User:Egm6321.f09.team5.GV/HW2


 * 1) Obtain h(y) for Case 2 from pages 6-3 and 6-4
 * 2) Show that result after integration of application of 8-2 is y= ((x^3)/4) + (c/x), where c is constant.
 * 3) Show that eq. 3 of 9-2 of 1st order linear ODE with Varying Coefficients (L1.ODE.VC) is exact
 * 4) Show that eq. 8 of 9-3 is an exact N1.ODE
 * 5) Show that the 2nd exactness condition is satisfied. Pg 10-3.
 * 6) P. 10-2 Eq. 5, Derive this relation by differentiating eq. (3) p. 10-1 wrt p=y'. i.e. Gpp=....... (2nd relation in exactness condition 2)
 * 7) P. 12-1 Use PHIxy=PHIyx to obtain eq. 4 from Pg. 10-2
 * 8) Show that application on pg. 12-1 is exact as determined by the 2nd exactness condition.
 * 9) Prove that Eq. 1 on pg. 12-2 in not exact. find f & g.

Solution
If the condition for exactness is not satisfied: $$ \frac{\partial M}{\partial y}$$ does not equal to $$ \frac{\partial N}{\partial x} $$ The functions M(x,y) and N(x,y) may be multiplied by another function H(x,y) which is yet unknown. The result of the multiplication is as follows:

$$ \ H(x,y)[M(x,y)dx + N(x,y)dy] = 0 $$

$$ \ H(x,y)M(x,y) = M* $$

$$ \ H(x,y)N(x,y) = N* $$

$$ \ M*_{y}=h_{y}M+hM_{y} $$

$$ \ N*_{x}=h_{x}N+hN_{x} $$

$$ \ h_{x}N-h_{y}M+h(N_{x}-M_{y})=0 $$

If we set $$ \ (h_{x}N=0$$, we are able to solve for H(x,y) as follows: $$ \ h_{y}M-h(N_{x}-M_{y})=0$$

$$ \frac{h_{y}}{h} = \frac{1}{M}(N_{x}-M_{y})dy $$

If we integrate both side: $$ \int_{0}^{y} \frac{h_{y)}{N} dy $$

Problem Statement
Consider:

$$ \ xyy^{''} + x(y^{'})^{2} + yy^{'} = 0 $$ Given that the 1st exactness condition is satisfied show that the 2nd exactness condition is satisfied

Solution
By the 1st exactness condition the equation can be written as follows:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ F(x,y,p)= xy $$

$$ G(x,y,p)= x(y^{'})^{2} + yy^{'} $$

It is then needed to identify the following partial derivatives with respect to the different variables:

$$ F_{x}=y  \qquad \qquad  G_{y}= p $$

$$ F_{xx}=0 \qquad \qquad  G_{yp}= 1 $$

$$ F_{xy}=1 \qquad \qquad  G_{x}= (y^{'})^{2} $$

$$ F_{y}=x  \qquad \qquad  G_{xp}= 2y^{'} $$

$$ F_{yy}=0 \qquad \qquad  G_{p}= 2xp + y $$

$$ F_{yp}=0 \qquad \qquad  G_{pp}= 2x $$

$$ F_{xp}=0 \qquad \qquad $$

The second exactness condition is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Eq. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Eq. 2 $$

Filling in for Equations 1 the results are as follows:

$$ \ 0 + 2p + 0 = 2p + p - p $$

$$ 2p=2p $$  '''Eq. 1''' is satisfied.

Filling in for Equations 2 the results are as follows:

$$ \ 0 + 0 + 2x = 2x $$

$$ \ 2x = 2x $$   '''Eq. 2''' is satisfied.

By satisfying Eq. 1 and Eq. 2, the 2nd order ODE is considered to be exact.

Problem Statement
The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

Derive the relation in Equation 1.

Solution
The form for an exact 2nd order ODE is as follows:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

Since $$ ( \phi_{p} )_{y} = ( \phi_{y})_{p} $$

Rearranging the definitions of F and G, the previous equation can be written as:

$$ \ F_{y} = \left[ \frac{1}{p}(G - \phi_{x}) \right]_{p} $$

This can be evaluated to obtain the following:

$$ \ F_{y} = \frac{-1}{p^{2}}(G - \phi_{x}) + \frac{1}{p}(G_{p} - \phi_{xp}) $$

Rearranging:

$$ \ F_{y}p^{2} = - (G - \phi_{x}) + p(G_{p} - \phi_{xp}) $$

$$ \phi_{x} = F_{y}p^{2} + G - p(G_{p} - F_{x}) \qquad \qquad Equation 3. $$

Taking a partial derivative of Equation 3. with respect to p

$$ \phi_{xp} = F_{y}2p + F_{yp}p^{2} + G_{p} - pG_{pp} - G_{p} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = 2pF_{y} + p^{2}F_{yp} - pG_{pp} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \ 0 = 2F_{y} + pF_{yp} - G_{pp} + F_{xp} $$

$$ \ G_{pp} = F_{xp} + pF_{yp} + 2F_{y} $$

The result is indeed the second equation of the 2nd exactness condition, Equation 2.

