General Solution of dy/dt + y = tDate: 10/25/1999 at 02:27:06 From: Jennifer Wilson Subject: Differential equations, linear first-order I am having a hard time with a basic differential equations problem in a second-semester calculus course for biology majors. Find the general solution: dy/dt + y = t using the equation: y = e^(kt) int[b(t) * e^(-kt) dt] + Ce^(kt) y equals e to the kt times the integral of b(t) times e to the negative kt all added to C e to the kt Thank you so much for your time. Jennifer Wilson Date: 10/25/1999 at 11:10:10 From: Doctor Anthony Subject: Re: Differential equations, linear first-order For an equation of the type dy/dt + P.y = Q where P and Q are functions of t you multiply throughout by the integrating factor e^(INT[P.dt]) Apply this to the equation dy/dt + y = t We have P = 1 so integrating factor is e^(INT[dt]) = e^t The equation becomes e^t.(dy/dt) + y.e^t = t.e^t The left-hand side is an exact differential = d[y.e^t]/dt d[y.e^t]/dt = t.e^t d[y.e^t] = t.e^t.dt and now integrate both sides y.e^t = INT[t.e^t.dt] ...............................[1] We use integration by parts for the right-hand side. INT[t.e^t.dt] = t.e^t - INT[e^t.dt] = t.e^t - e^t + C where C is a constant Equation [1] can be written y.e^t = t.e^t - e^t + C divide through by e^t y = t - 1 + C.e^(-t) This is the general solution of the differential equation. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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