Question

We saw that a mathematical model for the velocity v of a chain slipping off the edge of a high horizontal platform is

$\mathit{x}\mathit{v}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathit{v}}^{\mathbf{2}}\mathbf{=}\mathbf{32}\mathit{x}\mathbf{.}$

In that problem you were asked to solve the DE by converting it into an exact equation using an integrating factor. This time solve the DE using the fact that it is a Bernoulli equation.

Short answer

The equation is$v=8\sqrt{\frac{x}{3}-\frac{9}{x^2}}$

Step-by-step solution

Bernoulli’s equation

We know that a Bernoulli's equation is of the form

$\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)y^n$

where n is any real number. We substitute u = y^1-n to reduce any Bernoulli's equation to a linear equation.

For the given exercise, we first rewrite the given DE in Bernoulli's equation form by dividing the equation by x v.

$$\begin{gathered} xv\frac{dv}{dx}+v^2=32x \\ \frac{dv}{dx}+\frac{v}{x}=32v^{-1} \end{gathered}$$

Here, the variable v is used instead of with value of n = - 1.

We substitute or $u=v^{1-\left(-1\right)}=v^{2}orv=u^{1/2}$to reduce the equation to a linear one. Then, using the chain rule, we have

$\frac{dv}{dx}=\frac{dv}{du}.\frac{du}{dx}=\frac{1}{2}{u}^{-1/2}\frac{du}{dx}$

Upon substitution into the given equation, we get

$$\begin{gathered} \frac{1}{2}{u}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{du}{dx}+\frac{u^{1/2}}{x}=32{\left(u^{1/2}\right)}^{-1} \\ =32u^{-1/2} \\ \frac{1}{2}\frac{du}{dx}+\frac{1}{x}u=32 \\ \frac{du}{dx}+\frac{2}{x}u=64 \end{gathered}$$

We have reduced the Bernoulli's equation to a linear equation now. The integrating factor for this linear equation is

$\begin{array}{rcl}{P}^{Pdx}& =& e^{\frac{2}{x}dx}\\ & =& e^{2Inx}\\ & =& e^{In{x}^2}\\ & =& x^2\end{array}$

Multiplying both sides

Multiplying both sides of the equation with the integrating factor then gives the equation

$\frac{du}{dx}{x}^2+2ux=64x^2$

which is same as

$\frac{d}{dx}\left[x^{2}u\right]=64x^2$

Upon integration, we get

$$\begin{gathered} \frac{d}{dx}\left[x^{2}u\right]=64x^2 \\ {x}^{2}u=\frac{64x^3}{3}+c \\ u=\frac{64x}{3}+\frac{c}{x^2} \end{gathered}$$

Since u = v² the solution of the given equation will be

$$\begin{gathered} v^2=\frac{64x}{3}+\frac{c}{x^2} \\ or \\ v=\sqrt{\frac{64x}{3}+\frac{c}{x^2}} \end{gathered}$$

We have considered only positive value of v because chain is falling down and it is given that the positive direction is downward.

Since, length of the overhanging chain is 3ft and the chain is at rest initially, we have the initial conditions x = 3 and v = 0.

Substituting these in the above solution then gives

$$\begin{gathered} 0=\sqrt{\frac{64\left(3\right)}{3}+\frac{c}{3^2}}=\sqrt{64+\frac{c}{9}} \\ \frac{c}{9}=-64 \\ c=-576 \end{gathered}$$

Final proof

Therefore, the solution of the given equation is

$v=\sqrt{\frac{64x}{3}-\frac{576}{x^2}}$

or equivalently

$v=8\sqrt{\frac{x}{3}-\frac{9}{x^2}}$