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### First Order Bernoulli Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}\frac{dy}{dx} - 5y &= (8x + 5)y^{-4}\\
y(0) &= 4
\end{align}
$$
This is a Bernoulli equation. First we find the general solution following the paradigm.
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©2010, 2014 Andrew G. Bennett

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- We substitute $ y = v^{1/(1-(-4)} = v^{1/5}$, so $ dy/dx = (1/5)v^{-4/5}dv/dx$, and our equation becomes $$ (1/5)v^{-4/5}\frac{dv}{dx} - 5v^{1/5} = (8x + 5)v^{-4/5} $$
- Multiply by $5v^{4/5}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 25v = 40x + 25 $$
- Solve the linear equation.
- Find the integrating factor $$ \mu(x) = \exp\left(\int -25 dx \right) = \exp(-25x) $$
- Multiply through by the integrating factor $$ \exp(-25x)\frac{dv}{dx} - 25\exp(-25x)v = (40x + 25)\exp(-25x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-25x)v) =(40x + 25)\exp(-25x) $$
- Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$ \exp(-25x)v = (-(8/5)x - 133/125)\exp(-25x) + C $$
- Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(8/5)x - 133/125+ C\exp(25x) $$

- Back substitute for $ y.$ $$ y = (-(8/5)x - 133/125+ C\exp(25x))^{1/5} $$
- We check, but $ y = 0$ is not a solution to this equation.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett