Solving Separable Ordinary Differential Equations
Simple guide to solving separable differential equations, plus explanation how it works.
A separable differential equation is an ODE in the form
<$$ y\prime(x) = f(y(x))g(x),$$>where <$y\prime$> stands for <$\frac{dy}{dx}$> and <$y,f,g\colon \mathbb{R} \to \mathbb{R}$>. To solve this equation, we need to consider two cases. First, if <$y(x)$> is a solution of this equation and there exists <$\tilde{x}$> such that <$f(y(\tilde{x})) = 0$>, then necessarily <$y \equiv \tilde{x}$>, <$f(y(x)) = 0\; \forall x \in \mathbb{R}$>, and <$y\prime \equiv 0$>. This follows from the uniqueness of solutions.
Second, let us consider the case when <$f(y(x)) \neq 0\; \forall x \in \mathbb{R}$>. We can divide both sides of the equation by <$f(y(x))$>:
<$$ \frac{1}{f(y(x))}y\prime(x) = g(x). $$>Integrating both sides with respect to <$x$>, we obtain
<$$ \int \frac{y\prime(x)}{f(y(x))}\, dx = \int g(x)\, dx. $$>On the left hand side, it is now easy to substitute for <$y(x)$>, i.e. <$t := y(x),\; dt = y\prime(x)\,dx$>:
<$$ \int \frac{dt}{f(t)} = \int g(x)\, dx. $$>All that is left is working out the integrals on both sides and reverting the substitution, i.e. replacing <$t$> with <$y(x)$> again. Don’t forget to add the integration constant! Supposing that <$H(t)$> is the antiderivative of <$\frac{1}{f(t)}$> and <$G(x)$> the antiderivative of <$g(x)$>, we have the implicit form
<$$ H(y(x)) = G(x) + C,\; C \in \mathbb{R}. $$>If it is possible to find the inverse of <$H$>, we express <$y(x)$> explicitly as
<$$ y(x) = H^{-1}\left(G(x) + C\right). $$>Care must be taken to determine the interval(s) where <$y(x)$> is defined.
If we want to account for an initial condition <$y(x_0) = y_0$>, we require
<$$ H(y_0) = G(x_0) + C \quad \Rightarrow \quad C = H(y_0) - G(x_0). $$>Substituting <$C$> back into the implicit form, we get
<$$ H(y(x)) = G(x) + H(y_0) - G(x_0) $$>which can be conveniently rearranged as
<$$ H(y(x)) - H(y_0) = G(x) - G(x_0) $$>and rewritten according to the definition of Newton’s integral as
<$$ \int_{y_0}^{y(x)} \frac{1}{f(t)}\, dt = \int_{x_0}^x g(t)\, dt. $$>
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