User:Egm6321.f09.team6.breaux/hw2

HW p.7-1

Given the Equation hxN-hyM+h(Nx-My)=0

Consider the case where hxN=0

Since N can not be 0, then h(y)

So you get -hyM+h(Nx-My)=0

Rearranging you are left with hy/h=(1/M)(Nx-My)=-g(y)

HW p.8-2

Given the Equation y'+(1/x)y=x2

Solve for h(x), where h(x)=

$$\int$$a0dx

and a0=(1/x)

So h(x)=

$$\int$$1/x=elnx=x

Therefore h(x)=x

HW p.8-3

Given the equation hy'+h'y=hb where

h=x and b=x2

The resulting equation is xy'+y=x3

which can also be written as d/dx(xy)=x3

One can now find y(x)=

(1/x)$$\int(x)$$3

=1/x(x4/4+C)

So y(x)=x3/4+C/x