User:Egm6321.f10.team5.steinberg/hw2

Given
A first order nonlinear ordinary differential equation (N1-ODE) may be represented as follows:

$$ F(x,y,y') = M(x,y) + N(x,y)f(y') = 0 \! $$

Find
Find $$ f(y') \! $$ such that there is no analytical solution to $$ f(y') = - \frac{M}{N} \! $$ (i.e. such N1-ODE cannot be exact).

Solution
The first condition of exactness is defined as follows.

For $$ F(x,y,y') = 0 \! $$ to be exact, it must be in the form $$ F(x,y,y') = M(x,y) + N(x,y)y' = 0 \! $$.

From this condition, it is understood that if $$ f(y') \not= y' \!$$, then $$ f(y') \! $$ must be solved explicitly for $$ y' \!$$ in order to force the equation into the form shown in the first condition definition above.

For the present problem, we are interested in finding a function $$ f(y') \! $$ that cannot be solved explicitly. To state this another way, we desire $$ f(y') \!$$ such that $$ f^{-1}(y') \!$$ cannot be determined analytically.

Let $$ f(y') = y'(3+y') \!$$, for example.

$$ F(x,y,y') = M(x,y) + N(x,y)f(y') = M(x,y) + N(x,y)y'(3+y') = 0 \! $$

$$ y'(3+y') = - \frac{M}{N} \! $$

Clearly, this expression cannot be solved explicitly for $$ y' \! $$. Furthermore, the N1-ODE is not exact because this particular $$ f(y') \!$$ cannot satisfy the first condition of exactness.

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 $$ h_x = 0 \!$$

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

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

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

??????