Solve Homogeneous D.E. need help integrating

1. Mar 5, 2014

Jtechguy21

1. The problem statement, all variables and given/known data

Dy/Dx = (Y-x)/(Y+x)

2. Relevant equations

Y=ux
dy=udx+xdu

3. The attempt at a solution

Dy/Dx = (Y-x)/(Y+x)

Plug in my substitutions
udx+xdu(1/dx)=(ux/ux+x) - X/(ux+x)

Simplify
u+x(du/dx)=(ux)/x(u+1) - (x)/((x)(u+1))

u+x(du/dx)=u/(u+1) -(1)/(u+1))

u+x(du/dx)=u-1/(u+1)

This is where I think i begin to mess up

u+du=(u-1)/(u+1) dx/x

substract (u-1)/(u+1) to the other side

u-(u-1)/(u+1) du=dx/x

I know the right side integrates to Lnx +c

but on the left side if i do
(u^2-1)/(u+1)

I split it up into

the integral (u^2)/(u+1) minus integral of 1/(u+1)
(u^2)/(u+1)<-use long division

I get u+(1/u+1) minus the integral of 1/(u+1)
i am left with just the integral
of u

u^2/2= lnx+c

plug u back in.

((y/x)^2)/2 =lnx +c

is this sufficient of an answer?

according to the answer key im gonna end up with the arctan somewhere in my answer. so i may have already messed up :(

Last edited: Mar 6, 2014
2. Mar 6, 2014

pasmith

Why are you dividing by $y + x$ here? Is your equation actually
$$\frac{dy}{dx} = \frac{y - x}{y + x}$$
and not
$$\frac{dy}{dx} = (y - x)(y + x)$$
as you have written?

This should be "u dx/x + du" on the left hand side.

You can't; it's multiplied by dx/x.

What you have after replacing $y$ is
$$x\frac{du}{dx} + u = \frac{u - 1}{u + 1}$$
Subtracting $u$ from both sides and then dividing by $x$ puts this in the separable form
$$\frac{du}{dx} = \frac1x \left(\frac{u-1}{u+1} - u\right)$$
Continue.

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