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Separable Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= \frac{\exp(-3y)}{2x + 5}\\ y(0) &= 2 \end{align}$$This is a separable equation. First we find the general solution following the paradigm.

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

    We simplify the expression to obtain the general solution $$ y = (1/3)\log((3/2)\log|2x + 5| + C) $$

  4. Check for singular solutions

    Since we never divided by anything that could be 0, there are no singular solutions.

Now we plug in the initial values $x = 0$ and $y = 2$ and solve for the arbitrary constant, which we compute to be $\exp(6) - 3\log|5|/2.$ So the solution to the initial value problem is $$y = (1/3)\log((3/2)\log|2x + 5| + \exp(6) - 3\log|5|/2).$$

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