Toronto Math Forum
MAT2442013S => MAT244 MathTests => Quiz 1 => Topic started by: Victor Ivrii on January 16, 2013, 07:33:36 PM

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

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
$$
y(t) = e^{2t}(t+2).
$$

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