Kansas State University

Mathematics Department

Textbook Contents

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Separable Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= 5\cos(2x)(y^2 + 6y + 8)\\ y(0) &= -4 \end{align}$$This is a separable equation. First we find the general solution following the paradigm.

  1. Separate the variables $$ \frac{dy}{y^2 + 6y + 8} = 5\cos(2x) dx $$
  2. Integrate both sides $$ (1/2)\log|y + 2| - (1/2)\log|y + 4| = (5/2)\sin(2x) + C $$
  3. Solve for y (if possible)

    We simplify the expression to obtain the general solution $$ \frac{y + 2}{y + 4} = k \exp(5\sin(2x)) $$

  4. Check for singular solutions

    We divided by $y^2 + 6y + 8$ which is 0 when $y = -2$ or $y = -4.$ We can check these are both solutions. Now $y = -2$ is already in the general solution (when the constant is 0), but $y = -4$ is a singular solution.

We immediately observe that our initial condition matches the solution $y = -4$ that we noted in step 4, so the solution to the initial value problem is just $$y = -4.$$

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