Use the method discussed under “Bernoulli Equations” to solve problems21-28. \

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

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 P(x) and Q(x) are continuous on an interval (a, b) and n is a real number, is called a Bernoulli equation

Given:

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{y}}{\mathbf{x}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$

Compare with the general form of the Bernoulli equation.

n = 2, $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}$, and $\mathbf{Q}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$.

Now, divide by, we get,

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

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)

$\mathbf{-}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{v}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}$

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

Now, integrate the P(x) first. Where $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}$.

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

Then,

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

Multiply $\mu \left(\mathbf{x}\right)$with equation (2).

$\begin{array}{c}\frac{\mathbf{1}}{\mathbf{x}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{v}\mathbf{=}\mathbf{-}\mathbf{x}\\ \frac{\mathbf{d}}{\mathbf{dx}}\left[\frac{\mathbf{1}}{\mathbf{x}}\mathbf{v}\right]\mathbf{=}\mathbf{-}\mathbf{x}\end{array}$

Integrate both sides,

data-custom-editor="chemistry" $\begin{array}{c}\int \frac{\mathbf{d}}{\mathbf{dx}}\left[\frac{\mathbf{1}}{\mathbf{x}}\mathbf{v}\right]\mathbf{dx}\mathbf{=}\mathbf{-}\int \mathbf{x}\mathbf{dx}\\ \frac{\mathbf{1}}{\mathbf{x}}\mathbf{v}\mathbf{=}\mathbf{-}\int \mathbf{x}\mathbf{dx}\\ \frac{\mathbf{1}}{\mathbf{x}}\mathbf{v}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{v}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{2}}\mathbf{+}\mathbf{x}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{x}{\mathbf{C}}_{\mathbf{1}}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}{\mathbf{2}}\\ \mathbf{=}\frac{\mathbf{Cx}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}{\mathbf{2}}\end{array}$

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

$\begin{array}{c}{\mathbf{y}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{Cx}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}{\mathbf{2}}\\ \mathbf{y}\mathbf{=}{\left(\frac{\mathbf{Cx}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}{\mathbf{2}}\right)}^{\mathbf{-}\mathbf{1}}\\ \mathbf{y}\mathbf{=}\frac{\mathbf{2}}{\mathbf{Cx}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}\end{array}$

Hence the solution is$\mathbf{y}\mathbf{=}\frac{\mathbf{2}}{\mathbf{Cx}\mathbf{-}{\mathbf{x}}^{\mathbf{3}}}$