Kansas State University

Mathematics Department

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + 4y &= (-6x - 7)y^{2}\\ y(0) &= 4 \end{align} $$ This is a Bernoulli equation. First we find the general solution following the paradigm.

  1. We substitute $ y = v^{1/(1-2)} = v^{-1}$, so $ dy/dx = -v^{-2}dv/dx$, and our equation becomes $$ -v^{-2}\frac{dv}{dx} + 4v^{-1} = (-6x - 7)v^{-2} $$
  2. Multiply by $-v^{2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 4v = 6x + 7 $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -4 dx \right) = \exp(-4x) $$
    2. Multiply through by the integrating factor $$ \exp(-4x)\frac{dv}{dx} - 4\exp(-4x)v = (6x + 7)\exp(-4x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-4x)v) =(6x + 7)\exp(-4x) $$
    4. Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$ \exp(-4x)v = (-(3/2)x - 17/8)\exp(-4x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(3/2)x - 17/8+ C\exp(4x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(3/2)x - 17/8+ C\exp(4x))^{-1} $$
  5. We check that $ y = 0$ is indeed a singular solution.

Now we plug in the initial values $ x = 0$ and $ y = 4$ and solve for $ C = 19/8,$ to obtain the solution to the initial value problem $$ y = ( -(3/2)x - 17/8 + (19/8)\exp(4x) )^{-1} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett