User:Egm6321.f09.team3.diggs/HW1

1. Complete Case 2 details (7-1) to obtain expression for h as a function of g(y).
$$h_xN - h_yM + h(N_x - M_y) = 0 $$ may be used to solve for the integration factor h(x,y). Case 1, derived and discussed in class, is if $$ h_yM = 0 $$. Case 2, derived and discussed below, is if $$ h_xN = 0 $$

For $$ h_xN = 0 $$, $$ -h_yM + h(N_x - M_y) = 0 $$

rearrange to get: $$ \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) $$

integrate to get: $$ ln(h) = \int - \frac{1}{M} (N_x - M_y) dy $$

which yields, $$ h(x,y) = exp(- \int \frac{1}{M}(N_x - M_y) dy) $$

The case $$ h_xN = 0 $$ implies $$h_x = 0 $$ which implies h(y) only!

therefore, $$ \frac{-1}{M(x,y)}[\partial {N(x,y)}{x} - \partial{M(x,y)}{y}] = g(y) $$ only

if the equation above is satisfied, then $$ h(y) = exp[\int g(s) ds] $$

2. & 3. Given $$ y'+ \frac{1} {x} y = x^2 $$ where $$ a_0 = \frac{1} {x} $$ and $$ b = x^2 $$, find h(x) and solve for y(x).
Recall, $$ h(x) = exp( \int a_0(s) ds) $$, which yields $$ h(x) = exp(\int \frac{1}{s} ds = exp(ln(x)) = x $$, as expected from class notes.

Additionally, the solution to the ODE is obtained by:

multiplying h(x) to the ODE: $$ hy' + ha_0y = hb $$

which is simplified as: $$ (hy)' = hb = x \cdot x^2 = x^3 $$ (once it is recognized $$ha_0 = h'$$)

integrate both sides to get: $$hy = \frac{x^4}{4} + C$$

and solve for y: $$y(x) = \frac{x^3}{4} + \frac{C}{x}$$, as expected from class notes.

4. Given the form of an exact, linear, 1st order ODE as $$ \bar{b}(x)y' + [a(x)y + k] = 0 $$, where $$a(x) = x^4$$, $$b(x) = x$$, $$\bar{b}(x) = \frac{x^2}{2}$$, and $$k=10$$, show the ODE is exact.
ODE is: $$ \frac{x^2}{2}y' + [x^4y+10] = 0 $$

where $$M(x,y) + N(x,y)\cdot y' = 0$$ is the first condition of exactness, and the second condition of exactness is $$M_y = N_x$$

Here, $$M(x,y) = x^4y + 10$$ and $$N(x,y) = \frac{x^2}{2}$$

$$ M_y = x^4 $$ and $$ N_x = x $$

$$ M_y ~= N_x $$, therefore the equation is not exact.

The ODE is not exact, but can be solved bu integration factors method.

ODE is: $$ y' + 2x^2y = -20/x $$

here, $$ a_0 = 2x^2 $$ and $$ b = -\frac{20}{x} $$

$$ h(x) = exp \int a_0(s) ds = exp \int 2s^2ds = exp[ \frac{2}{3} x^3] $$

Multiply both sides of the ODE by the integrating factor: $$ hy' + ha_0y + hb $$

which is: $$ (yh)' = hb = exp[\frac{2}{3} x^3 (-\frac{20}{x}) $$

$$ hy = \int (\frac{2}{3} s^3)(-\frac{20}{s}) $$

solving for y, FINISH THE INTEGRAL!!

== 5. Given the form of an exact, linear, 1st order ODE as $$ a(x) \bar{c}(y) + \bar{b} (x) c(y) y' = 0 $$, where $$a(x) = 5x^3+2$$, $$b(x) = x^2$$, $$\bar{b}(x) = \frac{x^3}{3}$$, $$ c(y) =y^4 $$ and $$\bar{c}(y) = \frac{y^5}{5}$$, show the ODE is exact. ==

ODE is: $$(5x^3+2) \frac {y^5} {5} + \frac{x^3}{3} y^4 y' = 0 $$

where $$M(x,y) + N(x,y)\cdot y' = 0$$ is the first condition of exactness, and the second condition of exactness is $$M_y = N_x$$

Here, $$M(x,y) = (5x^3+2)\frac{y^5}{5} $$ and $$N(x,y) = \frac{x^3}{3}y^4$$

$$ M_y = 5x^3y^4 + 2y^4 $$ and $$ N_x = \frac{2}{3}x^2y^4 $$

$$ M_y ~= N_x $$, therefore the equation is not exact.

The ODE is not exact, but can be solved using integrating factor method.

The ODE is: $$ y' + (3 + \frac{6}{5x^3})y = 0 $$

where $$ a_0 = 3 + \frac{6}{5x^3} $$ and $$ b = 0 $$

$$ h(x,y) = exp \int a_0(s)ds = exp \int 3 + \frac{6}{5s^3} ds $$

$$ h(x.y) = exp(3x - \frac{3}{5x^2}) $$

Multiply both sides of the ODE by h(x) to solve, of the form:

$$ hy' + ha_0y = 0 $$

which is $$ (hy)' = 0 $$

integrate to get: $$ hy = C $$ where C is a constant

$$ y = \frac{C}{h} = \frac{C}{exp(3x - \frac{3}{5x^2}} $$

The solution is $$ y(x) = \frac{C}{exp[3x - \frac{3}{5x^2}} $$, where C is a constant

6. Is the following ODE exact? $$ xyy'' + x(y')^2 + yy' = 0 $$
The first condition is satisfied, as $$ f(x,y,p) = xy $$ and $$ g(x,y,p) = xp^2 + yp $$.

The second exactness conditions are:

(i). $$ f_{xx} = 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y $$

(ii). $$ f_{xp} + pf_{yp} + 2f_y = g_{pp} $$

Find the partial derivatives: $$ f_x = y $$

$$ f_{xx} = 0 $$

$$ f_{xy} = 1 $$

$$ f_{yy} = 0 $$

$$ f_p = f_{py} = f_{px} = f_{py} = f_{pp} = 0 $$

$$ g_x = p^2 $$

$$ g_{xy} = 0 $$

$$ g_{xp} = 2p $$

$$ g_y = p $$

$$ g_{yp} = 1 $$

$$ g_{yy} = g_{yx} = 0 $$

$$ g_p = 2xp + y $$

$$ g_{py} = 1 $$

$$ g_{px} = 2p $$

$$ g_{pp} = 2x $$

(i). $$ 0 + 2p + p^2 \cdot 0 = 2p + p - p $$, satisfied!

(ii). $$ 0 + p \cdot 0 + 2x = 2x $$, satisfied!

Yes, the ODE is exact!

7. Derive $$ f_{xy} + pf_{yp} + 2f_y = g_{pp} $$ (10-2, Eqn 5)
Differentiate $$ g = \phi _x + \phi _y p $$ with respect to p

$$ g_p = \phi _{xp} + \phi _{yp} p + \phi _y $$

Recall, $$ f = \phi _p $$, and $$ g = \phi _x + \phi_y p $$

$$ g_{pp} = \phi _{xpp} + \phi _{ypp}p + \phi _{yp} + \phi _{yp} $$

$$ g_{pp} = (\phi _p) _{xpp} + (\phi _p) _{yp}p + 2 (\phi _p)_y $$

$$ g_{pp} = f_{xp} + f_{yp}p + 2f_y $$, which is the equation above.

8. Use $$ \phi _{xy} = \phi _{yx} $$ to obtain $$ f_{xx} + 2pf_{xy} + p^2 f_{yy} = g_{xp} + pg_{yp} - g{y} $$ (10-2, Eqn 4)
$$ \phi _{xy} = \phi _{yx} $$ is also $$ (\phi _x)_y = (\phi _y)_x $$

$$ \phi_x = g - p (g_p) - f_x) + p^2f_y $$ $$ \phi_y = g_y = f_yp - f_x $$

$$ [\phi_x = g - p (g_p) - f_x) + p^2f_y]_y = (\phi_y = g_y = f_yp - f_x)_x $$

$$ g_y - p(g_{py} - f_{xy}) + p^2f_{yy} = g_{px} - f_{yx}p - f_{xx} $$ which is $$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{px} + pg_{yp} - g_y $$, as expected.

9. Show $$ (8x^5y')'' + 2x^2y' + 20x^4(y')^2 + 4xy = 0 $$ is exact by condition 2
$$ f(x,y,p) = 8x^5p $$ $$ g(x,y,p) = 2x^2 + 20x^4p^2 + 4xy $$

Take partial derivatives:

$$ f_x = 40x^4p $$

$$ f_{xx} = 160x^3p $$

$$ f_{xy} = 0 $$

$$ f_{xp} = 40x^4 $$

$$ f_y = f_{yy} = f_{yp} = 0 $$

$$ f_p = 8x^5 $$

$$ f_{pp} = 0 $$

$$ g_x = 4xp + 80x^3p^2 + 4y $$

$$ g_{xx} = 4p + 240x^2p$$

$$ g_{xy} = 4$$

$$ g_{xp} = 4x + 160 x^3 p $$

$$ g_y = 4x $$

$$ g_{yy} = 0 $$

$$ g_{yp} = 0 $$

$$ g_p = 2x^2 + 40x^4p $$

$$ g_{pp} = 40x^4 $$

(i). $$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y $$ $$ 160x^3p + 2p \cdot 0 + p^2 \cdot 0 = 4x + 160 x^3p + p \cdot 0 - 4x $$, which is satisfied.

(ii). $$ f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ $$ 40x^4 + p \cdot 0 + 2 \cdot 0 = 40x^4 $$, which is satisfied.

The equation is exact.

10. Verify exactness for $$ \sqrt {x} y'' + 2xy' + 3y = 0 $$
where $$ f(x,y,p) = \sqrt {x} $$ and $$ f(x,y,p) = 2xp + 3y $$

satisfies condition 1 of exactness.

Take the partial derivatives:

$$ f_x = \frac {1} {2 \sqrt{x}} $$

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

$$ f_{xy} = f_{xp} = 0 $$

$$ f_y = f_{yp} = f_{yy} = f_p = f_pp = 0 $$

$$ g_x = 2p $$

$$ g_{xx} = g_{xy} = 0 $$

$$ g_{xp} = 2 $$

$$ g_y = 3 $$

$$ g_{yp} = g_{yy} = 0 $$

$$ g_p = 2x $$

$$ g_{pp} = 0 $$

Exactness Conditions

(i). $$f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_y$$

$$ \frac {-1}{4x^{-3/2}} + 2p \cdot 0 + p^2 \cdot 0 = 2 + p \cdot 0 - 3 $$, which does not satisfy the condition.

(ii). $$ f_{xp} + pf_{yp} + 2f_y = g_{pp} $$

$$ 0 + p \cdot 0 + 2 \cdot 0 = 0 $$, which does satisfy the condition.

Since both equations do not satisfy the condition, the equation is not exact.