User:Egm6321.f10.team4.Yoon/HW12009

 See my comments or TA's comments below. Egm6321.f09 20:49, 19 September 2009 (UTC)

= Problem 2 - Method of Integrating Factor =

Given
A Linear 1st order ODE is given by


 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle y' + y = x$$
 * $$\displaystyle (Eq. 4)
 * $$\displaystyle (Eq. 4)


 * }
 * }

Find
Solve Eq. 4 for y(x) and show
 * {| style="width:100%" border="0" align="left"


 * $$\displaystyle y(x) = A e^{-x} + x - x$$
 * $$\displaystyle y(x) = A e^{-x} + x - x$$


 * }
 * }

Solution
To find the solution, we need to perform the exactness test which requires two steps:

1. Writing the ODE in the form of
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$
 * $$\displaystyle M(x,y)+N(x,y)y'=0$$

$$
 * $$\displaystyle (Eq. 5)
 * style= |
 * }

2. Checking the condition given by
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * $$\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

$$
 * $$\displaystyle (Eq. 6)
 * style= |
 * }.

Checking Condition 1

Re-writing Eq. 4 in the form of Eq. 5 gives


 * {| style="width:100%" border="0"


 * style="width:100%" |
 * $$\displaystyle (y-x)+ y' = 0$$
 * $$\displaystyle (y-x)+ y' = 0$$


 * style= |
 * }

where $$\displaystyle M=(y-x)$$ and $$\displaystyle N=1$$. It is shown that condition 1 is met.

Checking Condition 2

Now, we have to test if $$\displaystyle M_y=N_x$$


 * $$\displaystyle M_y = 1$$ and $$\displaystyle N_x = 0$$

Therefore, Eq. 4 is not an exact differential equation and an integrating factor is needed to force the ODE to be exact. The definition of the integrating factor is


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$
 * $$\displaystyle h(x)=\exp \int^x f(x)dx$$


 * style= |
 * }

where


 * $$\displaystyle f(x)=\frac{1}{N}\left[ M_p - N_x \right ]$$

Substiuting the values of M and N yield


 * $$\displaystyle f(x)== \frac{1}{1}\left[ 1 - 0 \right ] = 1 $$

Therefore the integrating factor is
 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle h(x)= e^x$$
 * $$\displaystyle h(x)= e^x$$


 * style= |
 * }

Multiplying Eq. 4 by the integrating factor gives


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle e^x y + e^x y' = xe^x$$
 * $$\displaystyle e^x y + e^x y' = xe^x$$


 * style= |
 * }

Now it can be seen that the left hand side is in the form of a derivative product. Recognizing this yields:


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle (e^xy)' = xe^x$$
 * $$\displaystyle (e^xy)' = xe^x$$


 * style= |
 * }

Integrating both sides results


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$
 * $$\displaystyle (e^xy) = e^x(x-1)+A$$

Finally, dividing by e^x results in
 * style= |
 * }
 * {| style="width:60%" border="0"

$$\displaystyle y = Ae^{-x}+(x-1)$$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }

= Problem #3 = Show that the first order ODE $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is nonlinear

Solution

Define a linear operator $$D\left ( \cdot \right )$$ such that $$D(\cdot )=2x^2+\sqrt{\left (\cdot  \right ) }+x^5\left (\cdot  \right ) ^3\frac{d}{dx}\left ( \cdot  \right )$$

The linearity condition then becomes$$\forall\alpha,\beta \in\mathbb{R}$$, $$D(\alpha u+\beta v)=\alpha D(u)+\beta D(v)$$ where u and v are any functions of x

Substituting in for the left side of the linearity condition gives:

$$D(\alpha u + \beta v)=2x^2+\sqrt{\alpha u + \beta v}+x^5\left ( \alpha u + \beta v \right )^3\frac{d}{dx}\left ( \alpha u+\beta v \right )$$

Substituting for the right side of the linearity condition:

$$\alpha D(u)+\beta D(v)=\alpha\left (2x^2+\sqrt{u}+ x^5u^3\frac{du}{dx} \right )+\beta\left ( 2x^2+\sqrt{v}+x^5v^3\frac{dv}{dx} \right )$$

Simplifying and comparing both sides of the linearity condition:

$$2x^2+\sqrt{\alpha u+\beta v}+\alpha x^5(\alpha u+ \beta v)^3\frac{du}{dx}+\beta x^5(\alpha u + \beta v)^3\frac{dv}{dx}\neq2x^2\alpha+\alpha\sqrt{u}+\alpha x^5u^3\frac{du}{dx}+2x^2\beta+\beta\sqrt{v}+\beta x^5v^3\frac{dv}{dx}$$

The condition is therefore not met, and $$(2x^2+\sqrt{y})+x^5y^3y^{\prime}=0$$ is a nonlinear 1st Order ODE.

= Problem 4 = Given:

$$\ F(x,y,y') = x^2y^5 + 6(y')^2 = 0 $$

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:

$$\frac{d}{dx}(x^2y^5+6(y')^2)= \frac{d}{dx}(x^2y^5) + \frac{d}{dx}(6(y')^2) = 2xy^5 $$

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

= Problem #5 = Part 1. Create an exact nonlinear 1st Order ODE of the form $$\Phi_x\left(x,y\right)+\Phi_y\left(x,y\right)y^{\prime}=0$$ using the equation $$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Solution:

If,

$$\Phi\left(x,y\right)=6x^4+2y^{3/2}$$

Then

$$\Phi_{x}\left(x,y\right)=24x^{3}=M$$

and

$$\Phi_{y}\left(x,y\right)=3y^{1/2}=N$$

For a nonlinear, 1st order, ODE:

$$M + Ny^{\prime}=0$$

Therefore,

$$24x^3 + 3y^{1/2}y^{\prime}=0$$

We have forced exactness

Part 2. Create three more exact nonlinear 1st Order ODEs by inventing new $$\Phi\left(x,y\right)$$ functions.

Ex. 1

Choose $$\Phi\left(x,y\right)=x^{1/2}+y^{3}\quad\text{:}$$

$$ \begin{align} \Phi_{x}\left(x,y\right)&=0.5x^{-1/2}=M\left(x,y\right) \\ \Phi_{y}\left(x,y\right)&=3y^{2}=N\left(x,y\right) \end{align} $$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = \frac{1}{2}x^{-1/2}+3y^{2}y' = 0$$

Ex. 2

Choose $$\Phi\left(x,y\right)=3y^2+2x^5\quad\text{:}$$

$$\Phi_x\left(x,y\right)=10x^4=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=6y=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 10x^4+(6y)y' = 0$$

Ex. 3

Choose $$\Phi\left(x,y\right)=3x^2y^3\quad\text{:}$$

$$\Phi_x\left(x,y\right)=6xy^3=M\left(x,y\right)$$

$$\Phi_y\left(x,y\right)=9x^2y^2=N\left(x,y\right)$$

This leads to the following exact, first-order ODE:

$$M\left(x,y\right) + N\left(x,y\right)y' = 6xy^3+\left(9x^{2}y^{2}\right)y' = 0$$