Mathematics Department

Textbook Contents

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Separable Equations

Additional Examples

Solve the following initial value problem, $$\begin{align} \frac{dy}{dx} &= (x + 3)\exp(-3y)\\ y(0) &= 1 \end{align}$$This is a separable equation. First we find the general solution following the paradigm.

  1. Separate the variables $$ \exp(3y) dy = (x + 3) dx $$
  2. Integrate both sides $$ (1/3)\exp(3y) = (1/2)x^2 + 3x + C $$
  3. Solve for y (if possible)

    We simplify the expression to obtain the general solution $$ y = (1/3)\log((3/2)x^2 + 9x + C) $$

  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 $y = 1$ and solve for the arbitrary constant, which we compute to be $\exp(3).$ So the solution to the initial value problem is $$y = (1/3)\log((3/2)x^2 + 9x + \exp(3)).$$

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