User:Egm6321.f12.team1.cla/report4

=Problem 4.1* - Exactness for N2-ODE=

Find:
Determine if ($$) is exact

Solution:
To check the first exactness condition, we need to verify whether ($$) conforms to the following form
 * $$G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0$$

We can clearly see that this condition is satisfied with

and

To check the second exactness condition, we need to verify whether ($$) and ($$) satisfies the second exactness conditions i.e.,

From ($$) and ($$), we can obtain
 * $$f_{xx}=-\frac{1}{4}x^{-3/2}$$
 * $$f_{xy}=0$$
 * $$f_{yy}=0$$
 * $$g_{xp}=2$$
 * $$g_{yp}=0$$
 * $$g_{y}=3$$

Plugging these values into ($$) yields the following
 * $$f_{xx}+2pf_{xy}+p^2f_{yy}=\frac{1}{4}x^{-3/2}+2p \times 0+p^2 \times 0 =\frac{1}{4}x^{-3/2} $$
 * $$g_{xp}+pg_{yp}-g_{y}=2+p \times 0 -3= -1$$
 * $$\Rightarrow \frac{1}{4}x^{-3/2} \neq -1$$

Which is clearly false. Since ($$) does not satisfy ($$), we don't need to check ($$). Pavel Bhowmik (talk) 01:30, 16 October 2012 (UTC)

=Problem 4.2* - Find the integrating factor for the differential equation in power form=

Find:
The $$m$$ and $$n$$ real numbers that make $$G$$ in ($$) exact. Show that the first integral equation is a linear first order ordinary differential equation with varying coefficients with the following form
 * $$\phi(x,y,p)=xp+(2x^\frac{3}{2}-1)y=k$$

Solve for $$y(x)$$ in ($$).

Part 1:

 * $$G=\underbrace{x^m y^n\sqrt{x}}_{f(x,y,p)}y''+\underbrace{x^m y^n(2xy'+3y)}_{g(x,y,p)}=0$$


 * $$f(x,y,p):=x^{m+\frac{1}{2}} y^n$$
 * $$g(x,y,p):=2x^{m+1} y^n p+3x^m y^{n+1}$$
 * $$p=y'$$

We need to find the $$m$$ and $$n$$ of equation ($$). We can use equations ($$) and ($$) to solve for these variables. First we will use ($$) to solve for one of the variables.
 * $$f_{xp}+2pf_{yp}+2f_{y}=g_{pp}$$

where
 * $$f_{xp}=0$$
 * $$f_{yp}=0$$
 * $$f_{y}=ny^{n-1}x^{\frac{1}{2}+m}$$
 * $$g_{pp}=0$$

Plugging the terms into ($$) we get,
 * $$0+2p(0)+2ny^{n-1}x^{\frac{1}{2}+m}=0$$
 * $$\Rightarrow 2ny^{n-1}x^{\frac{1}{2}+m}=0$$

The only way the right side of the above statement can be zero is if the $$n=0$$ therefore,
 * $$n=0$$
 * $$f(x,y,p)=x^{m+\frac{1}{2}}$$
 * $$g(x,y,p)=2x^{m+1} p+3x^m y$$

Now we need to find the $$m$$ but we need another equation. So we use the other exactness equation ($$).
 * $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_{y}$$

where
 * $$f_{xx}=x^{-\frac {3}{2}+m}(-\frac {1}{2}+m)(\frac {1}{2}+m)$$
 * $$f_{xy}=0$$
 * $$f_{yy}=0$$
 * $$g_{xp}=2(m+1)x^m$$
 * $$g_{yp}=0$$
 * $$g_{y}=3x^m$$

Now plug in the terms above into ($$)
 * $$x^{\frac {-3}{2}+m}(\frac {-1}{2}+m)(\frac {1}{2}+m)+2p(0)+p^2(0)=2(m+1)x^m+p(0)-3x^m$$
 * $$\Rightarrow x^{-\frac {3}{2}+m}(-\frac {1}{2}+m)(\frac {1}{2}+m)=2(m+1)x^m-3x^m$$
 * $$\Rightarrow (m^2-\frac {1}{4})x^{-\frac {3}{2}+m}=(2m-1)x^m$$
 * $$\Rightarrow m^2-\frac {1}{4}=2m-1=0$$
 * $$\Rightarrow m=\frac {1}{2}$$

.

Part 2:

 * $$G=[xy''+2x^{\frac{3}{2}}y'+3x^{\frac {1}{2}}y]=0$$
 * $$\phi(x,y,p)=h(x,y)+\int f(x,y,p)dp$$

The integral of $$f$$ is easy to find because we already have$$f$$.

