Question

Let x be the particular solution

Answer 1

`f(x)=\sqrt{(1)/(4x^2-3)}` for
`(-\infty, -\sqrt {\frac 3 4 })` and
`(\infty, \sqrt {\frac 3 4 })` .

Explanation:

The given differential equation is
`(dy)/(dx)=-4xy^3`

Rearranging it as

`y^(-3)dy=-4xdx \\\\(y^(-3+1))/(-3+1)=-4(x^2)/(2)+C \\\\(y^(-2))/(2)=4(x^2)/(2)+C \\\\y^(-2)=4x^2+C\cdots(i)`

where C is constant.

The initial condition is f(-1)=-1, i.e for x=-1, y=-1.

So, by using the initial condition to get the value of constant:

`(-1)^(-2)=4(-1)^2+C \\\\1=4+C \\\\C=1-4=-3`

From equation (i), we have

`y^(-2)=4x^2+(-3) \\\\y^2=(1)/(4x^2-3) \\\\y=\sqrt{(1)/(4x^2-3)} \\\\`

As y=f(x), so the required particular solution

`f(x)=\sqrt{(1)/(4x^2-3)}`

Now, determining the domain of the function

As for a real value output, the term under square root must be positive.

So,

`(1)/(4x^2-3) >0 \\\\\Rightarrow 4x^2-3 >0 \\\\\Rightarrow x^2 > \frac 3 4 \\\\\Rightarrow |x|>\sqrt {\frac 3 4 } \\\\\Rightarrow x> \sqrt {\frac 3 4 } \; and \; x< -\sqrt {\frac 3 4 }`

Therefore, the domain of the function is

`(-\infty, -\sqrt {\frac 3 4 }) \;and\; (\infty, \sqrt {\frac 3 4 })`

Hence,
`f(x)=\sqrt{(1)/(4x^2-3)}` for
`(-\infty, -\sqrt {\frac 3 4 })` and
`(\infty, \sqrt {\frac 3 4 })` .