MathDB
Today's calculation of Integral 195

Source: Waseda university entrance exam /Science and Engineering 1989

March 31, 2007
calculusintegrationfunctionreal analysisalgebrapartial fractionscalculus computations

Problem Statement

Find continuous functions x(t), y(t)x(t),\ y(t) such that          x(t)=1+∫0te−2(t−s)x(s)ds\ \ \ \ \ \ \ \ \ x(t)=1+\int_{0}^{t}e^{-2(t-s)}x(s)ds          y(t)=∫0te−2(t−s){2x(s)+3y(s)}ds\ \ \ \ \ \ \ \ \ y(t)=\int_{0}^{t}e^{-2(t-s)}\{2x(s)+3y(s)\}ds
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