We have to use equation ($$) and solve for $$h$$ with respect to $$x$$ and $$y$$ independently.
 * $$g=\phi_{x}+\phi_{y}p$$

Therefore by plugging in terms into equation ($$),
 * $$g=(h_{x}+p)+(h_{y}+0)y'=h_{x}+(h_{y}+1)y'=2x^{\frac{3}{2}}p+3x^{\frac {1}{2}}y$$

Therefore,
 * $$h_{x}=3x^{\frac {1}{2}}y$$
 * $$h_{y}=2x^{\frac{3}{2}}-1$$

Taking the integral of $$h_{y}$$ with respect to y we get
 * $$\Rightarrow h(x,y)=2yx^{\frac{3}{2}}-y+k_{1}(x)$$
 * $$h_{x}=3x^{\frac {1}{2}}y+k_{1}'(x)=3x^{\frac {1}{2}}y$$
 * $$\Rightarrow k_{1}'(x)=0$$
 * $$\Rightarrow k_{1}(x)=k_1$$
 * $$h(x,y)=(2x^{\frac{3}{2}}-1)y+k_{1}$$

Now we can plug the $$h(x,y)$$ into ($$),


 * {| border="1" cellspacing="0" cellpadding="5" align="left"

.
 * $$\phi(x,y,p)=(2x^{\frac{3}{2}}-1)y+xp=k_{2}$$
 * }
 * }

Part 3:

 * $$\phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k$$


 * $$\phi(x,y,p)=p+\frac{(2x^{\frac{3}{2}}-1)}{x}y=\frac{k}{x}$$

where $$a_{0}(x):=\frac{(2x^{\frac{3}{2}}-1)}{x}$$ and $$b(x):=\frac{k}{x}$$ We know that the integrating constant for this form of equation can be obtained as
 * $$h(x)=\exp \int^x a_{0}(x)dx$$

Plugging in the terms we get,
 * $$h(x)=\exp \int^x \frac{(2x^{\frac{3}{2}}-1)}{x}dx$$
 * $$\Rightarrow h(x)=\exp [\frac{4}{3}x^{\frac{3}{2}}-ln \, x]$$
 * $$\Rightarrow h(x)=\frac{\exp [\frac{4}{3}x^{\frac{3}{2}}]}{x}$$

We know we can get $$y(x)$$ from the equation below,
 * $$y(x)= \frac{\int^x h(s)b(s)ds+k_{2}}{h(x)}$$
 * {| border="1" cellspacing="0" cellpadding="5" align="left"

.
 * $$\displaystyle y(x)= \frac{x*\int^x \frac{\exp [\frac{4}{3}s^{\frac{3}{2}}]}{s}*\frac{k}{s}ds+k_{2}}{\exp [\frac{4}{3}x^{\frac{3}{2}}]}$$
 * }
 * }

=Problem 4.3* - Show the first integral function=

Find:
To show that ($$)and ($$) leads to the first integral
 * $$\phi = P(x)p + T(x)y + k$$

Solution:

 * $$G(x,y,p,y) = \phi_x +\phi_y p + \phi_p y$$
 * $$=\frac{d\phi(x,y,p)}{dx}$$
 * $$\phi_p = P(x) $$

Integral $$\phi_p $$, we get

Take partial derivative of $$\phi$$ with respect to $$x$$ and $$y$$ yields
 * $$\phi_x = P'(x)p +\frac{\partial k(x,y)}{\partial x}$$
 * $$\phi_y = \frac{\partial k(x,y)}{\partial y}$$

Because
 * $$G(x,y,p,y) = R(x)y + Q(x)y' + P(x)y$$
 * $$ = P(x) y'' + \phi_x(x,y,p) + \phi_y(x,y,p) y'$$
 * $$ = P(x) y'' + P'(x)p + \frac{\partial k(x,y,p)}{\partial x} + \frac{\partial k(x,y,p)}{\partial y}p$$
 * $$ = P(x) y'' + \underbrace_{Q(x)} +\underbrace{\frac{\partial k(x,y,p)}{\partial x}}_R p{}$$

where
 * $$\frac{\partial k(x,y,p)}{\partial x} = R(x)y$$

Take integral
 * $$k(x,y) = y \int\limits_{}^{x}R(s) ds + k_1(y)$$

Then ($$) becomes
 * $$\phi = P(x)p + y\int\limits_{}^{x}R(s) ds +k_1(y)$$


 * $$= P(x)p + T(x)y + k_1(y)$$


 * $$\phi_y = T(x) + k'(y)$$

From ($$)
 * $$\phi_y = Q(x)$$


 * $$\Rightarrow T(x)+ k'(y) = Q(x)$$


 * $$\Rightarrow k'_1(y) = 0$$


 * $$\Rightarrow k_1(y) = constant $$


 * {| border="1" cellspacing="0" cellpadding="5" align="left"


 * $$\phi = P(x)p + T(x)y + k_1$$
 * }
 * }

=Problem 4.4* - Solving a linear second order ordinary differential equation=

Given:
The necessary and sufficient conditions for exactness of a 2nd order linear differential equatino as

Find:
1. Show that $$G$$ below is exact.

