MTH 240 Lecture : Note 5
17 May 2011
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This is not a separable d. e. www. notesolution. com www. notesolution. com www. notesolution. com www. notesolution. com www. notesolution. com www. notesolution. com www. notesolution. com www. notesolution. com. De nition 1 a rst-order linear de is one of the form dy dx. + p (x)y = q(x), where p (x) and q(x) are continuous functions on a given interval. Step 1: integrating factor i(x) = er p (x)dx. Step 2: multiplying both sides of by i(x), we get dx dy. Step 3: integrating the above, we get yi(x) = z q(x)er p (x)dxdx + c. So, y = er p (x)dx(cid:18)c +z q(x)er p (x)dxdx(cid:19) Variation of constants (also called variation of pa- rameters) It is called homogeneous when q(x) 0. Otherwise, it is an inhomogeneous linear di erential equation. For the homogeneous d. e. arable d. e. , we can nd its general solution dx dy. + p (x)y = 0, since it is a sep- y = ce r p (x)dx.
