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

Answer :

lublana

Answer:

[tex]y(x)=\frac{1}{x^2-1}[/tex]

[tex]x\in (1,\infty)[/tex]

Step-by-step explanation:

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

[tex]y=\frac{1}{x^2+c}[/tex]

DE

[tex]y'+2xy^2=0[/tex]

[tex]y(2)=\frac{1}{3}[/tex]

Substitute x=2

[tex]\frac{1}{3}=\frac{1}{2^2+c}[/tex]

[tex]\frac{1}{3}=\frac{1}{4+c}[/tex]

[tex]4+c=3[/tex]

[tex]c=3-4=-1[/tex]

Substitute the value of c

[tex]y(x)=\frac{1}{x^2-1}[/tex]

Solution is defined for all  values of x except x=[tex]\pm 1[/tex]

Therefore, by unique existence theorem

The largest interval on which the solution is defined

[tex]x\in (1,\infty)[/tex]

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