- #1
Pr0x1mo
- 21
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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?
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?