

First, assume a solution of the form

taking the derivative

Noting that $$f$$ is the negative integral of $$a_0$$

which is the solution to the homogeneous equation

easily verified by substituting $$y'$$ and $$y$$

Note that the solution is the homogeneous solution, the first term in King's formulation from part 2.

Required
where,
 * Show the given problem is either exact or can be made exact
 * Find $$\phi\left(x,y\right)=k$$

Verify the Exactness
We can get the differential equation from the given conditions.

where,

First condition of exactness
Equation 9.4 satisfies the 1st condition of exactness because it has the form of equation 9.5

Second condition of exactness
Before we proceed, we can rewrite equation 9.4 as $$ {e}^{2y}\neq 0 $$

Checking whether the equation satisfies equation 9.7

It is found equation 9.6 is not exact.

Making the equation exact by multiply Euler integrating factor $$ h\left(x\right) $$.

Multiply equation 9.11 into equation 9.6

Alternative Approach
It is not necessary to find $$ h\left(x\right) $$ with such verbosity.

$$ N\left(x,y\right) $$ and $$ M\left(x,y\right) $$ are the functions of x only, by making $$ N\left(x\right)=const $$ we can make $$ {M}_{y}\left(x,y\right)={N}_{x}\left(x,y\right)$$

in other words, by just multiplying $${\color{blue}\frac{1}{\mathrm{sin}x}}$$, we have made $$ N\left(x,y\right)=1 $$

then, $$ {M}_{y}\left(x,y\right)={N}_{x}\left(x,y\right)=0$$

This is the same integrating factor that we had found at equation 9.11

Exactness of the new Equation
Let's check the exactness of equation 9.12.

First condition of exactness
Equation 9.12 satisfies the 1st condition of exactness because it has the form of equation 9.5

Second condition of exactness
As equation 9.12 satisfies both the 1st and 2nd condition of exactness, it is an EXACT ODE.

Find $$\phi(x,y)=k$$
As for $$\phi\left(x,y\right)$$

We can find $$\phi\left(x,y\right)=k$$ by integrating $$M\left(x,y\right)$$ for x, or $$N\left(x,y\right)$$ for y.

i.e.

Therefore,

We can find $$k\left(y\right)$$ by $$\frac{\partial \phi(x,y)}{\partial y}$$