Section 1.2 Autonomous DEs of the form 0 The DE y is mthemticl model for wide vriety of pplictions. Some of the pplictions re descried y sying the rte of chnge of y(t) is proportionl to the mount present. So specil cse is the DE of the form ky, where proportionlity constnt k my e positive or negtive So if we cn solve DE y then we cn solve DE y setting k = ky nd = 0. The solution process requires simple integrtion.
Popultion growth or decy; y(t) = popultion t time t ky,k 0 for growth k 0 for decy Rdioctive decy; y(t) = mount of rdioctive mteril t time t ky,k 0 Chemicl or mediction in lood strem; y(t) = mount of sustnce in the lood strem t time t (recll tht the o sors the sustnce) ky,k 0 Investment with simple interest; y(t) = mount of money t time t ky,k 0 Newton s Lw of Cooling/Heting; y(t) = temperture of the oject t time t k(y(t) T ),where T mient temperture Mixing prolems; y(t) = mount of sustnce in the tnk t time t Input Rte of y(t) Output Rte of y(t)
RL-circuit; y(t) = i(t) the current; in n RL-circuit, the differentil eqution formed using Kirchhoff's lw, is di Ri L V Flling or tossed ojects; y(t) = v(t), the velocity t time t dv m mg v m = mss, g = ccelertion of grvity (g > 0 for flling nd g < 0 for tossed), γ is the drg coefficient for ir resistnce The restrictions on g re the wy our ook trets things which my e different thn in other courses.
Solving some utonomous DE y solution. ( nd re constnt nd 0) ; the oject is to otin formul for the generl The solution technique is to rerrnge terms so tht we cn perform n integrtion. Bsiclly we seprte the vriles s shown next. Next integrte oth sides: Use some lger: Solve for y = y(t): 1 ln y t C C is n ritrry constnt t y - = ±Ce (Renme ±C to e C nd solve for y.) This is clled the generl solution of the DE nd it represents the set of ll possile solutions to the DE.
Exmple on Rdioctive Decy Specil Cse: y y Ce t In nture, rdium is found in urnium. Rdium is not necessry for living orgnisms, nd dverse helth effects re likely when it is incorported into iochemicl processes ecuse of its rdioctivity nd chemicl rectivity. Currently, other thn its use in nucler medicine, rdium hs no commercil pplictions. Rdium is highly toxic. When uying house it should e checked for rdium (gs) levels to see if n exhust system needs to e instlled. https://en.wikipedi.org/wiki/rdium The DE tht governs rdioctive decy is ky,k 0 Thus the solution of the DE is where = 0 nd = -k; so y Ce kt To hve this solution pply to rdium we need to determine (rte constnt) k. y Ce Scientist hve determined tht the hlf-life of rdium is 1600 yers. This mens if t time = 0 we hd 100 ls of rdium tht in 1600 yers it would hve decyed to 50 ls. We cn use this to determine the rte constnt k. Let y(0) = 100 nd y(1600) = 50. Since y(0) = 100 it is esy to show tht C = 100. Then the eqution is -kt y = 100e. Let t = 1600 nd y = 50 to solve for k. ln( 50 / 100) 1600 -k1600 50 = 100 e Use logrithms; we get 4 k 4. 332 * 10 t
Exmple on Popultions Specil Cse: y y Ce t Mice nd Owls Model Our Mice nd Owls model hs n initil vlue prolem (IVP). = 0.5 nd = 450 so the solution is (replcing y y p) 450 p Ce Ce 900 Ce 05. t 0. 5 t 0. 5 t Initil condition p(0) = 600 Apply the initil condition: Set t = 0 nd p = 600 nd solve for C. 600 = 900 + C C = -300 Solution of the IVP is p = 900-300e 0.5t How do we find the time (in months) tht the popultion of mice ecomes extinct?
Specil Cse: Exmple on Flling Bo Flling Bo Model Our flling o model hs n initil vlue prolem (IVP) Replce y with v. y y dv v = 9.8-5 v(0) = 0 Ce t = -1/5 = -0.2 nd = -9.8 so the solution is 98. v Ce Ce 49 Ce 02. t 0. 2 t 0. 2 t Apply the initil condition: Set t = 0 nd v = 0 nd solve for C. 0 = 49 + C so C = -49 Solution of the IVP is v = 49-49e -0.2t This gives the velocity of the flling oject t ny positive time. Wht is the limiting (terminl) velocity (the fstest the o could trvel)? How do you determine the time t which the o is 80% of its limiting velocity?
Exmple Newton s Lw of Cooling y t Specil Cse: y Ce Suppose tht uilding loses het in ccordnce with Newton's lw of cooling nd tht the rte constnt k hs the vlue -0.19 per hour. Assume tht the interior temperture is 68 F t the time the heting system fils. If the externl temperture is 15 F, how long will it tke for the interior temperture to fll to 40 F? Newton s Lw of Cooling sys tht the temperture of n oject chnges t rte proportionl to the difference etween the temperture of the oject itself nd the temperture of its surroundings (the mient temperture). Let Q(t) e the temperture of the oject t time t. The mient temperture is 15 F nd the rte constnt is -0.19 per hour (since we re cooling). We get initil vlue prolem dq dq = 0.19(Q - 15), Q(0) = 68 = 0.19(Q - 15) t -2.85-0.19 t -0.2 t Q = + Ce = + Ce = 15 + Ce -0.19 rerrnge Set t = 0 nd Q = 68 nd solve for C. C = 68 15 = 53 The solution of this IVP is Q(t) = 15 + 53e -0.19t. -0.19Q + (-0.19)(-15) = -0.19Q + 2.85 = -0.19, = -2.85 How do you determine the time (in hours) it tkes for the interior temperture to rech 40 F?
Exmple Tnk Prolem Specil Cse: y y Ce t A 500 gllon tnk initilly contins 200 gllons of slt solution which contins 25 pounds of slt. Wter contining 2ls of slt per gllon flows in t the rte of 5 gllons minute (the mixture is kept well stirred) nd flows out of the tnk t the sme rte. Let Q(t) e the mount of slt in the tnk t time t. Then the rte of chnge of the slt in the tnk is modeled y the DE dq = Input Rte of Slt - Output Rte of Slt Q(0) = 25 dq = 2l / gl * 5gl / min - Ql / 200gl * 5gl / min = 10 - Q / 40 t -10-0.19 t -1/ 40 t Q = + Ce = + Ce = 400 + Ce -1/ 40 Set t = 0 nd Q = 25 nd solve for C. C = 25 400 = -375 The solution of this IVP is Q(t) = 400-375e -1/40 t. = -1/40, = -10 Will the tnk ever empty? Explin.