Kansas State University

Mathematics Department

Textbook Contents

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First Order Bernoulli Equations

Additional Examples

Solve the following initial value problem $$ \begin{align}\frac{dy}{dx} + 6y &= (-5x - 5)y^{3}\\ y(0) &= 2 \end{align} $$ 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 $$ -(1/2)v^{-3/2}\frac{dv}{dx} + 6v^{-1/2} = (-5x - 5)v^{-3/2} $$
  2. Multiply by $-2v^{3/2}$ to obtain a linear equation in the usual form. $$ \frac{dv}{dx} - 12v = 10x + 10 $$
  3. Solve the linear equation.

    1. Find the integrating factor $$ \mu(x) = \exp\left(\int -12 dx \right) = \exp(-12x) $$
    2. Multiply through by the integrating factor $$ \exp(-12x)\frac{dv}{dx} - 12\exp(-12x)v = (10x + 10)\exp(-12x) $$
    3. Recognize the left-hand-side as $\displaystyle \frac{d}{dx}(\mu(x)v).$ $$ \frac{d}{dx}(\exp(-12x)v) =(10x + 10)\exp(-12x) $$
    4. Integrate both sides. In this case you will need to integrate by parts to evaluate the integral on the right. $$ \exp(-12x)v = (-(5/6)x - 65/72)\exp(-12x) + C $$
    5. Divide through by $\mu(x)$ to solve for $ v.$ $$ v = -(5/6)x - 65/72+ C\exp(12x) $$

  4. Back substitute for $ y.$
    5. $$ y = (-(5/6)x - 65/72+ C\exp(12x))^{-1/2} $$
  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 = 83/72,$ to obtain the solution to the initial value problem $$ y = ( -(5/6)x - 65/72 + (83/72)\exp(12x) )^{-1/2} $$ You may reload this page to generate additional examples.

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