User:Egm6321.f2010.team5/hw2p1

=Problem 1 - Verify that the given equation is a non-linear 1st-order ODE=

Given

 * $$\displaystyle M(x,y) + N(x,y)\underbrace{\frac{dy}{dx}}_{y'} = 0$$

Find
Verify that the given equation is an N1-ODE.

Solution
We first recognize that the given equation has 1 independent variable and its derivatives are with respect to that variable; we therefore determine that the given equation is an ordinary differential equation (ODE).

Secondly, the highest order derivative found in the given equation is 1; the given equation is thus a 1st-order ODE.

Linearity
Lastly, we check the linearity of the given equation. A linear equation $$F(y)$$ will satisfy the following two equalities:
 * 1) $$\displaystyle F(u+v) = F(u) + F(v)$$
 * 2) $$\displaystyle F(\alpha u) = \alpha F(u)$$

First condition
Let $$F(y)$$ equal the given equation,
 * $$\displaystyle F(y) := M(x,y) + N(x,y)y' = 0$$

The left side yields,
 * $$\displaystyle F(u+v) = M(x,u+v) + N(x,u+v)(u+v)'$$

The right side yields,
 * $$\displaystyle F(u) + F(v) = M(x,u) + N(x,u)u' + M(x,v) + N(x,v)v'$$

Thus,
 * $$\displaystyle F(u+v) \neq F(u) + F(v)$$

Second condition
The left side yields,
 * $$\displaystyle F(\alpha u) = M(x,\alpha u) + N(x,\alpha u)(\alpha u)'$$

The right side yields,
 * $$\displaystyle \alpha F(u) = \alpha (M(x,u) + N(x,u)u')$$

Thus,
 * $$\displaystyle F(\alpha u) \neq \alpha F(u)$$

The math above proves the non-linearity of the given equation. Thus, from the entire discussion above we can conclude that:


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The given equation is a non-linear 1st-order ODE.
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 * }

--Egm6321.f2010.team5.riveros 15:34, 21 September 2010 (UTC)