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### General Solution of dy/dt + y = t

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Date: 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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Associated Topics:
College Calculus
High School Calculus

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