Kansas State University

Mathematics Department

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + 5y &= -10\exp(-5x)y^{3}\\ y(0) &= 4 \end{align} $$ This is a Bernoulli equation. First we find the general solution following the paradigm.

  1. We substitute $ y = v^{1/(1-3)} = v^{-1/2}$, so $ dy/dx = -(1/2)v^{-3/2}dv/dx$, and our equation becomes $$ -(1/2)v^{-3/2}\frac{dv}{dx} + 5v^{-1/2} = -10\exp(-5x)v^{-3/2} $$
  2. Multiply by $-2v^{3/2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 10v = 20\exp(-5x) $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -10 dx \right) = \exp(-10x) $$
    2. Multiply through by the integrating factor $$ \exp(-10x)\frac{dv}{dx} - 10\exp(-10x)v = 20\exp(-5x)\exp(-10x) = 20\exp(-15x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-10x)v) =20exp(-15x) $$
    4. Integrate both sides. $$ \exp(-10x)v = -(4/3)\exp(-15x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(4/3)\exp(-5x) + C\exp(10x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(4/3)\exp(-5x) + C\exp(10x))^{-1/2} $$
  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 = 67/48,$ to obtain the solution to the initial value problem $$ y = ( -(4/3)\exp(-5x) + (67/48)\exp(10x) )^{-1/2} $$ You may reload this page to generate additional examples.

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