Essential MA266 Practice Problems for Mastering Differential
School
Epic Charter School**We aren't endorsed by this school
Course
EPIC ANSWERS
Subject
Mathematics
Date
Dec 10, 2024
Pages
6
Uploaded by UltraAtom7917
MA266 Practice Problems1.Ify′+1 +1ty=1tandy(1) = 0, theny(ln 2) =?A. ln 2−ln(ln 2)B. ln(ln 2)C. ln(ln 2) +12 ln 2D.1ln 21−e2E.1ln 2−12.What is the largest open interval for which a unique solution of the initial value problemty′+1t+ 1y=t−2t−3,y(1) = 0is guaranteed?A. 0< t <1B. 0< t <2C. 0< t <3D.−1< t <3E.−1< t <13.An explicit solution ofy′=y2−1 is?A.y=Ce2t1−Ce2tB.y=1 +Ce2t1−Ce2tC.y=11−Ce2tD.y=1 +Ce2t1−e2tE.y33−y=C4.Ify′=y3andy(0) = 1, theny(−1) =?A. 5−14B.√3C. 1D.1√3E. Does not exist5.Lety(x) be the solution to the initial value problemxy′= 3y+ 2x4,y(1) = 0.Then,y(2) isA. 4B. 8C. 16D. 20E. 326.A tank initially contains 40 ounces of salt mixed in 100 gallons of water. A solution containing 4 oz of saltper gallon is then pumped into the tank at the rate of 5 gal/min. The stirred mixture flows out of the tankat the same rate. How much salt is in the tank after 20 minutes?A. 400−360e−1B. 20C. 80D. 40 + 20eE. 400 + 360e27.Find the general solution of a homogeneous equation using substitutionv=yx.dydx=5x2+ 3y22xyA. 3y2+ 5x2=Cx2B.y2+ 5x2=Cx3C.x2+ 3y2=CxD. 2y−5x2=Cx4E.y2+ 3x2=Cx38.Suppose thatdydx= (x+y)2−1.What is the implicit general solution to this differential equation? (Hint: use the substitutionv(x) =x+y.)A.1x+y−x=CB.xy+x=CC.xy−x=CD.x(x+y) + 1 =CE.1x+y+x=C1
9.An implicit solution ofy2+ 1 + (2xy+ 1)dydx= 0is?A. 2(xy2+y) =CB.xy2+y=CC.xy2+x+y=CD.y33+y+x2y+x=CE.y=xy2+C10.Consider the autonomous differential equationdydt=−110(y−1)(y−4)2.Classify the stability of each equilibrium solution.A.y= 1 andy= 4 both unstableB.y= 1 unstable;y= 4 stableC.y= 0 andy= 1 stable;y= 4unstableD.y= 1 stable;y= 4 semistableE.y= 0 stable;y= 1 andy= 4 unstable11.Consider the following doomsday/extinction differential equation for a populationP(t) with the initialpopulationP(0) = 4.dPdt= 3P(P−2)At what timetdoes “Doomsday” occur (which means the population explodes)?A.ln (2)6B.ln (2)3C.ln (4)3D.ln (4)6E.∞12.Use Euler’s method with step sizeh= 1 to find the approximate value ofy(3), wherey(x) solves the initialvalue problemy′=x+y2,y(0) =−8.A.−17B.−22.5C.−23.5D.−24.5E.−2713.If the WronskianW(f, g) =−3e4tandf(t) = 4e2t, theng(t) could beA.34te2tB. 12e2tC.−32e2tD.−34te4tE.−34te2t14.The general solution ofy′′−4y′+ 4y= 0is?A.y=C1e2t+C2te2tB.y=C1e2t+C2e2tC.y=C1e2t+C2e−2tD.y=C1e−2t+C2te−2tE.y=C1t+C2t215.The general solution ofy′′′+ 4y′′+ 5y′= 0is?A.y=C1e−2tcost+C2e−2tsintB.y=C1+C2e−2tcost+C3e−2tsintC.y=C1+C2etcos 2t+C3etsin 2tD.y=C1+C2cost+C3sintE.y=C1+C2e2tcost+C3e2tsint16.Lety(x) be the solution to the reducible second-order differential equationy′′+ (y′)2= 0, y(0) = 0, y′(0) = 1.Findy(2). (Use the substitutionp=y′>0.)A. ln 3B.e−2C. ln 5D.e4E. 4Page 2
