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

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

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$\mathbf{P}\left(\mathbf{x}\right)$ 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.

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

Compare with the general form of the Bernoulli equation.

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

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

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

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^3$.

Differentiate with respect to t.

$\begin{array}{c}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}3{\mathbf{y}}^2\frac{\mathbf{dy}}{\mathbf{dx}}\\ \frac{\mathbf{1}}{3}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{=}{\mathbf{y}}^2\frac{\mathbf{dy}}{\mathbf{dx}}\end{array}$

Substitute it on equation (1)

$\begin{array}{c}\frac{\mathbf{1}}{\mathbf{3}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\mathbf{v}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\\ \frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{x}}······\left(2\right)\end{array}$

Now, integrate the first P(x). Where $P\left(x\right)=3$.

data-custom-editor="chemistry" $\begin{array}{c}\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}\mathbf{=}\int \mathbf{3}\mathbf{dx}\\ \mathbf{=}\mathbf{3}\mathbf{x}\end{array}$

Then,

$\begin{array}{c}\mu \left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\end{array}$

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

$\begin{array}{c}{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\frac{\mathbf{dv}}{\mathbf{dx}}\mathbf{+}\mathbf{3}{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\mathbf{v}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\\ \frac{\mathbf{d}}{\mathbf{dx}}\left[{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\mathbf{v}\right]\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\end{array}$

Integrate both sides,

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

Substitute $\mathbf{v}\mathbf{=}{\mathbf{y}}^{\mathbf{3}}$.

$\begin{array}{l}{\mathbf{y}}^{\mathbf{3}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{-}\mathbf{3}\mathbf{x}}\\ \mathbf{y}\mathbf{=}\sqrt[\mathbf{3}]{\frac{\mathbf{3}}{\mathbf{4}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{-}\mathbf{3}\mathbf{x}}}\end{array}$

Hence the solution is$\mathbf{y}\mathbf{=}\sqrt[\mathbf{3}]{\frac{\mathbf{3}}{\mathbf{4}}{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{-}\mathbf{3}\mathbf{x}}}$