- #1

Pr0x1mo

- 21

- 0

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 v

^{2}= 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?