DE Poll
$f(t)$ $F(s)$ $e^{at}$ $\frac{1}{s-a}$ $t^n$ $\frac{n!}{s^{n+1}}$ $\cos (at)$ $\frac{s}{s^2+a^2}$ $\sin(at)$ $\frac{a}{s^2+a^2}$ $u_c(t)$ $\frac{e^{-cs}}{s}$ $\delta(t-c)$ $e^{-cs}$ Notation $F=\mathcal{L}(f)$ means $F(s)=\int_0^\infty e^{-st}f(t)dt$. Heaviside function $u_c(t)=\begin{cases}0 & t\in[0,c) \\ 1& t\geq c\end{cases}$
$$\mathcal{L}(c_1f_1+c_2f_2)=c_1F_1(s)+c_2F_2(s)$$ $$ \mathcal{L}(f')=s F(s)-f(0)$$ $$ \mathcal{L}(f'')=s^2 F(s)-s f(0)-f'(0)$$ $$ \mathcal{L}(f(t-c)u_c(t))=e^{-cs} F(s)$$
Express $$f(t)=\begin{cases} 0 & t\in[0, 3) \\ 2 & t\in[3,4) \\ 12 &t\geq 4 \end{cases}$$ in terms of functions $u_c$.
- $f(t)=2 u_3(t)+12 u_4(t)$
- $f(t)=2 u_3(t)+10 u_4(t)$
- $f(t)=12 (u_3(t)-u_4(t))$
- None of the above
Answer: B.
Solution $f(t)=2 (u_3-u_4)+12 u_4$
Express $$f(t)=\begin{cases} 0 & t\in[0,1) \\ t-1 & t\in[1,2) \\ 3-t & t\in[2,3) \\ 0 & t\geq 3 \end{cases}$$ in terms of functions $u_c$.
- $f(t)=t u_1(t)- t u_2(t)+u_3(t)$
- $f(t)= t u_1(t)+ 2 u_2(t)+(2-t)u_3(t)$
- $f(t)=t u_1(t)- u_3(t)$
- None of the above
Answer: D.
Solution $f(t)=(t-1) (u_2-u_1)+(3-t)(u_3-u_2)= (1-t)u_1+ 2(t-2) u_2+(3-t)u_3 $
Express $$f(t)=\begin{cases} 0 & t\in[0,1) \\ \\ 3 & t\in[1,2)\\ \\ 0 & t\geq 2 \end{cases}$$ in terms of functions $u_c$.
- $f(t)= u_1(t)+2u_2(t)$
- $f(t)=2 u_1(t)-u_2(t)$
- $f(t)= 3 u_2(t)-3 u_1(t)$
- None of the above
Answer: C.
Solution $f(t)=3(u_1-u_2)$
Not available
Answer: ?
Solution
Determine the Laplace transform of $ (1-t)u_1(t)+ 2(t-2) u_2(t)+(3-t)u_3(t)$
- $\frac{e^{-s}+2 e^{-2s}+e^{-3s}}{s}$
- $\frac{-e^{-s}+2 e^{-2s}-e^{-3s}}{s^2}$
- $\frac{e^{-s}+2 e^{-2s}+e^{-3s}}{s^2}$
- None of these
Answer: B.
Solution
We need to re-write the function as $-(t-1)u_1(t)+2(t-2)u_2(t)-(t-3)u_3(t)$
Determine the Laplace transform of $ t u_1(t)+ 2 t u_2(t)$
- $\frac{e^{-s}+2e^{-2s}}{s^2}+\frac{e^{-s}+4 e^{-2s}}{s}$
- $\frac{e^{-s}+2e^{-2s}}{s^2}$
- $\frac{e^{-s}+2 e^{-2s}}{s}$
- $\frac{e^{-s}+2e^{-2s}}{s^2+s}$
- None of these
Answer: A.
Solution We need to rewrite the function: $ t u_1(t)+ 2 t u_2(t)=(t-1)u_1(t)+2(t-2)u_2(t)+u_1(t)+4 u_2(t)$
Determine the Laplace transform of $(1-u_\pi(t))\sin t $
- $ \frac{e^{-\pi s}}{s^2+1}$
- $ \frac{1-u_\pi(s)}{s^2+1}$
- $\frac{1-e^{-\pi s}}{s^2+1}$
- $\frac{1+e^{-\pi s}}{s^2+1}$
- None of these
Answer: D.
Solution The function is $\sin t + \sin(t-\pi)u_\pi(t)$
Determine the Laplace transform of $(1-u_\pi(t))\cos t $
- $ s\frac{e^{-\pi s}}{s^2+1}$
- $ s\frac{1-u_\pi(s)}{s^2+1}$
- $s\frac{1-e^{-\pi s}}{s^2+1}$
- $s\frac{1+e^{-\pi s}}{s^2+1}$
- None of these
Answer: D.
Solution Rewrite the function as $\cos t +\cos (t-\pi) u_\pi(t)$
Compute the Laplace transform $Y(s)$ for the solution of $$y''+ y=\begin{cases} \sin t & t\in[0,\pi) \\ 0 & t\geq \pi \end{cases}$$ with initial conditions $y(0)=0,y'(0)=0$. Simplify your answer!
