### Author Topic: Night section: 2.1 #17  (Read 5455 times)

#### Victor Ivrii ##### Night section: 2.1 #17
« on: January 16, 2013, 07:33:36 PM »

Find solution of the given initial value problem
$$yâ€²âˆ’2y=e^{2t},\qquad y(0)=2$$

#### Alexander Jankowski

• Full Member
•   • Posts: 23
• Karma: 19 ##### 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 ##### Re: Night section: 2.1 #17
« Reply #2 on: January 17, 2013, 02:06:54 AM »
OK. No need for absolute value in $\ln |\mu(t)|$