Problem Statement
The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

Derive the relation of Equation 2

Solution
The form for an exact 2nd order ODE is as follows:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

$$ \ (\phi_{x})_{p} = G_{p} - (p \phi_{y})_{p} $$

Since: $$ (\phi_{x})_{p} = (\phi_{p})_{x} $$

$$ \ F_{x}= G_{p} - (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = p \phi_{yp} + \phi_{y} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p \phi_{yp} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p (F_{y}) \qquad Equation 3 $$

Since $$ (\phi_{p})_{y} = F_{y} \qquad $$ By the definition of F.

From the definition of G, the following can be identified:

$$ \phi_{x} = G - p \phi_{y} \qquad Equation 4 $$

Using Eqs. 1 and 2 the following is found:

$$ \phi_{xy} = \phi_{yx} $$

$$ \left[G - p \phi_{y} \right]_{y} = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p (\phi_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - \left[ p (G_{py} - F_{xy} - pF_{yy}) \right] = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p(G_{py}) + pF_{xy} + p^{2}F_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

Rearranging Yields:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation 5. $$

Equation 5 is equivalent to the Equation 2 from the exactness condition.

Problem Statement
Given:

$$ (8x^{5}y^{'})y^{''} + 2x^{2}y^{'} + 20x^{4}(y^{'})^{2} + 4xy = 0 $$

Show that the Nonlinear 2nd-Order ODE is exact

Solution
Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = 8x^{5}p \qquad \qquad G(x,y,p) = 2x^{2}p + 20x^{4}(y^{'})^{2} + 4xy $$

It is concluded that it satisfies the first exactness condition.

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

The Following partial derivatives are then identified:

$$ F_{x}=40x^{4}p  \qquad \qquad  G_{x}= 4xp + 80x^{3}p^{2} + 4y $$

$$ F_{xx}=160x^{3}p \qquad \qquad  G_{xp}= 4x + 160x^{3}p $$

$$ F_{xy}=0 \qquad \qquad  G_{y}= 4x $$

$$ F_{xp}=40x^{4}  \qquad \qquad  G_{yp}= 0 $$

$$ F_{y}=0 \qquad \qquad  G_{p}= 2x^{2} + 40x^{4}p $$

$$ F_{yy}=0 \qquad \qquad  G_{pp}= 40x^{4} $$

$$ F_{yp}=0 \qquad \qquad $$

Applying the results in Eq. 1:

$$ \ 160x^{3}p + 2p(0) + p^{2}(0) = 4x + 160x^{3}p + 0 - 4x $$

$$ \ 160x^{3}p = 160x^{3}p $$

Eq. 1 is then satisfied.

Applying the results in Eq. 2:

$$ \ 40x^{4} + p(0) + 2(0) = 40x^{4} $$

$$ \ 40x^{4} = 40x^{4} $$

Eq. 2 is then satisfied.

Since Equation 1 and Equation 2 are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the N2.ODE is exact.

Problem Statement
Given

$$ \sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0 $$

Prove that the equation is not exact.

Solution
Exactness Condition 1:

$$ \ F(x,y,p)y^{''} + G(x,y,p) = 0, \qquad where \quad p = y^{'} $$

$$ \ F(x,y,p) = \sqrt[]{x} \qquad \qquad G(x,y,p) = 2xp + 3y $$

It is concluded that it satisfies the first exactness condition.

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

The following partial derivatives are found:

$$ F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}  \qquad \qquad   G_{x}= 2p $$

$$ F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}  \qquad \qquad  G_{xp}= 2 $$

$$ F_{xy}= 0 \qquad \qquad \qquad  G_{y}= 3 $$

$$ F_{xp}= 0  \qquad  \qquad \qquad  G_{yp}= 0 $$

$$ F_{y}= 0 \qquad \qquad \qquad  G_{p}= 2x $$

$$ F_{yy}=0 \qquad \qquad \qquad  G_{pp}= 0 $$

$$ F_{yp}=0 \qquad \qquad $$

Using these values in Eq. 1:

$$ \frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3 $$

$$ \frac{-1}{4}x^{\frac{-3}{2}} = -1 $$

Equation 1. is not satisfied therefore the ODE is not exact.

Equation 2. can be populated to find:

$$ \ 0 = 0 + p(0) + 2(0) $$

$$ \ 0 = 0 $$

Equation 2. is satisfied.

In order for the ODE to be exact it must satisfy both Eq. 1 and Eq. 2, since it fails to do this it is concluded that the ODE is not exact.