Answer: $Y(s)=\frac{1+e^{-\pi s}}{(s^2+1)^2}$
Solution The right hand side is $(1-u_\pi(t))\sin t = \sin t + \sin (t-\pi) u_\pi(t)$ with Laplace Transform $\frac{1+e^{-\pi s}}{s^2+1}$ so $Y(s)=\frac{1+e^{-\pi s}}{(s^2+1)^2}$.
Solution of the equation
According to WolframAlpha, $\frac{1}{(s^2+1)^2}=\frac12\mathcal{L}(\sin t-t \cos t)$. Indeed, once known, this answer can be confirmed by hand: $\frac{1}{(s^2+1)^2}= \frac12 \left(\frac{1}{s^2+1}+\frac{d}{ds}\frac{s}{s^2+1}\right)$ So the solution is $$ \begin{multline*} y(t)=\frac{1}{2}(\sin t-t \cos t)+u_\pi(t) \frac{1}{2}(\sin (t-\pi)-(t-\pi) \cos (t-\pi)) \\=\frac{1}{2}(\sin t-t \cos t)-u_\pi(t) \frac{1}{2}(\sin (t)-(t-\pi) \cos (t)) = \begin{cases} \frac{1}{2}(\sin t-t \cos t) & t\leq \pi \\ -\frac12 \pi \cos t & t\geq \pi \end{cases} \end{multline*} $$
Compute the Laplace transform $Y(s)$ for the solution of $$y''+4 y'+ 4 y=\begin{cases} 4 t & t\in[0,2)\\ 8 & t\geq 2 \end{cases}$$ with initial conditions $y(0)=0,y'(0)=0$. Simplify your answer!
Answer: $Y(s)=4\frac{1-e^{-2s}}{s^2(s+2)^2}$
Solution The right hand side is $4t (1-u_2)+8u_2=4t- 4(t-2)u_2$ so its Laplace transform is $4\frac{1-e^{-2s}}{s^2}$ $s^2Y(s)+4 s Y(s)+4 Y(s)=4\frac{1-e^{-2s}}{s^2}$ so $Y(s)=4\frac{1-e^{-2s}}{s^2(s+2)^2}$
Solution of the equation
Partial fraction decomposition is $\frac{4}{s^2(s+2)^2}=\frac{1}{s^2}+\frac{1}{(s+2)^2}+\frac{1}{s+2}-\frac{1}{s}$. So the solution is $y(t)=t+te^{-2t}+e^{-2t}-1+ u_2(t)\left(t-2+(t-2)e^{-2(t-2)}+e^{-2(t-2)}-1 \right)$
Compute the Laplace transform $Y(s)$ for the solution of $y''+ y=0,\; y(0)=1,y'(0)=1$. Simplify your answer!
- $Y(s)=0$
- $Y(s)=\frac{2}{1+s^2}$
- $Y(s)=\frac{1}{s+1}$
- $Y(s)=\frac{s+1}{s^2+1}$
- none of these
Answer: D.
SolutionThe Laplace transform of the equation is $s^2Y(s)-s-1+Y(s)=0$. So $Y(s)=\frac{s+1}{s^2+1}$
To solve the initial value problem $y''+4 y'+ 4 y=8 t^2,\; y(0)=2,y'(0)=0$ a student seeks a particular solution of the form $y_*=A t^2+Bt+C$. After some work he determines that $y_*=+2t^2-4t+3$ so the general solution is $y=C_1 e^{-2t}+C_2 t e^{-2t}+2t^2-4t+3$What value of $C_2$ will he get?
Answer: $C_2=2$
Solution
Money Type D:
A college graduate borrows 50,000 to buy a Tesla S at an interest rate of 6%. Anticipating steady salary increases, he expects to steady increase his monthly payments, with $m(t)=500+10 t$ after $t$ months.Assuming that this payment schedule can be maintained, what will be the amount of loan in 5 years? Will the loan be fully paid in 6 years?
SolutionLets use months as units of $t$. The loan amount changes at the rate $B'= \frac{.06}{12} B - (500+ 10 t)$ with the initial condition $B(0)=50000$. This is a linear but non-separable equation. Solution from WolframAlpha is $B(t)=500000. - 450000. e^{0.005 t} + 2000. t$. Answer: $B(60)=12563.5$ (From the graph of $B(t)$, the loan will be paid of in about 71.17 months.)
Type D: hard
A tank originally contains 100 gal of fresh water and has capacity of 200 gal. Then water containing ${\frac 12}$ lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixture is allowed to leave at a rate of 2 gal/min. Setup the differential equation for the amount $A(t)$ of salt in the tank at time $t$. What is the amount of salt in the tank at the time of overespil?l
Solution Water level increases at the rate of 1 gal/min, so the volume of water at time $t$ is $V(t)=100+t$. The water with overspill at $t=100$ min.
The "rate in - rate out" formula gives $A'(t)=3\times \frac12 -2\times \frac{A}{100+t}$ and $A(0)=0$ (fresh water initialy). Solving the equation by WolframAlpha we get $A(t)=\frac{t (30000 + 300 t + t^2)}{2 (100 + t)^2}$ and $A(100)=\frac{175}{2}=87.5 $