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Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(t)y = g(t) have jump discontinuities. If to is such a point of discontinuity, then it is necessary to solve the equation separately for t < to and t > to. Afterward, the two solutions are matched so that y is continuous at to; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at to. Solve the initial value problem. y' + 3y = g(t), y(0) = 0, where 1, 0≤t≤1 g(t) = { 0, 1>1. t> 0≤t≤1 y(t) = t>1 = "

Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(t)y = g(t) have jump discontinuities. If to is such a point of discontinuity, then it is necessary to solve the equation separately for t < to and t > to. Afterward, the two solutions are matched so that y is continuous at to; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at to. Solve the initial value problem. y' + 3y = g(t), y(0) = 0, where 1, 0≤t≤1 g(t) = { 0, 1>1. t> 0≤t≤1 y(t) = t>1 = "

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(t)y = g(t) have jump discontinuities. If to is such a point of discontinuity, then it is necessary to solve the equation separately for t < to and t > to. Afterward, the two solutions are matched so that y is continuous at to; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at to. Solve the initial value problem. y' + 3y = g(t), y(0) = 0, where 1, 0≤t≤1 g(t) = { 0, 1>1. t> 0≤t≤1 y(t) = t>1 = "
Transcribed Image Text:Linear differential equations sometimes occur in which one or both of the functions p(t) and g(t) for y' + p(t)y = g(t) have jump discontinuities. If to is such a point of discontinuity, then it is necessary to solve the equation separately for t < to and t > to. Afterward, the two solutions are matched so that y is continuous at to; this is accomplished by a proper choice of the arbitrary constants. The following problem illustrates this situation. Note that it is impossible also to make y' continuous at to. Solve the initial value problem. y' + 3y = g(t), y(0) = 0, where 1, 0≤t≤1 g(t) = { 0, 1>1. t> 0≤t≤1 y(t) = t>1 = "
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