Author Topic: Quiz 5 TUT0402  (Read 849 times)

Di Qiu

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Quiz 5 TUT0402
« on: December 06, 2019, 12:21:46 PM »
Question: Find a particular solution of the given inhomogeneous Euler's equation
$$(1-t)y'' + ty' - y = 2(t-1)^2e^{-t},\quad 0<t<1;\qquad y_1(t)=e^t$$
Solution:
Assume the solution is of the form $y = v(t)*e^t$, then we will have
\begin{align*}
y &= v(t)e^t\\
y' &= v'(t)e^t + v(t)e^t\\
y'' &= v''(t)e^t + 2v'(t)e^t + v(t)e^t
\end{align*}
Plug into the above equation, we have
\begin{align*}
(1-t)y'' + ty' - y &= (1-t)v''(t) + (2-t) v'(t)e^t
\end{align*}
Let r(t) = v'(t), We get
\begin{align*}
(1-t)r'(t)e^t +(2-t)r(t)e^t=2(t-1)^2e^{-t}
\end{align*}
\begin{align*}
r'(t)+\frac{2-t}{1-t}r(t)=2(1-t)e^{-2t}
\end{align*}
Use integrating factor method, we have
\begin{align*}
\mu(t) &= e^{\int\frac{2-t}{1-t}dt} = e^{t-ln(1-t)} = \frac{e^t}{1-t} \\
r(t) &= \frac{\int2(1-t)e^{-2t}*\mu dt}{\mu} = \frac{\int2e^{-t}dt}{e^t/(1-t)}=2(t-1)e^{-2t}
\end{align*}
Integrate r(t), we get
\begin{align*}
v(t) = \int r(t)dt=\int 2(t-1)e^{-2t}=-te^{-2t}+\frac{1}{2}e^{-2t}
\end{align*}
Finally, a particular solution of the given inhomogeneous Euler's equation is
\begin{align*}
y(t)=v(t)e^t=-te^{-t}+\frac{1}{2}e^{-t}
\end{align*}