# Integrating Factor Problem

• Sep 5th 2011, 05:08 PM
VitaX
Integrating Factor Problem
Find the solution of the initial value problem.

$ty' + (t+1)y = t$ ; $y(ln2) = 1$ ; $t > 0$

$\frac{dy}{dx} + p(x)y = Q(x)$

$y' + \left(1 + \frac{1}{t} \right)y = 1$

$p(t) = 1 + \frac{1}{t}$ ; $Q(t) = 1$

$\int p(x) = \int 1 + \frac{1}{t} dt = t + ln(t)$

Integrating Factor: $e^{t + ln(t)} = e^{t}e^{ln(t)} = te^t$

$te^t \left[y' + \left(1 + \frac{1}{t} \right)y \right] = te^t$

$te^t y' + \left(1 + \frac{1}{t} \right)y = te^t$

$u'v + v'u = \frac{d}{dx} (uv)$ ; $u = y$ ; $v = te^t$

$\frac{d}{dt}[yte^t] = te^t$

$\int d[yte^t] = \int te^t dt$

$yte^t = \int te^t dt$

Integrate by Parts: $\int te^t dt = te^t - e^t$

$yte^t = te^t - e^t$

$y(t) = 1 - \frac{1}{t} + C$

Initial Condition: $y(ln2) = 1$

$y(ln2) = 1 = 1 - \frac{1}{ln2} + C$

$C = \frac{1}{ln2}$

My Solution: $y(t) = 1 - \frac{1}{t} + \frac{1}{ln2}$

Book Solution: $y(t) = \frac{t - 1 + 2e^{-t}}{t}$

Where exactly did I go wrong?
• Sep 5th 2011, 05:37 PM
Chris L T521
Re: Integrating Factor Problem
Quote:

Originally Posted by VitaX
Find the solution of the initial value problem.

$ty' + (t+1)y = t$ ; $y(ln2) = 1$ ; $t > 0$

$\frac{dy}{dx} + p(x)y = Q(x)$

$y' + \left(1 + \frac{1}{t} \right)y = 1$

$p(t) = 1 + \frac{1}{t}$ ; $Q(t) = 1$

$\int p(x) = \int 1 + \frac{1}{t} dt = t + ln(t)$

Integrating Factor: $e^{t + ln(t)} = e^{t}e^{ln(t)} = te^t$

$te^t \left[y' + \left(1 + \frac{1}{t} \right)y \right] = te^t$

$te^t y' + \left(1 + \frac{1}{t} \right)y = te^t$

$u'v + v'u = \frac{d}{dx} (uv)$ ; $u = y$ ; $v = te^t$

$\frac{d}{dt}[yte^t] = te^t$

$\int d[yte^t] = \int te^t dt$

$yte^t = \int te^t dt$

Integrate by Parts: $\int te^t dt = te^t - e^t$

$yte^t = te^t - e^t$ This is where you made your mistake.

$y(t) = 1 - \frac{1}{t} + C$

Initial Condition: $y(ln2) = 1$

$y(ln2) = 1 = 1 - \frac{1}{ln2} + C$

$C = \frac{1}{ln2}$

My Solution: $y(t) = 1 - \frac{1}{t} + \frac{1}{ln2}$

Book Solution: $y(t) = \frac{t - 1 + 2e^{-t}}{t}$

Where exactly did I go wrong?

You made your mistake when you said $yte^t = te^t-e^t$. It should be $yte^t=te^t-e^t{\color{red} + C}$.

That way, when you divide through by $te^t$, you get $y=1-\frac{1}{t}+\frac{1}{t}Ce^{-t}$

Now apply your initial condition to get the book's solution.

I hope this helps.
• Sep 5th 2011, 05:41 PM
VitaX
Re: Integrating Factor Problem
Ah, OK. Thanks for clearing that up.