Answer with Step-by-step explanation:
We are given that a differential equation
[tex]y'=2xy^2[/tex] and [tex]y=\frac{1}{16-x^2}=(16-x^2)^{-1}[/tex]
We have to verify that the function y=[tex]\phi(x)[/tex] is an explicit function of the given first order differential equation.
Differentiate w.r.t x
[tex]y'=-(16-x^2)^{-2}(-2x)[/tex]
By using the formula
[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
[tex]y'=\frac{2x}{(16-x^2)^2}[/tex]
Substitute the value of y in given differential equation
[tex]y'=2x\times (\frac{1}{16-x^2})^2[/tex]
[tex]y'=\frac{2x}{(16-x^2)^2}[/tex]
LHS=RHS
Hence, the function y is an explicit function of the given first order differential equation.
