Use the method discussed under “Bernoulli Equations” to solve problems 21-28

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

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 $\left(\mathbf{a}\mathbf{,}\mathbf{b}\right)$and is a real number, is called a Bernoulli equation.

$\begin{array}{c}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}-2{\mathbf{y}}^{-3}\frac{\mathbf{dy}}{\mathbf{dx}}\\ -\frac{\mathbf{1}}{2}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{y}}^{-3}\frac{\mathbf{dy}}{\mathbf{dx}}\end{array}$Given, $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{-}{\mathbf{y}}^{\mathbf{3}}\mathbf{x}$

Compare with the general form of the Bernoulli equation.

n = 3, $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}1$, and $\mathbf{Q}\left(\mathbf{x}\right)\mathbf{=}\mathbf{x}$.

Now, divide by ${\mathbf{y}}^3$, we get,

${\mathbf{y}}^{\mathbf{-}\mathbf{3}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{y}}^{\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{x}······\left(1\right)$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^{-2}$.

Differentiate with respect to t.
$\begin{array}{c}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}-2{\mathbf{y}}^{-3}\frac{\mathbf{dy}}{\mathbf{dx}}\\ -\frac{\mathbf{1}}{2}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{y}}^{-3}\frac{\mathbf{dy}}{\mathbf{dx}}\end{array}$

Substitute it on equation (1)

$\begin{array}{c}-\frac{\mathbf{1}}{2}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\mathbf{v}\mathbf{=}-\mathbf{x}\\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{-}\mathbf{2}\mathbf{v}\mathbf{=}\mathbf{2}\mathbf{x}······\left(2\right)\end{array}$

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

$\begin{array}{c}\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}\mathbf{=}\int -2\mathbf{dx}\\ \mathbf{=}-2\mathbf{x}\end{array}$

Then,

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

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

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

Integrate both sides,

$\begin{array}{c}\int \frac{\mathbf{d}}{\mathbf{dx}}\left[{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{v}\right]\mathbf{dx}\mathbf{=}\int \mathbf{2}\mathbf{x}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{dx}\\ {\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{v}\mathbf{=}\mathbf{2}\int \mathbf{x}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{dx}······\left(3\right)\end{array}$

Integrate $\int {\mathbf{xe}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{dx}$separately by using the parts method.

$\begin{array}{c}\mathbf{u}\mathbf{=}\mathbf{x}\\ \mathbf{du}\mathbf{=}\mathbf{1}\\ \mathbf{dw}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\\ \mathbf{w}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}}{\mathbf{2}}\end{array}$

Then, the general form of integration by parts is$\int \mathbf{u}\mathbf{dw}\mathbf{=}\mathbf{uw}\mathbf{-}\int \mathbf{w}\mathbf{du}$ .

Substitute$\int {\mathbf{xe}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{dx}$ in equation (3)

$\begin{array}{c}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{v}\mathbf{=}\mathbf{2}\int {\mathbf{xe}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{dx}\\ \mathbf{=}\mathbf{2}\left(\mathbf{-}\frac{{\mathbf{xe}}^{\mathbf{-}\mathbf{2}\mathbf{x}}}{\mathbf{2}}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}}{\mathbf{4}}\right)\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{=}\mathbf{-}{\mathbf{xe}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{-}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}}{\mathbf{2}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\\ \mathbf{v}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\mathbf{x}}\end{array}$

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^{-2}$.

$\begin{array}{c}{\mathbf{y}}^{\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\mathbf{x}}\\ \mathbf{y}\mathbf{=}\mathbf{\pm }\frac{\mathbf{1}}{\sqrt{\mathbf{-}\mathbf{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{2}\mathbf{x}}}}\end{array}$

Hence the solution is" width="9" height="19" role="math">