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### First Order Bernoulli Equations

#### Additional Examples

Solve the following initial value problem
$$
\begin{align}\frac{dy}{dx} + 5y &= (-3x - 1)y^{2}\\
y(0) &= 1
\end{align}
$$
This is a Bernoulli equation. First we find the general solution following the paradigm.
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If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010 Andrew G. Bennett

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- We substitute $ y = v^{1/(1-2)} = v^{-1}$, so $ dy/dx = -v^{-2}dv/dx$, and our equation becomes $$ -v^{-2}\frac{dv}{dx} + 5v^{-1} = (-3x - 1)v^{-2} $$
- Multiply by $-v^{2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 5v = 3x + 1 $$
- Solve the linear equation.
- Find the integrating factor $$ \mu(x) = \exp\left(\int -5 dx \right) = \exp(-5x) $$
- Multiply through by the integrating factor $$ \exp(-5x)\frac{dv}{dx} - 5\exp(-5x)v = (3x + 1)\exp(-5x) $$
- Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-5x)v) =(3x + 1)\exp(-5x) $$
- Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$ \exp(-5x)v = (-(3/5)x - 8/25)\exp(-5x) + C $$
- Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(3/5)x - 8/25+ C\exp(5x) $$

- Back substitute for $ y.$ $$ y = (-(3/5)x - 8/25+ C\exp(5x))^{-1} $$
- 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 Andrew G. Bennett