Question

The differential equation $x(y\text{'}{)}^2-4y\text{'}-12x^3=0$has the form given in (4). Determine whether the equation can be put into the normal form $dy/dx=f(x,y)$.

Short answer

The original differential equation breaks down into two different differential equations, it can not be put into the normal form.

Step-by-step solution

Definition of differential equation.

A differential equation is defined as the derivative function of one or more unknown functions (dependable variables) with respect to one or more undependable variables.

Quadratic equation form

The differential equation,

$x(y\text{'}{)}^2-4y\text{'}-12x^3=0$is in the quadratic equation form and so, we can use the quadratic formula to find$y\text{'}$as,

$$\begin{gathered} y\text{'}=\frac{4\pm \sqrt{4^2-4\left(x\right)(-12x^3)}}{2x} \\ y\text{'}=\frac{4\pm \sqrt{16+48x^4}}{2x} \end{gathered}$$

Simplify,

$y\text{'}=\frac{2\pm 2\sqrt{1+3x^4}}{x}$

Hence, we get the differential equations,

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

Or,

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

Both of which are of the form$f(x,y)$.

But, since the original differential equation breaks down into two different differential equations, it can not be put into the normal form.