# User:Egm6321.f09.team5/HW1

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## Problem Statements

1. Derive Equations 1 & 2 on page 1-2
2. Derive Equation 2 on page 4-2 using the integrating factor method
3. Show that Equation 1 on page 4-3 is non-linear
4. Show that F(x,y,y') is a non-linear 1st order ode in Equation 3 page 5-2
5. Complete application on page 6-1 and invent three more

Having the problem statements up top here is good. Next time make the statements more specific; include the equations to be derived so that the problem statement can stand alone, i.e., without reference to transparencies.--Egm6321.f09.TA 03:17, 24 September 2009 (UTC)

## Problem 1

A Function is defined as follows:

$\ F(y(t),t)$ The first derivative is then defined using the definition of a Total Derivative

${\frac {dF}{dt}}={\frac {\partial f}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}=F_{t}+F_{y}y'$ Its second derivative can then be found by using the total derivative definition as well as the Chain Rule as follows:

${\frac {df}{dt}}=F_{t}(y(t),t)+F_{y}(y(t),t)y'$ ${\frac {(d^{2}F)}{dt^{2}}}={\frac {\partial F_{t}}{\partial t}}+{\frac {\partial F_{t}}{\partial y}}{\frac {dy}{dt}}+F_{y}{\frac {d^{2}y}{dt^{2}}}+\left[{\frac {\partial f}{\partial t}}+{\frac {\partial f}{\partial y}}{\frac {dy}{dt}}\right]{\frac {dy}{dt}}=F_{tt}+F_{ty}y'+F_{y}y''+F_{ty}y'+F_{yy}(y'')^{2}$ ${\frac {(d^{2}F)}{dt^{2}}}=F_{y}y''+2\left[f_{ty}y'\right]+F_{tt}+F_{yy}(y')^{2}$ Your steps are neat and clear. If you want to change the notation from the class notation that is fine, but make an explicit remark about this, and then make sure to express your final answer in the class notation so that it is easily compared and evident that you have in fact arrived at the proper result.--Egm6321.f09.TA 03:19, 24 September 2009 (UTC)

## Problem 2

The following initial 2nd-Order ODE is expressed:

$\ G(y'',y',y,x)=y''+y'-x=0$ We note that the previous equations has a missing variable, y. Creating a variable P as defined below:

$\ P(x)=y'(x)$ This definition allows for us to turn the 2nd-Order ODE into a 1st order ODE as follows:

$\ P'+P-x=0$ The objective is to identify P(x) using the integration factor method, note the equation is also in standard form:

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

Applying this to our equation we identify each part as follows:

$\ M(x,p)=(P-x)\qquad N(x,p)=1$ With this in mind we further identify that the equation is not exact therefore this method will allow us to make it exact.

$\ h(x,P)\left[[M(x,P)dx+dP\right]=0$ $\ hMdx+hNdp=0$ $\ M*=hMdx\qquad N*=hNdp$ In order for the function to be exact the following must be true:

$\ M*_{p}=N*_{x}$ $\ M*_{p}=(hM)_{p}=h_{p}M+hM_{p}$ $\ N*_{x}=(hN)_{x}=h_{x}N+hN_{x}$ $\ h_{p}M+hM_{p}-h_{x}N-hN_{x}=0=h_{p}M-h_{x}N+h(M_{p}-N{x})$ At this point a particular case will be used in order to move forward. The case is as follows:

If $\ h_{p}M=0$ which means the previous equation becomes:

$\ h_{x}N+h(N_{x}-M_{p})=0$ this then is followed by rearranging:

${\frac {h_{x}}{h}}={\frac {-1}{N}}\left[N_{x}-M_{p}\right]$ The left side of the equation will yield:

$\ h(x)=exp-\int _{0}^{x}{\frac {-1}{N}}\left[N_{x}-M_{p}\right]dx$ From the Previous Definitions of N and M the results are as follows:

$\ N=1\qquad N_{x}=0\qquad M_{p}=1$ $\ h(x)=exp(x)+C$ From the definition of Integrating factor method the value of P(x) can then be found as follows:

$\ P(x)={\frac {1}{h(x)}}\int _{0}^{x}h(s)b(s)ds$ in our case b(s)=x

$\ P(x)={\frac {1}{exp(x))}}\int _{0}^{x}exp(s)*sds$ this can be evaluated by looking for the result in List of Integrals

$\ P(x)={\frac {x*exp(x)-exp(x)+A}{exp(x))}}=A*exp(-x)+x-1$ integrating P(x) will yield the value of the missing value y

## Problem 3

Given:

$\ M(x,y)+N(x,y)y'=0$ If $\ M(x,y)=2x^{2}+{\sqrt {y}}$ and $\ N(x,y)=x^{5}y^{3}$ $\ (2x^{2}+{\sqrt {y}})+x^{5}y^{3}y'=0$ Show that the above 1st order ODE is non-linear:

An ODE is linear if it has the form: an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = Q(x) (see )

So we should only have "x" values and coefficients in each "y" term. If the "y" terms are dependent on other "y" terms, then it cannot be linear.

By inspection, the above equation cannot be linear because the ${\sqrt {y}}$ and $x^{5}y^{3}$ terms make it nonlinear.

