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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

(a) The explicit formulas for cosh t and sinh t by solving the differential equation y ″ − y = 0 and show that d d t cosh t = sinh t and d d t sinh t = cosh t . The explicit formulas for cosh t and sinh t are cosh t = e t + e − t 2 and sinh t = e t − e − t 2 _ . The given differential equation is y ″ − y = 0 . The initial values for the solution cosh t are y ( 0 ) = 1 , y ′ ( 0 ) = 0 . The auxiliary equation associated with y ″ − y = 0 is r 2 − 1 = 0 . Simplify as, r 2 = 1 r = ± 1 That is, the roots r = − 1 and r = 1 are distinct and real. The solutions will be e − t and e t . Thus, the general solution will be, y ( t ) = c 1 e − t + c 2 e t . Apply the initial condition y ( 0 ) = 1 in y ( t ) = c 1 e − t + c 2 e t , y ( 0 ) = c 1 e − ( 0 ) + c 2 e 0 1 = c 1 ( 1 ) + c 2 ( 1 ) c 1 + c 2 = 1 Differentiate y ( t ) = c 1 e − t + c 2 e t with respect to t as, y ′ ( t ) = − c 1 e − t + c 2 e t . Apply the initial condition y ′ ( 0 ) = 0 in y ′ ( t ) = − c 1 e − t + c 2 e t , y ′ ( 0 ) = − c 1 e − ( 0 ) + c 2 e ( 0 ) 0 = − c 1 ( 1 ) + c 2 ( 1 ) c 1 = c 2 Substitute c 1 = c 2 in c 1 + c 2 = 1 , c 2 + c 2 = 1 2 c 2 = 1 c 2 = 1 2 Then, c 1 = c 2 becomes c 1 = 1 2 . Since c 1 = c 2 = 1 2 , the general solution y ( t ) = c 1 e − t + c 2 e t becomes, y ( t ) = 1 2 e − t + 1 2 e t = 1 2 ( e t + e − t ) = e t + e − t 2 Thus, cosh t = e t + e − t 2 . The initial values for the solution sinh t are y ( 0 ) = 0 , y ′ ( 0 ) = 1 . Apply the initial condition y ( 0 ) = 0 in y ( t ) = c 1 e − t + c 2 e t , y ( 0 ) = c 1 e − ( 0 ) + c 2 e 0 0 = c 1 ( 1 ) + c 2 ( 1 ) c 1 = − c 2 Apply the initial condition y ′ ( 0 ) = 1 in y ′ ( t ) = − c 1 e − t + c 2 e t , y ′ ( 0 ) = − c 1 e − ( 0 ) + c 2 e ( 0 ) 1 = − c 1 ( 1 ) + c 2 ( 1 ) − c 1 + c 2 = 1 Substitute c 1 = − c 2 in − c 1 + c 2 = 1 , − ( − c 2 ) + c 2 = 1 2 c 2 = 1 c 2 = 1 2 Then, c 1 = − c 2 becomes c 1 = − 1 2 . Since c 1 = − 1 2 and c 2 = 1 2 , the general solution y ( t ) = c 1 e − t + c 2 e t becomes, y ( t ) = − 1 2 e − t + 1 2 e t = 1 2 ( e t − e − t ) = e t − e − t 2 Thus, sinh t = e t − e − t 2 . Therefore, the explicit formulas for cosh t and sinh t are cosh t = e t + e − t 2 and sinh t = e t − e − t 2 _ . The next step is to show that, d d t cosh t = sinh t and d d t sinh t = cosh t . Differentiate cosh t = e t + e − t 2 with respect to t as, d d t cosh t = d d t ( e t + e − t 2 ) = 1 2 d d t ( e t + e − t ) = 1 2 ( e t − e − t ) = sinh t Differentiate sinh t = e t − e − t 2 with respect to t as, d d t sinh t = d d t ( e t − e − t 2 ) = 1 2 d d t ( e t − e − t ) = 1 2 ( e t + e − t ) = cosh t Therefore, d d t cosh t = sinh t and d d t sinh t = cosh t . (b) To prove: A general solution of the equation y ″ − y = 0 is y = c 1 cosh t + c 2 sinh t . From part (a), d d t cosh t = sinh t and d d t sinh t = cosh t . Differentiate y = c 1 cosh t + c 2 sinh t with respect to t as, y ′ = c 1 sinh t + c 2 cosh t . Differentiate y ′ = c 1 sinh t + c 2 cosh t with respect to t as, y ″ = c 1 cosh t + c 2 sinh t . Substitute y ″ = c 1 cosh t + c 2 sinh t and y = c 1 cosh t + c 2 sinh t in y ″ − y = 0 , ( c 1 cosh t + c 2 sinh t ) − ( c 1 cosh t + c 2 sinh t ) = 0 0 = 0 Since the equation is satisfied, y = c 1 cosh t + c 2 sinh t is a general solution of y ″ − y = 0 . (c) To show: A general solution of the equation a y ″ + b y ′ + c y = 0 is y = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) , if the roots are in the form α − β and α + β . Given that, roots of the equation a y ″ + b y ′ + c y = 0 are in the form α − β and α + β . Then, the solutions will be e ( α − β ) t and e ( α + β ) t . That is, the general solution will be, y ( t ) = a e ( α − β ) t + b e ( α + β ) t = a e α t e − β t + b e α t e β t = e α t ( a e − β t + b e β t ) From part (a), cosh t = e t + e − t 2 and sinh t = e t − e − t 2 . Since 2 cosh t = e t + e − t and 2 sinh t = e t − e − t , 2 cosh ( β t ) = e β t + e − β t and 2 sinh ( β t ) = e β t − e − β t . Adding the equations, 2 cosh ( β t ) + 2 sinh ( β t ) = ( e β t + e − β t ) + ( e β t − e − β t ) 2 [ cosh ( β t ) + sinh ( β t ) ] = 2 e β t cosh ( β t ) + sinh ( β t ) = e β t Subtracting the equations, 2 cosh ( β t ) − 2 sinh ( β t ) = ( e β t + e − β t ) − ( e β t − e − β t ) 2 [ cosh ( β t ) − sinh ( β t ) ] = 2 e − β t cosh ( β t ) − sinh ( β t ) = e − β t Substitute e β t = cosh ( β t ) + sinh ( β t ) and e − β t = cosh ( β t ) − sinh ( β t ) in y ( t ) = e α t ( a e − β t + b e β t ) , y ( t ) = e α t [ a ( cosh ( β t ) + sinh ( β t ) ) + b ( cosh ( β t ) − sinh ( β t ) ) ] = e α t [ a cosh ( β t ) + a sinh ( β t ) + b cosh ( β t ) − b sinh ( β t ) ] = e α t [ ( a + b ) cosh ( β t ) + ( a − b ) sinh ( β t ) ] = e α t [ c 1 cosh ( β t ) + c 2 sinh ( β t ) ] ( a + b = c 1 and a − b = c 2 ) = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) Therefore, a general solution of the equation a y ″ + b y ′ + c y = 0 is y = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) , if the roots are in the form α − β and α + β . (d) The solution to the initial value problem y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 . The solution to the initial value problem y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 is y ( t ) = 2 e − t 2 cosh ( − 5 2 t ) + 3 e − t 2 sinh ( − 5 2 t ) _ . The given initial value problem is y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 . The auxiliary equation associated with y ″ + y ′ − 6 y = 0 is r 2 + r − 6 = 0 . Facrorize the polynomial r 2 + r − 6 = 0 as, r 2 + 3 r − 2 r − 6 = 0 r ( r + 3 ) − 2 ( r + 3 ) = 0 ( r − 2 ) ( r + 3 ) = 0 r = − 3 , 2 From part (a), a general solution of the equation a y ″ + b y ′ + c y = 0 is y = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) , if the roots are in the form α − β and α + β . Let α − β = 2 and α + β = − 3 . That is, α = 2 + β . Substitute α = 2 + β in α + β = − 3 , 2 + β + β = − 3 2 + 2 β = − 3 2 β = − 5 β = − 5 2 Substitute β = − 5 2 in α = 2 + β , α = 2 − 5 2 = − 1 2 That is, α = − 1 2 and β = − 5 2 . Thus, the general solution y ( t ) = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) becomes, y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) . Apply the condition y ( 0 ) = 2 in y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) , y ( 0 ) = c 1 e − 1 2 ( 0 ) cosh ( − 5 2 × 0 ) + c 2 e − 1 2 ( 0 ) sinh ( − 5 2 × 0 ) 2 = c 1 ( 1 ) cosh ( 0 ) + c 2 ( 1 ) sinh ( 0 ) 2 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 2 Differentiate y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) with respect to t as, y ′ ( t ) = − 5 2 c 1 e − 1 2 t sinh ( − 5 2 t ) − 1 2 c 1 e − 1 2 t cosh ( − 5 2 t ) − 5 2 c 2 e − 1 2 t cosh ( − 5 2 t ) − 1 2 c 2 e − 1 2 t sinh ( − 5 2 t ) = ( − 5 2 c 1 − 1 2 c 2 ) e − 1 2 t sinh ( − 5 2 t ) + ( − 1 2 c 1 − 5 2 c 2 ) e − 1 2 t cosh ( − 5 2 t ) Apply the condition y ′ ( 0 ) = − 17 2 in y ′ ( t ) = ( − 5 2 c 1 − 1 2 c 2 ) e − 1 2 t sinh ( − 5 2 t ) + ( − 1 2 c 1 − 5 2 c 2 ) e − 1 2 t cosh ( − 5 2 t ) , y ′ ( 0 ) = ( − 5 2 c 1 − 1 2 c 2 ) e − 1 2 ( 0 ) sinh ( − 5 2 × 0 ) + ( − 1 2 c 1 − 5 2 c 2 ) e − 1 2 ( 0 ) cosh ( − 5 2 × 0 ) − 17 2 = ( − 5 2 c 1 − 1 2 c 2 ) sinh ( 0 ) + ( − 1 2 c 1 − 5 2 c 2 ) cosh ( 0 ) − 17 2 = 0 + ( − 1 2 c 1 − 5 2 c 2 ) ( 1 ) − 17 2 = − 1 2 c 1 − 5 2 c 2 Substitute c 1 = 2 in − 17 2 = − 1 2 c 1 − 5 2 c 2 , − 17 2 = − 1 2 ( 2 ) − 5 2 c 2 − 17 2 = − 1 − 5 2 c 2 5 2 c 2 = 17 2 − 1 5 2 c 2 = 15 2 c 2 = 3 That is, c 1 = 2 and c 2 = 3 . Thus, the general solution y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) becomes, y ( t ) = 2 e − 1 2 t cosh ( − 5 2 t ) + 3 e − 1 2 t sinh ( − 5 2 t ) . Therefore, the solution to the initial value problem y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 is y ( t ) = 2 e − t 2 cosh ( − 5 2 t ) + 3 e − t 2 sinh ( − 5 2 t ) _ .

(a) The explicit formulas for cosh t and sinh t by solving the differential equation y ″ − y = 0 and show that d d t cosh t = sinh t and d d t sinh t = cosh t . The explicit formulas for cosh t and sinh t are cosh t = e t + e − t 2 and sinh t = e t − e − t 2 _ . The given differential equation is y ″ − y = 0 . The initial values for the solution cosh t are y ( 0 ) = 1 , y ′ ( 0 ) = 0 . The auxiliary equation associated with y ″ − y = 0 is r 2 − 1 = 0 . Simplify as, r 2 = 1 r = ± 1 That is, the roots r = − 1 and r = 1 are distinct and real. The solutions will be e − t and e t . Thus, the general solution will be, y ( t ) = c 1 e − t + c 2 e t . Apply the initial condition y ( 0 ) = 1 in y ( t ) = c 1 e − t + c 2 e t , y ( 0 ) = c 1 e − ( 0 ) + c 2 e 0 1 = c 1 ( 1 ) + c 2 ( 1 ) c 1 + c 2 = 1 Differentiate y ( t ) = c 1 e − t + c 2 e t with respect to t as, y ′ ( t ) = − c 1 e − t + c 2 e t . Apply the initial condition y ′ ( 0 ) = 0 in y ′ ( t ) = − c 1 e − t + c 2 e t , y ′ ( 0 ) = − c 1 e − ( 0 ) + c 2 e ( 0 ) 0 = − c 1 ( 1 ) + c 2 ( 1 ) c 1 = c 2 Substitute c 1 = c 2 in c 1 + c 2 = 1 , c 2 + c 2 = 1 2 c 2 = 1 c 2 = 1 2 Then, c 1 = c 2 becomes c 1 = 1 2 . Since c 1 = c 2 = 1 2 , the general solution y ( t ) = c 1 e − t + c 2 e t becomes, y ( t ) = 1 2 e − t + 1 2 e t = 1 2 ( e t + e − t ) = e t + e − t 2 Thus, cosh t = e t + e − t 2 . The initial values for the solution sinh t are y ( 0 ) = 0 , y ′ ( 0 ) = 1 . Apply the initial condition y ( 0 ) = 0 in y ( t ) = c 1 e − t + c 2 e t , y ( 0 ) = c 1 e − ( 0 ) + c 2 e 0 0 = c 1 ( 1 ) + c 2 ( 1 ) c 1 = − c 2 Apply the initial condition y ′ ( 0 ) = 1 in y ′ ( t ) = − c 1 e − t + c 2 e t , y ′ ( 0 ) = − c 1 e − ( 0 ) + c 2 e ( 0 ) 1 = − c 1 ( 1 ) + c 2 ( 1 ) − c 1 + c 2 = 1 Substitute c 1 = − c 2 in − c 1 + c 2 = 1 , − ( − c 2 ) + c 2 = 1 2 c 2 = 1 c 2 = 1 2 Then, c 1 = − c 2 becomes c 1 = − 1 2 . Since c 1 = − 1 2 and c 2 = 1 2 , the general solution y ( t ) = c 1 e − t + c 2 e t becomes, y ( t ) = − 1 2 e − t + 1 2 e t = 1 2 ( e t − e − t ) = e t − e − t 2 Thus, sinh t = e t − e − t 2 . Therefore, the explicit formulas for cosh t and sinh t are cosh t = e t + e − t 2 and sinh t = e t − e − t 2 _ . The next step is to show that, d d t cosh t = sinh t and d d t sinh t = cosh t . Differentiate cosh t = e t + e − t 2 with respect to t as, d d t cosh t = d d t ( e t + e − t 2 ) = 1 2 d d t ( e t + e − t ) = 1 2 ( e t − e − t ) = sinh t Differentiate sinh t = e t − e − t 2 with respect to t as, d d t sinh t = d d t ( e t − e − t 2 ) = 1 2 d d t ( e t − e − t ) = 1 2 ( e t + e − t ) = cosh t Therefore, d d t cosh t = sinh t and d d t sinh t = cosh t . (b) To prove: A general solution of the equation y ″ − y = 0 is y = c 1 cosh t + c 2 sinh t . From part (a), d d t cosh t = sinh t and d d t sinh t = cosh t . Differentiate y = c 1 cosh t + c 2 sinh t with respect to t as, y ′ = c 1 sinh t + c 2 cosh t . Differentiate y ′ = c 1 sinh t + c 2 cosh t with respect to t as, y ″ = c 1 cosh t + c 2 sinh t . Substitute y ″ = c 1 cosh t + c 2 sinh t and y = c 1 cosh t + c 2 sinh t in y ″ − y = 0 , ( c 1 cosh t + c 2 sinh t ) − ( c 1 cosh t + c 2 sinh t ) = 0 0 = 0 Since the equation is satisfied, y = c 1 cosh t + c 2 sinh t is a general solution of y ″ − y = 0 . (c) To show: A general solution of the equation a y ″ + b y ′ + c y = 0 is y = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) , if the roots are in the form α − β and α + β . Given that, roots of the equation a y ″ + b y ′ + c y = 0 are in the form α − β and α + β . Then, the solutions will be e ( α − β ) t and e ( α + β ) t . That is, the general solution will be, y ( t ) = a e ( α − β ) t + b e ( α + β ) t = a e α t e − β t + b e α t e β t = e α t ( a e − β t + b e β t ) From part (a), cosh t = e t + e − t 2 and sinh t = e t − e − t 2 . Since 2 cosh t = e t + e − t and 2 sinh t = e t − e − t , 2 cosh ( β t ) = e β t + e − β t and 2 sinh ( β t ) = e β t − e − β t . Adding the equations, 2 cosh ( β t ) + 2 sinh ( β t ) = ( e β t + e − β t ) + ( e β t − e − β t ) 2 [ cosh ( β t ) + sinh ( β t ) ] = 2 e β t cosh ( β t ) + sinh ( β t ) = e β t Subtracting the equations, 2 cosh ( β t ) − 2 sinh ( β t ) = ( e β t + e − β t ) − ( e β t − e − β t ) 2 [ cosh ( β t ) − sinh ( β t ) ] = 2 e − β t cosh ( β t ) − sinh ( β t ) = e − β t Substitute e β t = cosh ( β t ) + sinh ( β t ) and e − β t = cosh ( β t ) − sinh ( β t ) in y ( t ) = e α t ( a e − β t + b e β t ) , y ( t ) = e α t [ a ( cosh ( β t ) + sinh ( β t ) ) + b ( cosh ( β t ) − sinh ( β t ) ) ] = e α t [ a cosh ( β t ) + a sinh ( β t ) + b cosh ( β t ) − b sinh ( β t ) ] = e α t [ ( a + b ) cosh ( β t ) + ( a − b ) sinh ( β t ) ] = e α t [ c 1 cosh ( β t ) + c 2 sinh ( β t ) ] ( a + b = c 1 and a − b = c 2 ) = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) Therefore, a general solution of the equation a y ″ + b y ′ + c y = 0 is y = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) , if