User:Egm6321.f09.Team06.sada/HW2

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

Show that equations (4) & (5) on 10-2 for exactness are satisfied for the above equations?

Solution:

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

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The partial derivatives are identified as,

$$ F_{x}=40x^{4}p  \qquad $$

$$ F_{xx}=160x^{3}p \qquad $$

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

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

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

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

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

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

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

$$ G_{y}= 4x \qquad $$

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

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

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

Substituting in Equ.(4), we get

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

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

Eq. (4) is satisfied.

Applying the results in Eq. (5), we get

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

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

Eq. (5) is also satisfied.

Since Equation (4) and Equation (5) are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the NL.ODE is exact.

Problem 9
$$ \sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0 $$

Verify exactness of above ODE.

Solution

Exactness Condition 1:

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

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

The 2nd exactness condition for a second order ODE is,

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. (4) $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. (5) $$

The following partial derivatives are found:

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

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

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

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

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

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

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

$$  G_{x}= 2p  \qquad $$

$$ G_{xp}= 2 \qquad $$

$$ G_{y}= 3 \qquad $$

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

$$ G_{p}= 2x \qquad $$

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

Substituting these values in Eq. (4):

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

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

Equation (4) is not satisfied therefore the ODE is not exact.

But it does satisfy Eq.(5), which is shown below

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

$$ \ 0 = 0 $$

But in order for the ODE to be exact it must satisfy both Eq.(4) and Eq.(5), since it doesn't satisfy equ.(4) it is concluded that the ODE is not exact.