2. Find the first integral using equation below if exact.

3. Find the solution to $$y(x)$$ for equation ($$).

Part 1:
First we have to find out if equation ($$) is exact. The first exactness condition is satisfied because equation ($$) was given in the correct form. The second exactness condition is satisfied if equations ($$) and ($$) are satisfied. So let's find $$f$$ and $$g$$ and then all the terms for equations ($$) and ($$).

By definition $$f$$ and $$g$$ are as follows,


 * $$f=cos(x)$$
 * $$g=(x^2-sin(x))p+2xy$$

where $$p$$ is equal to  $$y'$$.


 * $$f_{xx}=-cos(x)$$
 * $$f_{xy}=0$$
 * $$f_{yy}=0$$
 * $$f_{y}=0$$
 * $$f_{xp}=0$$
 * $$f_{yp}=0$$
 * $$g_{pp}=0$$
 * $$g_{xp}=2x-cos(x)$$
 * $$g_{yp}=0$$
 * $$g_{y}=2x$$

Now we plug these terms into ($$),
 * $$f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp}+pg_{yp}-g_{y}$$
 * $$-cos(x) + 2p(0) + p^2(0) = 2x-cos(x)+p(0)-2x$$

Canceling terms you are left with,
 * $$-cos(x)=-cos(x)$$

Plugging in terms for equation($$) we get,
 * $$f_{xp}+pf_{yp}+2f{y}= g_{pp}$$
 * $$0+p(0)+2(0)= 0$$
 * $$0=0$$

Part 2:

 * $$\phi(x,y,p)=h(x,y)+\int f(x,y,p)dp$$

The integral of $$f$$ is easy to find because we already have$$f$$.
 * $$\phi(x,y,p)=h(x,y)+\int cos(x)dp$$

However finding $$h(x,y)$$ is not so easy because we can not simply pull the term from $$G$$. We have to use equation ($$) and solve for $$h$$ with respect to $$x$$ and $$y$$ independently.
 * $$g=\phi_{x}+\phi_{y}p$$

Therefore by plugging in terms into equation ($$),
 * $$g=[h_{x}-sin(x)p]+(h_{y}+0)y'=x^2p-(sin(x)p)+2xy$$

Therefore,
 * $$h_{x}=2xy+k_{1}(y)$$
 * $$\Rightarrow h(x,y)=x^2 y+k_{1}(y)$$
 * $$h_{y}=x^2 +k_{1}'(y)=x^2$$
 * $$\Rightarrow k_{1}'(y)=0$$
 * $$\Rightarrow k_{1}(y)=k_1$$
 * $$h(x,y)=x^2 y+k_{1}$$

Now we can plug the $$h(x,y)$$ into ($$),


 * {| border="1" cellspacing="0" cellpadding="5" align="left"


 * $$\phi(x,y,p)=x^2y+cos(x)p=k_{2}$$
 * }
 * }

Part 3:
Now we need to find $$y(x)$$ and we can do this by solving the first integral for $$y$$.
 * $$\phi(x,y,p)=yx^2+cos(x)p=k_{2}$$

First put differential equation into the form,
 * $$\frac{x^2}{cos(x)}y+y'=\frac{k_{2}}{cos(x)}$$

where $$a_{0}(x):=\frac{x^2}{cos(x)}$$ and $$b(x):=\frac{k_{2}}{cos(x)}$$ We know that the integrating constant for this form of equation can be obtained as
 * $$h(x)=\exp \int^x a_{0}(x)dx$$

Plugging in the terms we get,
 * $$h(x)=\exp \int^x \frac{x^2}{cos(x)}dx$$

We know we can get $$y(x)$$ from the equation below,
 * $$y(x)= \frac{\int^x h(s)b(s)ds+k_{3}}{h(x)}$$
 * {| border="1" cellspacing="0" cellpadding="5" align="left"

We were unable to come up with a closed form solution because the integrating factor $$h(x)=\exp \int^x \frac{x^2}{cos(x)}dx$$ does not provide a closed form integral.
 * $$y(x)= \frac{\int^x \exp [\frac{s^2}{cos(s)}]*\frac{k_{2}}{cos(s)}ds+k_{3}}{\exp[\int^x \frac{x^2}{cos(x)}dx]}$$
 * }
 * }