17.An object weighting 8 pounds attached to a spring will stretch it 6 inches beyond its natural length. Thereis a damping force with a damping constantc= 6 lbs-sec/ft and there is no external force. If att= 0 theobject is pulled 2 feet below equilibrium and then released, the initial value problem describing the verticaldisplacementx(t) becomes?A. 8x′′+ 6x′+ 16x= 0, x(0) =−2, x′(0) = 0B. 8x′′+ 6x′+ 16x= 0, x(0) = 2, x′(0) = 0C.14x′′+6x′+ 16x= 0, x(0) = 2, x′(0) = 0D.14x′′+ 6x′+ 8x= 0, x(0) = 2, x′(0) = 0E. 256x′′+ 6x′+ 16x=0, x(0) = 2, x′(0) = 018.A particular solution,yp, ofy′′−4y′+ 3y= 2t+etis?A.−12tet+13t+12B.−12tet+12t+12C.−12et+13t+12D.t2+etE.−12tet+23t+8919.Determine the appropriate form for a particular solutionyp(x) to the third-order differential equationy(3)+y′′−y′−y= cosx+xe−x.A.Acosx+Bsinx+x2(Cx+D)e−xB.Acosx+x(Bx+C)e−xC.x2(Acosx+Bsinx)+(Cx+D)e−xD.Acosx+Bxe−xE.Acosx+Bsinx+ (Cx+D)e−x20.Ify′′+ 5y′+ 6y= 24et,y(0) = 0,y′(0) = 0, theny(1) =?A.e−e−2+ 6e−3B. 2e−8e−2+ 6e−3C.e−8e−2+ 6e−3D.e+ 8e−2+e−3E. 021.The differential equationy′′−2ty′+2t2y= 0has solutionsy1(t) =tandy2(t) =t2. Ify′′−2ty′+2t2y= 2;y(1) = 0, y′(1) = 0theny(2) =?A. 8 ln 2−4B. 0C.−6D. 8 ln 2 + 4E. 8 ln 222.A spring-mass system is governed by the initial value problemx′′+ 4x′+ 4x= 4 cosωtx(0) = 9,x′(0) =−2.For what value(s) ofωwill resonance occur?A. 0B. 2C. 4D. no value ofωE. 2< ω <inf23.Rewrite the second order equation2u′′+ 3u′+ku= cos 2tas a system of first order equations.A.(x′=yy′=12(−3x−ky+ cos 2t)B.(x′=xy′=12(−3y−kx+ cos 2t)C.(x′=yy′=12(−3y−kx+ cos 2t)D.(x′=yy′= 2y+kx+ cos 2tE.(x′= 2y+kx+ cos 2ty′=xPage 3
24.The solution ofx′=1141x,x(0) =32is?A. 2e3t12+e−t1−2B. 2e3t10+e−t12C.e3t21+e−t11D. 3e3t1−2−e−t1−2E. 3e3t12+e−t0−425.Solvex′=11−11x,x(0) =−12.A.x(t) = 2etsintcost−etcostsintB.x(t) = 2etsintcost+etcostsintC.x(t) = 2etsintcost−etcost−sintD.x(t) =etsintcost−etcostsintE.x(t) =et−sintcost−etcostsint26.Solve the initial value problemx′=Ax,x(0) =11,whereA=1101.A.et11−2tet10.B.et11+tet10.C.et11+ 2tet10.D.et01+ 2tet10.E.et11−2tet10.27.What values of the parameterαin the system below make the origin a saddle point in the phase plane:x′=11α2xA.α >2B.α >−14C.α <−14D. 2> α >−14E.α <−228.Find a particular solution ofx1x2′=0110x1x2−23.A.xp=23B.xp=−23C.xp=−32D.xp=32E.xp=1129.Find the general solution ofx1x2′=2011x1x2−6e−t1.A.c101et+c211e2t+2−1e−t+01B.c101et+c211e2tC.c101et+c211e2t−6e−t1D.c101et+c211e2t+60e−t+01Page 4
E.c110et+c211e2t+2−1e−t+0130.L{et(1 + cos 2t)}=?A.1s−1+1(s−1)2+ 4B.1s−11s+s−1(s−1)2+ 4C.1s−1s−1s2−2s+ 5D.1s+s(s−1)2+ 4E.1s−1+s−1s2−2s+ 531.Find the Laplace transform off(t) =t,0≤t <10,1≤t <∞.A.e−s1s+1s−2B.1s2−e−s1s2C.1s2−e−s1s+1s2D.1s2+2e−s1s+1s2E.e−s1s+1s232.Solvey′′+ 3y′+ 2y= 4u1(t)y(0) = 0,y′(0) = 1.A.u1(t)(2−4e−(t−1)+ 2e−2(t−1))B.u1(t)(2−4e−(t−1)+ 2e−2(t−1))+e−t−e−2tC.u0(t)(2−4e−(t−1)+ 2e−2(t−1))+e−t−e−2tD.(2−4e−(t−1)+ 2e−2(t−1))+e−t−e−2tE.e−t−e−2t33.Find the solution of the initial value problemy′′+y=δ(t−π)y(0) = 0,y′(0) = 1.A.y= sint+u0(t) sin(t−π)B.y= sint+uπ(t) sin(πt)C.y=uπ(t)(sint+ sin(t−π))D.y=uπ(t) sintE.y= sint+uπ(t) sin(t−π)34.The inverse Laplace transform ofF(s) =se−ss2+ 2s+ 5is?A.u1(t)(et−1cos 2(t−1)−12et−1sin 2(t−1))B.u1(t) (e−tcos 2t)−12e−tsin 2tC.u1(t)(e−t+1cos 2(t−1)−12e−t+1sin 2(t−1))D.u1(t)(e−tcos 2(t−1)−12e−tsin 2(t−1))E.e−t+1cos 2(t−1)−12e−t+1sin 2(t−1)35.LZt0sin 2(t−τ) cos(3τ)dτ=?A.1s2+ 4+ss2+ 9B.2s(s2+ 4)(s2+ 9)C.2s2+ 4+ss2+ 9D.2(s2+ 4)(s2+ 9)E.s(s2+ 4)(s2+ 9)Page 5
Answer Key:1. D2. C3. B4. D5. C6. A7. B8. E9. C10. D11. A12. C13. E14. A15. B16. A17. C18. E19. A20. B21. A22. D23. C24. A25. C26. B27. A28. D29. A30. E31. C32. B33. E34. C35. BPage 6