Finding function expression Given that the function f (x) f '(x) = KF (x) + b k < 0, f (0) = 0, f' (0) = 3 is the tangent slope of (0, 0), and the asymptote of F (x) is y = 2, can we find f (x)

Finding function expression Given that the function f (x) f '(x) = KF (x) + b k < 0, f (0) = 0, f' (0) = 3 is the tangent slope of (0, 0), and the asymptote of F (x) is y = 2, can we find f (x)

Because f '(x) = KF (x) + B, and f (0) = 0, f' (0) = 3
So 3 = f '(0) = KF (0) + B = b
Because the asymptote of F (x) is y = 2, when x approaches positive or negative infinity, f '(x) = 0, f (x) = 2
F '(x) = KF (x) + 3
K = - 3 / 2
So f '(x) = - 3 / 2 * f (x) + 3
So we can write dy / DX = 3-3y / 2
So dy / (3-3y / 2) = DX
The integral on both sides gives - 2ln (3-3y / 2) / 3 = x + C (C is a constant)
Because f (0) = 0,
So C = - (2ln3) / 3
So - 2ln (3-3y / 2) / 3 = x - (2ln3) / 3
So ln (3-3y / 2) = - 3x / 2 + Ln3
So e ^ (- 3x / 2 + Ln3) = 3-3y / 2
So y = 2-2e ^ (- 3x / 2 + Ln3) / 3
Checking calculation f '(x) = - 2E ^ (- 3x / 2 + Ln3) / 3 * (- 3 / 2) = e ^ (- 3x / 2 + Ln3)
-3/2 *f(x)+3=-3/2 *（2-2e^(-3x/2 +ln3)/3）+3=e^(-3x/2 +ln3)
f（0）=2-2e^(ln3)/3=0
Because f '(x) = e ^ (- 3x / 2 + Ln3)
So f '(0) = e ^ (Ln3) = 3
So the function y = 2-2e ^ (- 3x / 2 + Ln3) / 3
If f '(x) = KF (x) + b k < 0
f（0）=0
f’（0）=3
The asymptote of F (x) is y = 2