dy/dx = y/(x^2+4)
dy/y = dx/(x^2+4)
lny = 1/2 arctan(x/2) + c
y = c*e^(1/2 arctan(x/2)) (different c ... e^c is just a constant, so call it c)
arctan(∞ ) = π/2, so
ce^(1/2 * π/2) = 1
c = e^(-π/4)
so we could say that
y = e^(1/2 arctan(x/2) - π/4)
The slope of a curve is equal to y divided by 4 more than x2 at any point (x, y) on the curve.
A. Find a differential equation describing this curve.
B. Solve the differential equation from part A.
C. Suppose it’s known that as x goes to infinity on the curve, y goes to 1. Find the equation for the curve by using part B and determining the constant.
I got dy/dx = y/x^2+4 for A and |y| = e^c e^(arctan(x/2)/2) for B but I'm not sure
3 answers
oobleck, wouldn’t it be 1/4arctan since you factor out a 1/4 to get (x/2)^2 +1
That would make c = e^(-pi/8)
3 answers