DE Poll
f(t) F(s) eat 1s−a tn n!sn+1 cos(at) ss2+a2 sin(at) as2+a2 uc(t) e−css δ(t−c) e−cs Notation F=L(f) means F(s)=∫∞0e−stf(t)dt. Heaviside function uc(t)={0t∈[0,c)1t≥c
L(c1f1+c2f2)=c1F1(s)+c2F2(s) L(f′)=sF(s)−f(0) L(f″)=s2F(s)−sf(0)−f′(0) L(f(t−c)uc(t))=e−csF(s)
Express f(t)={0t∈[0,3)2t∈[3,4)12t≥4 in terms of functions uc.
- f(t)=2u3(t)+12u4(t)
- f(t)=2u3(t)+10u4(t)
- f(t)=12(u3(t)−u4(t))
- None of the above
Answer: B.
Solution f(t)=2(u3−u4)+12u4
Express f(t)={0t∈[0,1)t−1t∈[1,2)3−tt∈[2,3)0t≥3 in terms of functions uc.
- f(t)=tu1(t)−tu2(t)+u3(t)
- f(t)=tu1(t)+2u2(t)+(2−t)u3(t)
- f(t)=tu1(t)−u3(t)
- None of the above
Answer: D.
Solution f(t)=(t−1)(u2−u1)+(3−t)(u3−u2)=(1−t)u1+2(t−2)u2+(3−t)u3
Express f(t)={0t∈[0,1)3t∈[1,2)0t≥2 in terms of functions uc.
- f(t)=u1(t)+2u2(t)
- f(t)=2u1(t)−u2(t)
- f(t)=3u2(t)−3u1(t)
- None of the above
Answer: C.
Solution f(t)=3(u1−u2)
Not available
Answer: ?
Solution
Determine the Laplace transform of (1−t)u1(t)+2(t−2)u2(t)+(3−t)u3(t)
- e−s+2e−2s+e−3ss
- −e−s+2e−2s−e−3ss2
- e−s+2e−2s+e−3ss2
- None of these
Answer: B.
Solution
We need to re-write the function as −(t−1)u1(t)+2(t−2)u2(t)−(t−3)u3(t)
Determine the Laplace transform of tu1(t)+2tu2(t)
- e−s+2e−2ss2+e−s+4e−2ss
- e−s+2e−2ss2
- e−s+2e−2ss
- e−s+2e−2ss2+s
- None of these
Answer: A.
Solution We need to rewrite the function: tu1(t)+2tu2(t)=(t−1)u1(t)+2(t−2)u2(t)+u1(t)+4u2(t)
Determine the Laplace transform of (1−uπ(t))sint
- e−πss2+1
- 1−uπ(s)s2+1
- 1−e−πss2+1
- 1+e−πss2+1
- None of these
Answer: D.
Solution The function is sint+sin(t−π)uπ(t)
Determine the Laplace transform of (1−uπ(t))cost
- se−πss2+1
- s1−uπ(s)s2+1
- s1−e−πss2+1
- s1+e−πss2+1
- None of these
Answer: D.
Solution Rewrite the function as cost+cos(t−π)uπ(t)
Compute the Laplace transform Y(s) for the solution of y″+y={sintt∈[0,π)0t≥π with initial conditions y(0)=0,y′(0)=0. Simplify your answer!
Answer: Y(s)=1+e−πs(s2+1)2
Solution The right hand side is (1−uπ(t))sint=sint+sin(t−π)uπ(t) with Laplace Transform 1+e−πss2+1 so Y(s)=1+e−πs(s2+1)2.
Solution of the equation
According to WolframAlpha, 1(s2+1)2=12L(sint−tcost). Indeed, once known, this answer can be confirmed by hand: 1(s2+1)2=12(1s2+1+ddsss2+1) So the solution is y(t)=12(sint−tcost)+uπ(t)12(sin(t−π)−(t−π)cos(t−π))=12(sint−tcost)−uπ(t)12(sin(t)−(t−π)cos(t))={12(sint−tcost)t≤π−12πcostt≥π
Compute the Laplace transform Y(s) for the solution of y″+4y′+4y={4tt∈[0,2)8t≥2 with initial conditions y(0)=0,y′(0)=0. Simplify your answer!
Answer: Y(s)=41−e−2ss2(s+2)2
Solution The right hand side is 4t(1−u2)+8u2=4t−4(t−2)u2 so its Laplace transform is 41−e−2ss2 s2Y(s)+4sY(s)+4Y(s)=41−e−2ss2 so Y(s)=41−e−2ss2(s+2)2
Solution of the equation
Partial fraction decomposition is 4s2(s+2)2=1s2+1(s+2)2+1s+2−1s. So the solution is y(t)=t+te−2t+e−2t−1+u2(t)(t−2+(t−2)e−2(t−2)+e−2(t−2)−1)
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)=21+s2
- Y(s)=1s+1
- Y(s)=s+1s2+1
- none of these
Answer: D.
SolutionThe Laplace transform of the equation is s2Y(s)−s−1+Y(s)=0. So Y(s)=s+1s2+1
To solve the initial value problem y″+4y′+4y=8t2,y(0)=2,y′(0)=0 a student seeks a particular solution of the form y∗=At2+Bt+C. After some work he determines that y∗=+2t2−4t+3 so the general solution is y=C1e−2t+C2te−2t+2t2−4t+3What value of C2 will he get?
Answer: C2=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+10t 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′=.0612B−(500+10t) with the initial condition B(0)=50000. This is a linear but non-separable equation. Solution from WolframAlpha is B(t)=500000.−450000.e0.005t+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 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)=3×12−2×A100+t and A(0)=0 (fresh water initialy). Solving the equation by WolframAlpha we get A(t)=t(30000+300t+t2)2(100+t)2 and A(100)=1752=87.5