MAT244-2013S > Quiz 1

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**Victor Ivrii**:

Please post solution:

Find solution of the given initial value problem

$$yâ€²âˆ’2y=e^{2t},\qquad y(0)=2$$

**Alexander Jankowski**:

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).

$$

**Victor Ivrii**:

OK. No need for absolute value in $\ln |\mu(t)|$

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