the roots are in the form α − β and α + β . (d) The solution to the initial value problem y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 . The solution to the initial value problem y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 is y ( t ) = 2 e − t 2 cosh ( − 5 2 t ) + 3 e − t 2 sinh ( − 5 2 t ) _ . The given initial value problem is y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 . The auxiliary equation associated with y ″ + y ′ − 6 y = 0 is r 2 + r − 6 = 0 . Facrorize the polynomial r 2 + r − 6 = 0 as, r 2 + 3 r − 2 r − 6 = 0 r ( r + 3 ) − 2 ( r + 3 ) = 0 ( r − 2 ) ( r + 3 ) = 0 r = − 3 , 2 From part (a), a general solution of the equation a y ″ + b y ′ + c y = 0 is y = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) , if the roots are in the form α − β and α + β . Let α − β = 2 and α + β = − 3 . That is, α = 2 + β . Substitute α = 2 + β in α + β = − 3 , 2 + β + β = − 3 2 + 2 β = − 3 2 β = − 5 β = − 5 2 Substitute β = − 5 2 in α = 2 + β , α = 2 − 5 2 = − 1 2 That is, α = − 1 2 and β = − 5 2 . Thus, the general solution y ( t ) = c 1 e α t cosh ( β t ) + c 2 e α t sinh ( β t ) becomes, y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) . Apply the condition y ( 0 ) = 2 in y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) , y ( 0 ) = c 1 e − 1 2 ( 0 ) cosh ( − 5 2 × 0 ) + c 2 e − 1 2 ( 0 ) sinh ( − 5 2 × 0 ) 2 = c 1 ( 1 ) cosh ( 0 ) + c 2 ( 1 ) sinh ( 0 ) 2 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 2 Differentiate y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) with respect to t as, y ′ ( t ) = − 5 2 c 1 e − 1 2 t sinh ( − 5 2 t ) − 1 2 c 1 e − 1 2 t cosh ( − 5 2 t ) − 5 2 c 2 e − 1 2 t cosh ( − 5 2 t ) − 1 2 c 2 e − 1 2 t sinh ( − 5 2 t ) = ( − 5 2 c 1 − 1 2 c 2 ) e − 1 2 t sinh ( − 5 2 t ) + ( − 1 2 c 1 − 5 2 c 2 ) e − 1 2 t cosh ( − 5 2 t ) Apply the condition y ′ ( 0 ) = − 17 2 in y ′ ( t ) = ( − 5 2 c 1 − 1 2 c 2 ) e − 1 2 t sinh ( − 5 2 t ) + ( − 1 2 c 1 − 5 2 c 2 ) e − 1 2 t cosh ( − 5 2 t ) , y ′ ( 0 ) = ( − 5 2 c 1 − 1 2 c 2 ) e − 1 2 ( 0 ) sinh ( − 5 2 × 0 ) + ( − 1 2 c 1 − 5 2 c 2 ) e − 1 2 ( 0 ) cosh ( − 5 2 × 0 ) − 17 2 = ( − 5 2 c 1 − 1 2 c 2 ) sinh ( 0 ) + ( − 1 2 c 1 − 5 2 c 2 ) cosh ( 0 ) − 17 2 = 0 + ( − 1 2 c 1 − 5 2 c 2 ) ( 1 ) − 17 2 = − 1 2 c 1 − 5 2 c 2 Substitute c 1 = 2 in − 17 2 = − 1 2 c 1 − 5 2 c 2 , − 17 2 = − 1 2 ( 2 ) − 5 2 c 2 − 17 2 = − 1 − 5 2 c 2 5 2 c 2 = 17 2 − 1 5 2 c 2 = 15 2 c 2 = 3 That is, c 1 = 2 and c 2 = 3 . Thus, the general solution y ( t ) = c 1 e − 1 2 t cosh ( − 5 2 t ) + c 2 e − 1 2 t sinh ( − 5 2 t ) becomes, y ( t ) = 2 e − 1 2 t cosh ( − 5 2 t ) + 3 e − 1 2 t sinh ( − 5 2 t ) . Therefore, the solution to the initial value problem y ″ + y ′ − 6 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = − 17 2 is y ( t ) = 2 e − t 2 cosh ( − 5 2 t ) + 3 e − t 2 sinh ( − 5 2 t ) _ .

Solution Summary: The author analyzes the explicit formulas for mathrmcoshtandmathrsinh.

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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Chapter 4 Solutions

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

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