Kansas State University

Mathematics Department

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + 9y &= -3\exp(-10x)y^{4}\\ y(0) &= 1 \end{align} $$ This is a Bernoulli equation. First we find the general solution following the paradigm.

  1. We substitute $ y = v^{1/(1-4)} = v^{-1/3}$, so $ dy/dx = -(1/3)v^{-4/3}dv/dx$, and our equation becomes $$ -(1/3)v^{-4/3}\frac{dv}{dx} + 9v^{-1/3} = -3\exp(-10x)v^{-4/3} $$
  2. Multiply by $-3v^{4/3}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 27v = 9\exp(-10x) $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -27 dx \right) = \exp(-27x) $$
    2. Multiply through by the integrating factor $$ \exp(-27x)\frac{dv}{dx} - 27\exp(-27x)v = 9\exp(-10x)\exp(-27x) = 9\exp(-37x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-27x)v) =9exp(-37x) $$
    4. Integrate both sides. $$ \exp(-27x)v = -(9/37)\exp(-37x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(9/37)\exp(-10x) + C\exp(27x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(9/37)\exp(-10x) + C\exp(27x))^{-1/3} $$
  5. We check that $ y = 0$ is indeed a singular solution.

Now we plug in the initial values $ x = 0$ and $ y = 1$ and solve for $ C = 46/37,$ to obtain the solution to the initial value problem $$ y = ( -(9/37)\exp(-10x) + (46/37)\exp(27x) )^{-1/3} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett