Differential equation:
first, separate C:
c = y + 2x
Differentiate:
0 = y' + 2
y' = -2
Slope must be negative reciprocal. therefore m = 1/2;
dy/dx = 1/2
S2dy = Sdx (S - integration sign)
2y = x + C
C = 2y - x;
Thanks,
The rest of my question states
Find the differential equation of the family of curves and of the
orthogonal trajectories. Then solve the differential equation to find the
formula of the orthogonal trajectories.
I have to find these things below and I dont have to graph it.
Given the family
y = c - 2x
Differential equation of the family?
y' = 2
Differential equation of the orthogonal trajectories?
y' = -1 / 2
Orthogonal trajectories?
y = -x / 2 + c
This our my answers above
I think it is correct but I want to make sure
ok, the only reason I had positive 2 is because the given family is Y = c - 2x.
When you showed me the steps you had
Differential equation:
first, separate C:
c = y + 2x
Does that change anything in the answer I had since it is y = c - 2x instead of plus???
Sorry, but I don't quite undrstand what you mean. The equation is y = c - 2x, how could you have positive 2 in there? When you differentiate you get y' = -2, and that is the slope of that family. If you graph both families that might help. The orthogonal family must have a negative reciprocal slope because by deinition its curves intersect the original family at right angles. Sorry, if that doesn't help, ask again.