=Problem 4.5* - Equivalence of two exactness criterion=

Given:
The equation providing the second exactness condition of the second order linear differential equation $$G(x,y,y') = 0$$ as

Where

Prove:
Prove that ($$) is equivalent to the following conditions stated earlier

Where
 * $$p=y'$$

and$$,\; f_{xy} = \frac {\partial^2}{\partial x \partial y} f(x,y,p)$$

Solution:
We have


 * $$g_0 = \frac{\partial G(x,y,p)}{\partial y}$$


 * $$g_1 = \frac{\partial G(x,y,p)}{\partial y'} = \frac{\partial G(x,y,p)}{\partial p}$$
 * $$\Rightarrow g_1 = g_{p}+f_{p}p'$$
 * $$\Rightarrow \frac {\mathrm{d}g_1}{\mathrm{d}x} = \frac {\mathrm{d}}{\mathrm{dx}} \left(g_{p}+f_{p}p'\right) = \frac {\mathrm{d}}{\mathrm{dx}} \left(g_{p}\right) +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$
 * $$\Rightarrow \frac {\mathrm{d}g_1}{\mathrm{d}x} = \left (g_{px} + g_{py}y' + g_{pp}p'\right) +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$

$$g_2 = \frac{\partial G(x,y,p)}{\partial y''} = \frac{\partial G(x,y,p)}{\partial p'}$$ $$\Rightarrow g_2 = f$$ $$\Rightarrow \frac {\mathrm{d}g_2}{\mathrm{d}x} = \frac {\mathrm{d}f}{\mathrm{dx}} = f_{x} + f_{y} y' + f_{p} p'$$ $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \frac {\mathrm{d}}{\mathrm{dx}} \left( \frac {\mathrm{d}g_2}{\mathrm{d}x} \right) = \frac {\mathrm{d}}{\mathrm{dx}} \left( f_{x} + f_{y} y' + f_{p} p' \right)$$ $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \frac {\mathrm{d}}{\mathrm{dx}} \left( f_{x} \right)+ \frac {\mathrm{d}}{\mathrm{dx}} \left(  f_{y} y'  \right)+ \frac {\mathrm{d}}{\mathrm{dx}} \left(  f_{p} p' \right)$$ $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \left( f_{xx} + f_{xy}y' + f_{xp}p'\right)+ \frac {\mathrm{d}}{\mathrm{dx}} \left( f_{y} \right) y' + f_{y} y'' +\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$ $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \left( f_{xx} + f_{xy}y' + f_{xp}p'\right)+ \left(  f_{yx} + f_{yy}y' + f_{yp}p'\right) y' + f_{y} y'' + \frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$ $$\Rightarrow \frac {\mathrm{d^2}g_2}{\mathrm{d^2}x} = \left( f_{xx} + f_{xy}p + f_{xp}p'\right)+ \left(  f_{yx} + f_{yy}p + f_{yp}p'\right) p + f_{y} p' + \frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)$$

Substituting the terms obtained in ($$), ($$) and ($$) into ($$)
 * $$ \left [ g_{y}+f_{y}p' \right ] - \left [ \left (g_{px} + g_{py}p + g_{pp}p'\right) +\cancel{\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)} \right ] + \left [ \left( f_{xx} + f_{xy}p + f_{xp}p'\right)+ \left(  f_{yx}p + f_{yy}p^2 + f_{yp}pp'\right) + f_{y} p' + \cancel{\frac {\mathrm{d}}{\mathrm{dx}} \left(f_{p}p'\right)} \right ] = 0$$

Rearranging terms to segregate into the form $$a(x,y,p)p'+b(x,y,p) = 0$$
 * $$ \left ( f_{y} - g_{pp} + f_{xp} + f_{yp}p + f_{y}\right ) p' + \left ( g_{y} - g_{px} - g_{py}p + f_{xx} + f_{xy}p + f_{yx}p + f_{yy}p^2 \right ) = 0$$

Assuming $$f_{xy}=f_{yx}$$ and $$f$$ and $$g$$ are invariant to the order of differentiation with respect to other variables

Since ($$) is an identity, it should hold for all values of $$p'$$. So each of the two coefficients should be $$0$$ in order to satisfy this identity, i.e. in order to satisfy ($$). So
 * $$ 2f_{y} - g_{pp} + f_{xp} + pf_{yp} = 0$$

And$$,\; g_{y} - g_{xp} - pg_{yp}p + f_{xx} + 2pf_{xy} + p^2f_{yy} = 0$$

($$) and ($$) are identical to ($$) and ($$) respectively. Which implies that ($$) is equivalent to ($$) and ($$) together$$.\qquad \blacksquare$$ Pavel Bhowmik (talk) 01:30, 16 October 2012 (UTC)

=Contribution Table=

= References =