Question

Let x be the particular solution

Answer 1

Answer:

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

Step-by-step explanation:

The given differential equation is $\frac{dy}{dx}=-4xy^3$

Rearranging it as

$y^{-3}dy=-4xdx \\\\\frac{y^{-3+1}}{-3+1}=-4\frac{x^2}{2}+C \\\\\frac{y^{-2}}{2}=4\frac{x^2}{2}+C \\\\y^{-2}=4x^2+C\cdots(i)$

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

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

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

$f(x)=\sqrt{\frac{1}{4x^2-3}}$

Now, determining the domain of the function

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

So,

$\frac{1}{4x^2-3} >0 \\\\\Rightarrow 4x^2-3 >0 \\\\\Rightarrow x^2 > \frac 3 4 \\\\\Rightarrow |x|>\sqrt {\frac 3 4 } \\\\\Rightarrow x> \sqrt {\frac 3 4 } \; and \; x< -\sqrt {\frac 3 4 }$

Therefore, the domain of the function is

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

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