Kansas State University

Mathematics Department

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + y &= -6\exp(-2x)y^{2}\\ y(0) &= 2 \end{align} $$ 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 $$ -v^{-2}\frac{dv}{dx} + v^{-1} = -6\exp(-2x)v^{-2} $$
  2. Multiply by $-v^{2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - v = 6\exp(-2x) $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -1 dx \right) = \exp(-x) $$
    2. Multiply through by the integrating factor $$ \exp(-x)\frac{dv}{dx} - \exp(-x)v = 6\exp(-2x)\exp(-x) = 6\exp(-3x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-x)v) =6exp(-3x) $$
    4. Integrate both sides. $$ \exp(-x)v = -2\exp(-3x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -2\exp(-2x) + C\exp(x) $$

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

Now we plug in the initial values $ x = 0$ and $ y = 2$ and solve for $ C = 5/2,$ to obtain the solution to the initial value problem $$ y = ( -2\exp(-2x) + (5/2)\exp(x) )^{-1} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett