User:Egm6321.f10.team5.steinberg/hw3

Given
A second order nonlinear ordinary differential equation (N2-ODE) may be represented as follows (First Cond. of Exactness):

$$ F(x,y,y',y) = g(x,y,p) + f(x,y,p)y = 0 \! $$

where $$ p = y' \! $$

If both the first and second conditions of exactness are satisfied,

Find
Define the method for determining $$ h(x,y) \! $$ such that for the case when $$ h(x,y) \not= const \! $$

Solution
Using Eq. (8.3), the following partial derivatives can be obtained

Recalling the definition of $$ g \! $$ given in Eq. (8.1) and substituting (8.4) and (8.5) into (8.1) yields

Because we already know $$ g \! $$ from the N2-ODE given, we can set like terms equal in order to solve for $$ h(x,y) \! $$.

For example, should (8.6) simplify into the following form,

$$ ay' + b = h_x + h_y y' \! $$

then we may set $$ h_x = b \! $$ and $$ h_y = a \! $$ and solve for $$ h(x,y) \! $$.

Given
$$ a(x) = sin(x^3) \! $$

$$ b(x) = cos(x) \! $$

$$ c(y) = exp(2y) \! $$

Find
1. Using $$ \bar b(x)c(y)y'+a(x) \bar c(y) = 0 \! $$, find a N1-ODE that is either exact or can be made exact.

2. Find $$ \phi(x,y) = k \! $$.

Solution
$$ \bar b(x) = \int^x b(s) ds = \int^x cos(s) ds = sin(x) \! $$

$$ \bar c(y) = \int^y c(s) ds = \int^y exp(2s) ds = \frac{1}{2} exp(2y) \! $$

Substituting these along with $$a(x) \!$$ and $$c(y) \!$$ into the given equation yields the following N1-ODE:

$$ \frac{1}{2} sin(x^3)exp(2y) + sin(x)exp(2y)y' = 0 \! $$

In this form, the N1-ODE is not exact since the second condition of exactness is not satisfied i.e. $$ \frac{\partial M}{\partial y} \not= \frac{\partial N}{\partial x} $$.

However, we may employ the Euler integrating factor method to force the equation to be exact.

$$ hM + hNy' = 0, \bar M_y = \bar N_x \! $$

$$ \bar M_y = h_y M + h M_y \! $$

$$ \bar N_x = h_x N + h N_x \! $$ $$ h_x N - h_y M + h(N_x - M_y) = 0 \! $$

$$ M = \frac{1}{2} sin(x^3) exp(2y) \! $$

$$ N = sin(x) exp(2y) \! $$

$$ M_y = sin(x^3) exp(2y) \! $$

$$ N_x = cos(x) exp(2y) \! $$

Let $$ h_x = 0 \!$$

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

$$ = exp \Bigg ( \int \frac{2}{sin(x^3)exp(2y)} exp(2y) [ cos(x) - sin(x^3) ] dy \Bigg) $$

$$ = exp\Bigg (y \bigg [\frac{cos(x)}{sin(x^3)}-1 \bigg] \Bigg) $$

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