Question: In problems 33-40, solve the equation given in: Problem 6.

Referring to Problem 6: $\left({\mathbf{ye}}^{-2x}\mathbf{+}{\mathbf{y}}^3\right)\mathbf{dx}\mathbf{-}{\mathbf{e}}^{-2x}\mathbf{dy}\mathbf{=}\mathbf{0}\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Let us convert the equation in the form of Bernoulli equation.

$\begin{array}{c}\left({\mathbf{ye}}^{-2x}\mathbf{+}{\mathbf{y}}^3\right)\mathbf{dx}\mathbf{-}{\mathbf{e}}^{-2x}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{ye}}^{-2x}\mathbf{+}{\mathbf{y}}^3}{{\mathbf{e}}^{-2x}}\\ \mathbf{=}\frac{{\mathbf{ye}}^{-2x}}{{\mathbf{e}}^{-2x}}\mathbf{+}\frac{{\mathbf{y}}^3}{{\mathbf{e}}^{-2x}}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\left(\mathbf{-}\frac{{\mathbf{e}}^{-2x}}{{\mathbf{e}}^{-2x}}\right)\mathbf{y}\mathbf{=}\left(\frac{1}{{\mathbf{e}}^{-2x}}\right){\mathbf{y}}^3\\ \frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{+}\mathbf{P}\left(x\right)\mathbf{y}\mathbf{=}\mathbf{Q}\left(x\right){\mathbf{y}}^n\end{array}$

Compare with general form of Bernoulli equation.

$\mathbf{n}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\mathbf{1}\text{\hspace{0.33em}and\hspace{0.33em}}\mathbf{Q}\left(x\right)\mathbf{=}\frac{1}{{\mathbf{e}}^{-2x}}$

Now, divide by y³, we get,

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

Substitute v = y^-2.

Differentiate with respect to x.

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

Substitute it on equation (1)

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

Now, integrate the P (x) first. Where P (x) = 2.

Then,

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

Multiply$\mu \left(x\right)$ with equation (3)

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

Integrate both sides,

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[{\mathbf{e}}^{2x}\mathbf{v}\right]\mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{2}\int {\mathbf{e}}^{4x}\mathbf{dx}\\ {\mathbf{e}}^{2x}\mathbf{v}\mathbf{=}\mathbf{-}\mathbf{2}\int {\mathbf{e}}^{4x}\mathbf{dx}\\ {\mathbf{e}}^{2x}\mathbf{v}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{4x}}{2}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{v}\mathbf{=}\mathbf{-}\frac{{\mathbf{e}}^{2x}}{2}\mathbf{+}\frac{C}{{\mathbf{e}}^{2x}}\end{array}$

Substitute v = y^--2

$\begin{array}{c}{\mathbf{y}}^{-2}\mathbf{=}\frac{{\mathbf{e}}^{-2x}\mathbf{C}\mathbf{-}{\mathbf{e}}^{2x}}{2}\\ {\mathbf{y}}^2\mathbf{=}\frac{2}{{\mathbf{e}}^{-2x}\mathbf{C}\mathbf{-}{\mathbf{e}}^{2x}}\\ \mathbf{y}\mathbf{=}\mathbf{\pm }\sqrt{\frac{2}{{\mathbf{e}}^{-2x}\mathbf{C}\mathbf{-}{\mathbf{e}}^{2x}}}\end{array}$

Hence, the solution is$\mathbf{y}\mathbf{=}\mathbf{\pm }\sqrt{\frac{2}{{\mathbf{e}}^{-2x}\mathbf{C}\mathbf{-}{\mathbf{e}}^{2x}}}$.