# User:Egm6321.f09.team5.risher

## Problem 3

Given:

$M(x,y) + N(x,y)y' =0$

If $M(x,y)=2x^2 + \sqrt{y}$

and $N(x,y)=x^5y^3$

$(2x^2+ \sqrt{y}) + x^5y^3y'=0$

Show that the above 1st order ODE is non-linear:

An ODE is linear if it has the form: an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = Q(x) (see [1])

So we should only have "x" values and coefficients in each "y" term. If the "y" terms are dependent on other "y" terms, then it cannot be linear.

By inspection, the above equation cannot be linear because the $\sqrt{y}$ and $x^5y^3$ terms make it nonlinear.

Taking the derivative with respect to "x" yields an equation with complicated "y" terms:

$\frac{d}{dx}((2x^2+ \sqrt{y}) + x^5y^3y')= \frac{d}{dx}(2x^2) + \frac{d}{dx}(\sqrt{y}) + \frac{d}{dx}(x^5y^3y') = 4x + 5x^4y^3y'$

The ODE is nonlinear

## Problem 4

Given:

$F(x,y,y') = x^2y^5 + 6(y')^2 = 0$

Show that F(x,y,y')=0 in the above equation is a nonlinear 1st order ODE:

Like in homework problem #3, an ODE is linear if it has the form: an(x)y(n) + an-1(x)y(n-1) + ... + a1(x)y' + a0(x)y = Q(x) (see [2])

So we should only have "x" values and coefficients in each "y" term. If the "y" terms are dependent on other "y" terms, then it cannot be linear.

By inspection, the above equation cannot be linear because the $y^5$ and $(y')^2$ terms make it nonlinear.

Taking the derivative with respect to "x" yields an equation a high order "y" term:

$\frac{d}{dx}(x^2y^5+6(y')^2)= \frac{d}{dx}(x^2y^5) + \frac{d}{dx}(6(y')^2) = 2xy^5$

F(x,y,y') is nonlinear