DE Poll
$f(t)$ $F(s)$ $e^{at}$ $\frac{1}{s-a}$ $t^n$ $\frac{n!}{s^{n+1}}$ $\cos (at)$ $\frac{s}{s^2+a^2}$ $\sin(at)$ $\frac{a}{s^2+a^2}$ Notation $F=\mathcal{L}(f)$ means $F(s)=\int_0^\infty e^{-st}f(t)dt$.
$$\mathcal{L}(c_1f_1+c_2f_2)=c_1F_1(s)+c_2F_2(s)$$ $$ \mathcal{L}(f')=s F(s)-f(0)$$ $$ \mathcal{L}(f'')=s^2 F(s)-s f(0)-f'(0)$$
Determine the Laplace transform of $f(t)=\begin{cases} 0 & t\leq 3 \\ \\ 12 &t>3 \end{cases}$.
- $\displaystyle\frac{12}{s e^{s}}$
- $12 e^{-s}$
- $ 12 \frac{e^{-3s}}{s}$
- $e^{-12s}$
- None of the above
Answer: C.
Solution $F(s)=\int_3^\infty 12 e^{-st}dt=12 \frac{e^{-st}}{-s}|_{t=3}^{t=\infty}=12\frac{e^{-3s}}{s}$
Find the Laplace transform of $f(t)=\begin{cases} 0 & t\leq 1 \\ \\ 12 & t\in(0,1) \\ \\ 0 & t\geq 2 \end{cases}$
- $\displaystyle\frac{12}{s e^{2s}}$
- $12 e^{-2s}$
- $e^{-12s}$
- $ 12 \frac{e^{-2s}}{s}$
- None of the above
Answer: E.
Solution $F(s)=\int_1^2 12e^{-st}dt=12 \frac{e^{-st}}{-s}|_{t=1}^{t=2}=12\frac{e^{-2s}-e^{-s}}{s}$
Find the Laplace transform of $f(t)=\begin{cases} 0 & t\leq 2 \\ \\ 12 & t\geq 2 \end{cases}$
- $\displaystyle\frac{12}{s e^{2s}}$
- $12 e^{-2s}$
- $e^{-12s}$
- $ 2 \frac{e^{-12s}}{s}$
- None of the above
Answer: A.
Solution $F(s)=\int_2^\infty 12e^{-st}dt=12 \frac{e{-st}}{-s}|_{t=1}^{t=\infty}=\frac{12e^{-2s}}{s}=\frac{12}{se^{2s}}$
Not available
Answer: ?
Solution
Function $\frac{2}{s^2-1}$ is the Laplace transform of
- $\frac{2}{1-t^2}$
- $ e^{t}-e^{-t}$
- $ e^{t}+e^{-t}$
- $2 \sin t$
- $2 \cos t$
Answer: B.
Solution
$\frac{2}{s^2-1}=\frac{2}{(s-1)(s+1)}=\frac{1}{s-1}-\frac{1}{s+1}=\mathcal{L}[e^{t}]-\mathcal{L}[e^{-t}]$
Function $\frac{2s}{s^2-1}$ is the Laplace transform of
- $\frac{2}{1-t^2}$
- $ e^{t}-e^{-t}$
- $ e^{t}+e^{-t}$
- $2 \sin t$
- $2 \cos t$
Answer: C.
Solution $\frac{2s}{s^2-1}=\frac{2s}{(s-1)(s+1)}=\frac{1}{s-1}+\frac{1}{s+1}=\mathcal{L}[e^{t}]+\mathcal{L}[e^{-t}]$
Function $\frac{2s}{s^2+1}$ is the Laplace transform of
- $\frac{2}{1+t^2}$
- $ e^{t}-e^{-t}$
- $ e^{t}+e^{-t}$
- $2 \sin t$
- $2 \cos t$
Answer: E.
Solution Use the table or WolframAlpha
Function $\frac{2}{s^2+1}$ is the Laplace transform of
- $\frac{2}{1+t^2}$
- $ e^{t}-e^{-t}$
- $ e^{t}+e^{-t}$
- $2 \sin t$
- $2 \cos t$
Answer: D.
Solution
Compute the Laplace transform $Y(s)$ for the solution of $y''+ y=\cos t,\; y(0)=0,y'(0)=0$. Simplify your answer!
Answer: $Y(s)=\frac{s}{(s^2+1)^2}$
Solution $s^2Y(s)+Y(s)=\frac{s}{s^2+1}$ so $Y(s)=\frac{s}{(s^2+1)^2}$
Compute the Laplace transform $Y(s)$ for the solution of $y''+4 y'+ 4 y=8 t^2,\; y(0)=0,y'(0)=0$. Simplify your answer!
Answer: $Y(s)=\frac{16}{s^3(s+2)^2}$
Solution $s^2Y(s)+4 s Y(s)+4 Y(s)=\frac{16}{s^3}$ so $Y(s)=\frac{16}{s^3(s+2)^2}$
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 $