Differential Equation problem, I think i solved it, just need it to be check

[Pr0x1mo]

My teacher gave me this as an extra credit question:

solve: (x+y)dx-(x-y)dy=0

I first used the substitution of y=vx or v=y/x

Taking the partial derivative of y = vx yields:

dy/dx = v + x(dv/dx)

So then i rearranged the equation like: dy/dx = (x+y)/(x-y) which equals:

v + xv' = (x+y)/(x-y)

Then i divided everything on the right side by x to obtain:

v + xv' = (1 + (y/x))/(1 - (y/x)), and since y/x = v then i get:

v + xv' = (1 + v)/(1 - v)

I then multiply both sides of the equation by (1 - v) to get:

v - xvv' = 1 + v

-xvv' = 1

-vv' = 1/x which is really:

-v (dv/dx) = 1/x, so to separate i mutliply both sides by dx:

-v dv = 1/x dx

then integrate both sides of the equation:

-1/2 v² = ln|x| + c

and since v = y/x i get:

-1/2(y/x)² = ln|x| + c

Is this correct, or am i completely off track?

 

[Quinzio]

I haven't followed your text,
but I'd try going in polar coordinates. Looks promising.

$$(x+y)dx+(x-y)dy =0\\<br />$$

$$<br /> (rcos\theta+rsin\theta)(drcos\theta-rsin\theta d\theta)+(rcos\theta-rsin\theta)(drsin\theta+rcos\theta d\theta)=0 \\$$

$$((cos\theta)^2-(sin\theta)^2+2sin\theta cos\theta)rdr + r^2((cos\theta)^2-(sin\theta)^2-2sin\theta cos\theta)d\theta =0\\$$

$$<br /> ((cos\theta)^2-(sin\theta)^2+2sin\theta cos\theta)rdr + r^2((cos\theta)^2-(sin\theta)^2-2sin\theta cos\theta)d\theta=0<br />$$

 

[Quinzio]

$$<br /> \frac{dr}{d\theta} = r\frac{1-sin4\theta}{cos4\theta}\\<br />$$

I get this but I don't know what to do next.