Question

Differential equations

Answer 1

QUESTION A
Solve the initial value problem using the Laplace transform\[y''+16y=25e^{-3t}\\y(0)=y(0)=0\]


QUESTION B
Let 0<a<b be given. Write the function f in terms of the Heaviside function and calculate the Laplace transformation thereof\[f(t)=e^{-t}; a<t<b\\f(t)=0 ;otherwise\]

Answer 2

Here is what I did in Q A:
\[s^2Y-sy(0)-y'(0)+16Y=\frac{25}{s+3}\\Y(s^2+16)=\frac{25}{s+3}\\Y=\frac{25}{(s+3)(s^2+16)}\\Y=\frac{1}{s+3}+\frac{3-s}{s^2+16}\\Y=\frac{1}{s+3}+3\frac{1}{s^2+16}-\frac{s}{s^2+16}\\4Y=4\frac{1}{s+3}+3\frac{4}{s^2+16}-4\frac{s}{s^2+16}\]
and that gives\[4Y=4e^{3t}+3 \sinh(4t)-4 \cosh(4t)\\Y=e^{3t}+\frac{3}{4} \sinh(4t)- \cosh(4t)\]

Answer 3

Is that right?




For question b, I got to this part\[f(t)=e^{-t}[u(t-a)-u(t-b)]\] but I don't know how to do the Laplace on that

Answer 4

For part a), you shouldn't be getting the hyperbolic trig functions; just the regular ones. Also, the inverse transform of \(\dfrac{1}{s+3}\) is \(e^{-3t}\), not \(e^{3t}\).

For part b), try using the definition of the transform:
\[\begin{align*}\mathscr{L}\left\{e^{-t}\left[u(t-a)-u(t-b)\right]\right\}&=\mathscr{L}\left\{e^{-t}u(t-a)\right\}-\mathscr{L}\left\{e^{-t}u(t-b)\right\}\\
&=\color{red}{\int_0^\infty e^{-t}u(t-a)e^{-st}~dt}\\
&~~~~~~~~-\int_0^\infty e^{-t}u(t-b)e^{-st}~dt
\end{align*}\]
By the definition of \(u(t-a)\), you have \(u(t-a)=0\) for \(t<a\) and \(u(t-a)=1\) for \(t\ge a\), so you get
\[\begin{align*}\color{red}{\int_0^\infty e^{-t}u(t-a)e^{-st}~dt}&=\int_0^a e^{-t}(0)e^{-st}~dt+\int_a^\infty e^{-t}(1)e^{-st}~dt\\
&=\int_a^\infty e^{-(s+1)t}~dt\\
&=-\frac{1}{s+1}\lim_{b\to\infty}\left[e^{-(s+1)t}\right]_a^b\\
&=-\frac{1}{s+1}\lim_{b\to\infty}\left[e^{-(s+1)b}-e^{-a(s+1)}\right]\\
&=-\frac{1}{s+1}\left[0-e^{-a(s+1)}\right]\\
&=\color{red}{\frac{e^{-a(s+1)}}{s+1}}
\end{align*}\]
(\(s\) is assumed to be positive, so the first exponential term goes to 0)
Do the same for the transform of \(e^{-t}u(t-b)\).

Answer 5

Thanks! I now have question a right.

For b, I got this\[\mathscr{L}\left\{ e^{−t}u(t−a)-e^{−t}u(t−b) \right\}=[e^{-a(s+1} \times \frac{1}{s+1}]-[-\frac{1-e^{b-bs}}{1-s}]\\=[e^{-a(s+1} \times \frac{1}{s+1}]-[(1-e^{b-bs}) \times \frac{1}{s-1}]\\=e^{-a}e^{-t}-(1-e^b)e^{t}\]

Answer 6

The last line isn't right. You're transforming the function from the \(t\)-domain to the \(s\)-domain, so there should be no \(t\)'s anywhere in the final answer.

The line before it is right, but can be rewritten to look nicer.

Answer 7

Okay thanks so much for all the help!

Answer 8

You're welcome!