Taking the derivative with respect to "x" yields an equation with complicated "y" terms:

True, but why do you need to do this?--Egm6321.f09.TA 03:27, 24 September 2009 (UTC)

${\frac {d}{dx}}((2x^{2}+{\sqrt {y}})+x^{5}y^{3}y')={\frac {d}{dx}}(2x^{2})+{\frac {d}{dx}}({\sqrt {y}})+{\frac {d}{dx}}(x^{5}y^{3}y')=4x+5x^{4}y^{3}y'$ Additionally in order for the equation to be considered linear it must meet the principle of superposition, Pg. 1 equations (2), (3) and (4)

By rearranging the equation as follows:

$\ -2x^{2}=y^{1/2}+x^{5}y^{3}y'$ The function can then be written as:

$\ D(.)=(.)^{1/2}+x^{5}(.)^{3}(.)'=-2x^{2}\qquad$ where "." represents the variable y

Close, write your differential operator as $D(.)=(.)^{1/2}+x^{5}(.)^{3}(.)'+2x^{2}$ You also need to show the other property of linearity.--Egm6321.f09.TA 03:27, 24 September 2009 (UTC)

Applying the principle of superposition as follows:

$\ D(u+v)=D(u)+D(v)\qquad$ requirement for superposition. If linear it will meet the requirement.

$\ D(u+v)=(u+v)^{1/2}+x^{5}(u+v)^{3}(u+v)'=-2x^{2}$ $\ D(u)=(u)^{1/2}+x^{5}(u)^{3}(u)'=-2x^{2}$ $\ D(v)=(v)^{1/2}+x^{5}(v)^{3}(v)'=-2x^{2}$ $\ D(u)+D(v)=(u)^{1/2}+x^{5}(u)^{3}(u)'+v)^{1/2}+x^{5}(v)^{3}(v)'$ From here we can see that the principle of superposition is not met therefore:

The ODE is nonlinear

## Problem 4

Given:

$\ F(x,y,y')=x^{2}y^{5}+6(y')^{2}=0$ Show that F(x,y,y')=0 in the above equation is a nonlinear 1st order ODE:

Like in homework problem #3, an ODE is linear if it has the form: an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = Q(x) (see )

So we should only have "x" values and coefficients in each "y" term. If the "y" terms are dependent on other "y" terms, then it cannot be linear.

By inspection, the above equation cannot be linear because the $y^{5}$ and $(y')^{2}$ terms make it nonlinear.

Taking the derivative with respect to "x" yields an equation a high order "y" term:

${\frac {d}{dx}}(x^{2}y^{5}+6(y')^{2})={\frac {d}{dx}}(x^{2}y^{5})+{\frac {d}{dx}}(6(y')^{2})=2xy^{5}$ Additionally by testing the equation against the principle of superposition in a similar fashion as it was done in problem 3, the results are as follows:

$\ D(.)=x^{2}(.)^{5}+6((.)')^{2}$ $\ D(u+v)=x^{2}(u+v)^{5}+6((u+v)')^{2}=x^{2}(u+v)^{5}+6(u'+v')^{2}$ $\ D(u)=x^{2}(u)^{5}+6((u)')^{2}$ $\ D(v)=x^{2}(.)^{5}+6((.)')^{2}$ $\ D(u)+D(v)=x^{2}(u^{5}+v^{5})+6(u'^{2}+v'^{2})$ From the results presented it is concluded that the function does not meet the superposition requirement. Therefore:

F(x,y,y') is nonlinear

Good solution.--Egm6321.f09.TA 03:28, 24 September 2009 (UTC)

## Problem 5

Show that φ(x,y) = 6x4 + 2y3/2 is an exact equation. Also invent three additional exact equations.

φ(x,y) = 6x4 + 2y3/2

dφ = 24x3 dx + 3y1/2 dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:
${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}$ Applying the exactness test to the equation φ(x,y) yields:
${\frac {\partial M}{\partial y}}={\frac {\partial (24x^{3})}{\partial y}}=0$ ${\frac {\partial N}{\partial x}}={\frac {\partial (3y^{.5})}{\partial x}}=0$ ${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}=0$ Therefore, φ(x,y) = 6x4 + 2y3/2 is an exact equation.

Additional exact equations are as follows:

1) φ(x,y) = 5x5 + 7y2

dφ = 25x4 dx + 14y dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:
${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}$ Applying the exactness test to the equation φ(x,y) yields:
${\frac {\partial M}{\partial y}}={\frac {\partial (25x^{4})}{\partial y}}=0$ ${\frac {\partial N}{\partial x}}={\frac {\partial (14y)}{\partial x}}=0$ ${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}=0$ This example is exact, but it is linear, not nonlinear.--Egm6321.f09.TA 03:30, 24 September 2009 (UTC)

Therefore, φ(x,y) = 5x5 + 7y2 is an exact equation.

2) φ(x,y) = x2y + 6x –y3

dφ = (2xy +6) dx + (x2 – 3y2) dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:
${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}$ Applying the exactness test to the equation φ(x,y) yields:
${\frac {\partial M}{\partial y}}={\frac {\partial (2xy+6)}{\partial y}}=2x$ ${\frac {\partial N}{\partial x}}={\frac {\partial (x^{2}-3y^{2})}{\partial x}}=2x$ ${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}=2x$ Therefore, φ(x,y) = x2y + 6x –y3 is an exact equation.

3) φ(x,y) = 100x2 + 75y4

dφ = 200x dx + 300y3 dy

dφ = M(x,y) dx + N(x,y) dy

The test for exactness is:
${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}$ Applying the exactness test to the equation φ(x,y) yields:

${\frac {\partial M}{\partial y}}={\frac {\partial (200x)}{\partial y}}=0$ ${\frac {\partial N}{\partial x}}={\frac {\partial (300y^{3})}{\partial x}}=0$ ${\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}=0$ Therefore, φ(x,y) = 100x2 + 75y4 is an exact equation.

## Contributing Authors

--Egm6321.f09.team5.GV 00:33, 16 September 2009 (UTC)
Egm6321.f09.team5.bear 13:30, 16 September 2009 (UTC)
Egm6321.f09.team5.risher 13:43, 16 September 2009 (UTC)