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### First Order Bernoulli Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}\frac{dy}{dx} + 8y &= -4\exp(-9x)y^{3}\\
y(0) &= 1
\end{align}
$$
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/(1-3)} = v^{-1/2}$, so $ dy/dx = -(1/2)v^{-3/2}dv/dx$, and our equation becomes $$ -(1/2)v^{-3/2}\frac{dv}{dx} + 8v^{-1/2} = -4\exp(-9x)v^{-3/2} $$
- Multiply by $-2v^{3/2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 16v = 8\exp(-9x) $$
- Solve the linear equation.
- Find the integrating factor $$ \mu(x) = \exp\left(\int -16 dx \right) = \exp(-16x) $$
- Multiply through by the integrating factor $$ \exp(-16x)\frac{dv}{dx} - 16\exp(-16x)v = 8\exp(-9x)\exp(-16x) = 8\exp(-25x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-16x)v) =8exp(-25x) $$
- Integrate both sides. $$ \exp(-16x)v = -(8/25)\exp(-25x) + C $$
- Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(8/25)\exp(-9x) + C\exp(16x) $$

- Back substitute for $ y.$ $$ y = (-(8/25)\exp(-9x) + C\exp(16x))^{-1/2} $$
- We check that $ y = 0$ is indeed a singular solution.

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

©2010, 2014 Andrew G. Bennett