Question: Derive the following general formula for the solution to

the Bernoulli equation (9):

$\begin{array}{c}\mathbf{y}\mathbf{=}{\left[\frac{\left(1-n\right){\displaystyle \int e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)}dx+C_1}{e^{\left(1-n\right){\displaystyle \int P\left(x\right)dx}}}\right]}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{${}^{\mathbf{1}\mathbf{-}\mathbf{n}}$}\right.}\mathbf{,}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{n}\mathbf{\ne }\mathbf{1}\\ \mathbf{y}\mathbf{=}{\mathbf{C}}_{\mathbf{2}}{\mathbf{e}}^{\int \left(Q\left(x\right)-P\left(x\right)\right)dx}\mathbf{,}\mathbf{f}\mathbf{o}\mathbf{r}\mathbf{n}\mathbf{=}\mathbf{1}\end{array}$

One has to find the solutions for the Bernoulli equation (9) $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y^n$

The Bernoulli equation (9) is,

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y^n\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let $v=y^{1-n}$

Differentiating both sides with respect to x,

$\frac{dv}{dx}=\left(1-n\right)y^{-n}\frac{dy}{dx}$

Therefore, (1) becomes,

$\begin{array}{c}\frac{1}{\left(1-n\right)y^{-n}}\frac{dv}{dx}+P\left(x\right)y=Q\left(x\right)y^n\\ \frac{dv}{dx}+\left(1-n\right)y^{1-n}P\left(x\right)=\left(1-n\right)Q\left(x\right)\\ \frac{dv}{dx}+\left(1-n\right)P\left(x\right)v=\left(1-n\right)Q\left(x\right)\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

Integrating factor $=e^{\left(1-n\right)\int P\left(x\right)dx}$

Multiplying $e^{\left(1-n\right)\int P\left(x\right)dx}$on both sides of (2),

$\begin{array}{c}{e}^{\left(1-n\right)\int P\left(x\right)dx}\frac{dv}{dx}+\left(1-n\right)e^{\left(1-n\right)\int P\left(x\right)dx}P\left(x\right)v=\left(1-n\right)e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)\\ \frac{d}{dx}\left(v{e}^{\left(1-n\right)\int P\left(x\right)dx}\right)=\left(1-n\right)e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)\end{array}$

Integrating both sides,

$v\cdot e^{\left(1-n\right)\int P\left(x\right)dx}=\int \left(1-n\right)e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)dx+C_1\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Where C₁ is the constant of integration

Substitute $v=y^{1-n}$in equation (3),

$\begin{array}{c}{y}^{1-n}=\frac{\left(1-n\right){\displaystyle \int e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)}dx+C_1}{e^{\left(1-n\right){\displaystyle \int P\left(x\right)dx}}}\\ y={\left[\frac{\left(1-n\right){\displaystyle \int e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)}dx+C_1}{e^{\left(1-n\right){\displaystyle \int P\left(x\right)dx}}}\right]}^{\frac{1}{1-n}}\end{array}$

Hence, the solution for the Bernoulli equation (9) is $\mathit{y}\mathbf{=}{\left[\frac{\left(1-n\right){\displaystyle \int e^{\left(1-n\right)\int P\left(x\right)dx}Q\left(x\right)}dx+C_1}{e^{\left(1-n\right){\displaystyle \int P\left(x\right)dx}}}\right]}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{${}^{\mathbf{1}\mathbf{-}\mathbf{n}}$}\right.}$ when $\mathit{n}\mathbf{\ne }\mathbf{1}$.

Substitute n = 1 in equation (1),

$\begin{array}{c}\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)y\\ \frac{dy}{dx}=\left(Q\left(x\right)-P\left(x\right)\right)y\\ \frac{1}{y}\frac{dy}{dx}=\left(Q\left(x\right)-P\left(x\right)\right)\end{array}$

Integrating both sides,

$\begin{array}{c}\ln y=\int \left(Q\left(x\right)-P\left(x\right)\right)dx+C_3\\ y=e^{\int \left(Q\left(x\right)-P\left(x\right)\right)dx+C_3}\end{array}$

Where, C₃ is the constant of integration.

Take $C_2=e^{C_3}$

Therefore,

$y=C_2{e}^{\int \left(Q\left(x\right)-P\left(x\right)\right)dx}$

Hence, the solution for the Bernoulli equation (9) is $\mathit{y}\mathbf{=}{\mathit{C}}_{\mathbf{2}}{\mathit{e}}^{\int \left(Q\left(x\right)-P\left(x\right)\right)dx}$ whenn = 1.