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### First Order Bernoulli Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}\frac{dy}{dx} + 3y &= (-7x - 7)y^{4}\\
y(0) &= 1
\end{align}
$$
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/(1-4)} = v^{-1/3}$, so $ dy/dx = -(1/3)v^{-4/3}dv/dx$, and our equation becomes $$ -(1/3)v^{-4/3}\frac{dv}{dx} + 3v^{-1/3} = (-7x - 7)v^{-4/3} $$
- Multiply by $-3v^{4/3}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 9v = 21x + 21 $$
- Solve the linear equation.
- Find the integrating factor $$ \mu(x) = \exp\left(\int -9 dx \right) = \exp(-9x) $$
- Multiply through by the integrating factor $$ \exp(-9x)\frac{dv}{dx} - 9\exp(-9x)v = (21x + 21)\exp(-9x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-9x)v) =(21x + 21)\exp(-9x) $$
- Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$ \exp(-9x)v = (-(7/3)x - 70/27)\exp(-9x) + C $$
- Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(7/3)x - 70/27+ C\exp(9x) $$

- Back substitute for $ y.$ $$ y = (-(7/3)x - 70/27+ C\exp(9x))^{-1/3} $$
- 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