Kansas State University

Mathematics Department

Textbook Contents

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + 4y &= -8\exp(-2x)y^{3}\\ y(0) &= 2 \end{align} $$ This is a Bernoulli equation. First we find the general solution following the paradigm.

  1. We substitute $ y = v^{1/(1-3)} = v^{-1/2}$, so $ dy/dx = -(1/2)v^{-3/2}dv/dx$, and our equation becomes $$ -(1/2)v^{-3/2}\frac{dv}{dx} + 4v^{-1/2} = -8\exp(-2x)v^{-3/2} $$
  2. Multiply by $-2v^{3/2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 8v = 16\exp(-2x) $$
  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 $$ \exp(-8x)\frac{dv}{dx} - 8\exp(-8x)v = 16\exp(-2x)\exp(-8x) = 16\exp(-10x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-8x)v) =16exp(-10x) $$
    4. Integrate both sides. $$ \exp(-8x)v = -(8/5)\exp(-10x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(8/5)\exp(-2x) + C\exp(8x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(8/5)\exp(-2x) + C\exp(8x))^{-1/2} $$
  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 = 37/20,$ to obtain the solution to the initial value problem $$ y = ( -(8/5)\exp(-2x) + (37/20)\exp(8x) )^{-1/2} $$ You may reload this page to generate additional examples.

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