Question

Y = 1/(x² + c)

Answer 1

Answer:

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

$x\in (1,\infty)$

Step-by-step explanation:

We are given that one parameter family of solution of the first order DE

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

DE

$y'+2xy^2=0$

$y(2)=\frac{1}{3}$

Substitute x=2

$\frac{1}{3}=\frac{1}{2^2+c}$

$\frac{1}{3}=\frac{1}{4+c}$

$4+c=3$

$c=3-4=-1$

Substitute the value of c

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

Solution is defined for all  values of x except x=$\pm 1$

Therefore, by unique existence theorem

The largest interval on which the solution is defined

$x\in (1,\infty)$