Math 240 Home,
Textbook Contents,
Online Homework Home

**Warning: MathJax
requires JavaScript to process the mathematics on this page.**

If your browser supports JavaScript, be sure it is enabled.**
****
**

### First Order Bernoulli Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}\frac{dy}{dx} - 7y &= 3\sin(7x)y^{-3}\\
y(0) &= 2
\end{align}
$$
This is a Bernoulli equation. First we find the general solution following the paradigm.
*You may reload this page to generate additional examples.*

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

©2010, 2014 Andrew G. Bennett

If your browser supports JavaScript, be sure it is enabled.

- We substitute $ y = v^{1/(1-(-3)} = v^{1/4}$, so $ dy/dx = (1/4)v^{-3/4}dv/dx$, and our equation becomes $$ (1/4)v^{-3/4}\frac{dv}{dx} - 7v^{1/4} = 3\sin(7x)v^{-3/4} $$
- Multiply by $4v^{3/4}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 28v = 12\sin(7x) $$
- Solve the linear equation.
- Find the integrating factor $$ \mu(x) = \exp\left(\int -28 dx \right) = \exp(-28x) $$
- Multiply through by the integrating factor $$ \exp(-28x)\frac{dv}{dx} - 28\exp(-28x)v = 12\sin(7x)\exp(-28x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-28x)v) =12\sin(7x)\exp(-28x) $$
- Integrate both sides. In this case you will need to integrate by parts twice and then solve for the unknown integral to evaluate the integral on the right (or you can use a table of integrals). $$ \exp(-28x)v = (-(48/119)\sin(7x) - (12/119)\cos(7x))\exp(-28x) + C $$
- Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(48/119)\sin(7x) - (12/119)\cos(7x) + C\exp(28x) $$

- Back substitute for $ y.$ $$ y = (-(48/119)\sin(7x) - (12/119)\cos(7x) + C\exp(28x))^{1/4} $$
- 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