Author Topic: Night section: 2.1 #17  (Read 4154 times)

Victor Ivrii

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Night section: 2.1 #17
« on: January 16, 2013, 07:33:36 PM »
Please post solution:

Find solution of the given initial value problem
$$y′−2y=e^{2t},\qquad y(0)=2$$

Alexander Jankowski

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Re: Night section: 2.1 #17
« Reply #1 on: January 16, 2013, 10:21:46 PM »
Let $\mu(t)$ be the integrating factor. Then,

$$
y' - 2y = e^{2t} \Leftrightarrow y'\mu(t) - 2y\mu(t) = e^{2t}\mu(t).
$$

We require that $\frac{d\mu(t)}{dt} = -2\mu(t)$. Thus,

$$
\frac{d\mu(t)}{dt} = -2\mu(t) \Rightarrow \frac{d\mu(t)}{\mu(t)} = -2dt \Rightarrow \ln|\mu(t)| = -2t + K,
$$

where $K$ is some arbitrary constant that we null to acquire the simplest integrating factor:

$$
\ln|\mu(t)| = -2t \Leftrightarrow \mu(t) = e^{-2t}.
$$


Now, the differential equation is

$$
y'e^{-2t} - 2ye^{-2t} = e^{2t}e^{-2t} = 1.
$$

Applying the product rule gives $\frac{d}{dt}(e^{-2t}y) = 1$, to which the general solution is

$$
e^{-2t}y = t + C \Leftrightarrow y(t) = e^{2t}(t + C).
$$

Using the initial condition, we find that $C=2$. Conclusively, the desired solution is

$$
y(t) = e^{2t}(t+2).
$$
« Last Edit: April 06, 2013, 10:38:35 AM by Alexander Jankowski »

Victor Ivrii

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Re: Night section: 2.1 #17
« Reply #2 on: January 17, 2013, 02:06:54 AM »
OK. No need for absolute value in $\ln |\mu(t)|$