Separable first-order equations
For y′=f(t)g(y) on a region where g(y)≠0, integrating dy/g(y)=f(t)dt gives an implicit solution.
Key knowledge
For y′=f(t)g(y) on a region where g(y)≠0, integrating dy/g(y)=f(t)dt gives an implicit solution. Zeros of g must be checked separately because division can discard constant equilibrium solutions.
Conditions and scope
- Use continuous f,g on the relevant intervals; differentiated solutions must satisfy the original equation.
- An implicit formula may only define a local branch.
Sources · 1
References
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Evidence and review
Section 1.3.1 derivation and Example 1.3.1 zero-solution warning.
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