MATH 255 Lecture Notes - Lecture 13: System Of Linear Equations, Linear Equation, Phase Portrait
Document Summary
One can easily encounter a di eren- tial equation which no one knows how to solve. When we are given a rst-order equation, we judge rst if it is linear. (i) linear equation: y (x) + p(x)y(x) = f (x), where p(x) and f (x) are given continuous functions. This can be achieved by choosing r(x) which satis es. And r(x) = e p(x)dx will do the job. Integrating equation (4) yields r (x) = p(x)r(x). r(x)y(x) = r(x)f (x)dx + c. r(x)f (x)dx + c] = e p(x)dx[ e p(x)dxf (x) dx + c]. If the rst-order equation is not linear, then we see if it is separable or exact. (ii) separable equation: We then try to integrate the equation and get dy dt h(y) This gives a relation between y and t. one can then express y as a function of t explicitly or implicitly. (iii) exact equations: M (t, y) + n (t, y) dy dt.
