Kansas State University

Mathematics Department

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + 8y &= -5\sin(2x)y^{2}\\ y(0) &= 3 \end{align} $$ 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 $$ -v^{-2}\frac{dv}{dx} + 8v^{-1} = -5\sin(2x)v^{-2} $$
  2. Multiply by $-v^{2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 8v = 5\sin(2x) $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -8 dx \right) = \exp(-8x) $$
    2. Multiply through by the integrating factor $$ \exp(-8x)\frac{dv}{dx} - 8\exp(-8x)v = 5\sin(2x)\exp(-8x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-8x)v) =5\sin(2x)\exp(-8x) $$
    4. Integrate both sides. In this case you will need to integrate by parts twice and then solve for the unknown integral to evaluate the integral on the right (or you can use a table of integrals). $$ \exp(-8x)v = (-(10/17)\sin(2x) - (5/34)\cos(2x))\exp(-8x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(10/17)\sin(2x) - (5/34)\cos(2x) + C\exp(8x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(10/17)\sin(2x) - (5/34)\cos(2x) + C\exp(8x))^{-1} $$
  5. We check that $ y = 0$ is indeed a singular solution.

Now we plug in the initial values $ x = 0$ and $ y = 3$ and solve for $ C = 49/102,$ to obtain the solution to the initial value problem $$ y = ( -(10/17)\sin(2x) - (5/34)\cos(2x) + (49/102)\exp(8x) )^{-1} $$ You may reload this page to generate additional examples.

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©2010, 2014 Andrew G. Bennett