Differential Equation

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.

Solution

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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 =

e2x; M = e2x. Multiplying equation by M: d/dx(ve2x) = e2x(5 + 2x); ve2x = ∫e2x(5 + 2x)dx; ve2x = (5 + 2x)e2x/2 –

e2x/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: (- ∞, ∞).