differential equation

Question

Let y≠0 for the differential equation

2ydy ⁄dx = 5 + 2x − 2y²

(a) Find the implied general solution.

(b) Find the particular solution that passes through the point (0,√5) and the maximum range where it is

defined.

Answer 1

1.Let rewrite this equation: dy/dx + y = (5/2 + x)y^-1. This is Bernoulli ODE with n = - 1. Substitution: v = y²;

dv/dx = 2ydydx; dy/dx = (dv/dx)/(2y).

(dv/dx)/(2y) + y = (5/2 + x)y^-1; dv/dx + 2v = 5 + 2x. Now we have linear equation. Integrated factor M = e^∫2dx =

e^2x; M = e^2x. Multiplying equation by M: d/dx(ve^2x) = e^2x(5 + 2x); ve^2x = ∫e^2x(5 + 2x)dx; ve^2x = (5 + 2x)e^2x/2 -

e^2x/2 + C; v = (5 + 2x)/2 - 1/2 + Ce^-2x = 2 + x + Ce^-2x; y² = 2 + x + Ce^-2x;

2.If x =0 then y = √5; 5 = 2 + 0 + C, C = 3. So, y² = 2 + x + 3e-^2x_..

Maximum interval where y is exist we get from inequality 2 + x + 3e^-2x ≥ 0. As show graf of y²(TI- 84) it works for all real numbers: (- ∞, ∞).