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### First Order Bernoulli Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}\frac{dy}{dx} - 9y &= 7\exp(x)y^{-2}\\
y(0) &= 3
\end{align}
$$
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-(-2)} = v^{1/3}$, so $ dy/dx = (1/3)v^{-2/3}dv/dx$, and our equation becomes $$ (1/3)v^{-2/3}\frac{dv}{dx} - 9v^{1/3} = 7\exp(x)v^{-2/3} $$
- Multiply by $3v^{2/3}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 27v = 21\exp(x) $$
- Solve the linear equation.
- Find the integrating factor $$ \mu(x) = \exp\left(\int -27 dx \right) = \exp(-27x) $$
- Multiply through by the integrating factor $$ \exp(-27x)\frac{dv}{dx} - 27\exp(-27x)v = 21\exp(x)\exp(-27x) = 21\exp(-26x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-27x)v) =21exp(-26x) $$
- Integrate both sides. $$ \exp(-27x)v = -(21/26)\exp(-26x) + C $$
- Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(21/26)\exp(x) + C\exp(27x) $$

- Back substitute for $ y.$ $$ y = (-(21/26)\exp(x) + C\exp(27x))^{1/3} $$
- 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