Question

Use the method discussed under “Bernoulli Equations” to solve problems.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{2}\mathbf{y}}{\mathbf{x}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$

Short answer

Equation of the form of Bernoulli equation for the given equation is $\mathbf{y}\mathbf{=}\frac{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^5\mathbf{+}\mathbf{C}}$.

Step-by-step solution

General form of Bernoulli equation

Bernoulli’s equation

A first-order equation that can be written in the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{y}}^{\mathbf{n}}$, where$\mathbf{P}\left(\mathbf{x}\right)$ and$\mathbf{Q}\left(\mathbf{x}\right)$ are continuous on an interval$\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)$ and n is a real number, is called a Bernoulli equation.

Evaluate the given equation

Given, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{2}\mathbf{y}}{\mathbf{x}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$.

Compare with the general form of the Bernoulli equation.

n = 2,$\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}$, and $\mathbf{Q}\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}$.

Now, divide by ${\mathbf{y}}^{\mathbf{2}}$, we get,

${\mathbf{y}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}······\left(1\right)$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}$.

Differentiate with respect to x.

$\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}\mathbf{-}{\mathbf{y}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{dy}}{\mathbf{dx}}$

Substitute it on equation (1)

$\mathbf{-}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{v}\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}$

$\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{x}}\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}······\left(2\right)$

Integrate the equation

Now, integrate $\mathbf{P}\left(\mathbf{x}\right)$the first. Where $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\frac{2}{\mathbf{x}}$.

$\begin{array}{c}\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}\mathbf{=}\mathbf{2}\int \frac{\mathbf{1}}{\mathbf{x}}\mathbf{dx}\\ \mathbf{=}\mathbf{2}\mathbf{ln}\left|\mathbf{x}\right|\end{array}$

Then,

$\begin{array}{c}\mu \left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{ln}\left|\mathbf{x}\right|}\\ \mathbf{=}{\mathbf{x}}^{\mathbf{2}}\end{array}$

Multiply with equation (2).

$\begin{array}{c}{\mathbf{x}}^{\mathbf{2}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{x}}\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{4}}\\ \frac{\mathbf{d}}{\mathbf{dx}}\left[{\mathbf{x}}^{\mathbf{2}}\mathbf{v}\right]\mathbf{=}{\mathbf{x}}^{\mathbf{4}}\end{array}$

Integrate both sides,

$\begin{array}{c}\int \frac{\mathbf{d}}{\mathbf{dx}}\left[{\mathbf{x}}^{\mathbf{2}}\mathbf{v}\right]\mathbf{dx}\mathbf{=}\int {\mathbf{x}}^{\mathbf{4}}\mathbf{dx}\\ {\mathbf{x}}^{\mathbf{2}}\mathbf{v}\mathbf{=}\int {\mathbf{x}}^{\mathbf{4}}\mathbf{dx}\\ {\mathbf{x}}^{\mathbf{2}}\mathbf{v}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{5}}}{\mathbf{5}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{v}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{=}\frac{{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{5}{\mathbf{C}}_{\mathbf{1}}}{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}\\ \mathbf{=}\frac{{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{C}}{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}\end{array}$

Substitution method

Substitute . $\mathbf{v}\mathbf{=}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}$

$\begin{array}{c}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{C}}{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}\\ \mathbf{y}\mathbf{=}{\left(\frac{{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{C}}{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}\right)}^{\mathbf{-}\mathbf{1}}\\ \mathbf{y}\mathbf{=}\frac{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^5\mathbf{+}\mathbf{C}}\end{array}$

Hence the solution is$\mathbf{y}\mathbf{=}\frac{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^5\mathbf{+}\mathbf{C}}$