Answer to Question #122447 in Differential Equations for Joseph Se

Question

In this problem, y = 1/(x2 + c) is a one-parameter family of solutions of the first-order
DE y' + 2xy2 = 0.Find a solution of the first-order IVP consisting of this differential equation and the given initial condition.

y(3) = 1/5

y = ___

Give the largest interval I over which the solution is defined.

A (0, infinity)
B. (2, infinity)
C. ( - infinity, infinity)
D. (0, 2)
E. (-2, infinity)

Answer 1

Given,

$$y=\frac{1}{x^2+c}$$

is family solution of

$$y'+2xy^2=0$$

Initial conditions is $y(3)=\frac{1}{5}$ ,Thus, from General solution we get,

$$\frac{1}{5}=\frac{1}{(3^2+c)}\implies c=-4$$

Thus, particular solution will be

$$y(x)=\frac{1}{x^2-4}$$

Clearly, $y$ is not defined at $x=\pm 2$ , thus larges interval in which solution is defined is

$$(-\infty,\infty)\backslash\{-2, 2\}$$

Acceptable given option will be (B).