Kansas State University

Mathematics Department

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} - 5y &= (6x + 8)y^{-4}\\ y(0) &= 1 \end{align} $$ This is a Bernoulli equation. First we find the general solution following the paradigm.

  1. We substitute $ y = v^{1/(1-(-4)} = v^{1/5}$, so $ dy/dx = (1/5)v^{-4/5}dv/dx$, and our equation becomes $$ (1/5)v^{-4/5}\frac{dv}{dx} - 5v^{1/5} = (6x + 8)v^{-4/5} $$
  2. Multiply by $5v^{4/5}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 25v = 30x + 40 $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -25 dx \right) = \exp(-25x) $$
    2. Multiply through by the integrating factor $$ \exp(-25x)\frac{dv}{dx} - 25\exp(-25x)v = (30x + 40)\exp(-25x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-25x)v) =(30x + 40)\exp(-25x) $$
    4. Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$ \exp(-25x)v = (-(6/5)x - 206/125)\exp(-25x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(6/5)x - 206/125+ C\exp(25x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(6/5)x - 206/125+ C\exp(25x))^{1/5} $$
  5. We check, but $ y = 0$ is not a solution to this equation.

Now we plug in the initial values $ x = 0$ and $ y = 1$ and solve for $ C = 331/125,$ to obtain the solution to the initial value problem $$ y = ( -(6/5)x - 206/125 + (331/125)\exp(25x) )^{1/5} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett