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Math Advanced Math Student's Solutions Manual for Fundamentals of Differential Equations and Fundamentals of Differential Equations and Boundary Value Problems

(a) The behavior of the solution to the initial value problem y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 is oscillatory _ as t → + ∞ . Procedure used: The second-order constant-coefficient differential equation a y ″ + b y ′ + c y = 0 can be interpreted as a mass-spring system with mass a , damping coefficient b , spring stiffness c and displacement y , if these a is positive and b and c are nonnegative. With a = m and c = k , the conclusions are: (a) The oscillations will not occur for b > 4 m k this is called over damping phenomenon. (b) Negative damping coefficient b will give rise to a friction force F friction = − b y ′ that imparts energy to the system instead of draining it. The increase in energy over time reveal itself in oscillations of ever-greater amplitude. (c) For negative stiffness k , as the mass is moved away from the spring’s equilibrium point, the spring repels the mass farther with a force F spring = − k y that intensifies as the displacement increases. The spring must exile the mass to ± ∞ and expect all solutions y ( t ) to approach ± ∞ as t increases. Calculation: The given initial value problem is y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the positive coefficients are m = 1 , b = 0 and k = 16 . Then, 4 m k will be, 4 ( 1 ) ( 16 ) = 8 . Since 0 < 8 , it is obtained that b < 4 m k . As there is no damping, the solution to y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 will be oscillatory. Solve y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 + 16 = 0 . Factorize the polynomial r 2 + 16 = 0 as, r 2 = − 16 r = ± − 16 r = ± 4 i ( − 1 = i ) That is, r = 0 ± 4 i . Here, α = 0 and β = 4 . Thus, the general solution will be, y ( t ) = c 1 e 0 cos ( 4 t ) + c 2 e 0 sin ( 4 t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) Apply the condition y ( 0 ) = 2 in y ( t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) , y ( 0 ) = c 1 cos ( 4 × 0 ) + c 2 sin ( 4 × 0 ) 2 = c 1 cos ( 0 ) + c 2 sin ( 0 ) 2 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 2 Differentiate y ( t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) with respect to t as, y ′ ( t ) = − 4 c 1 sin ( 4 t ) + 4 c 2 cos ( 4 t ) . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 4 c 1 sin ( 4 t ) + 4 c 2 cos ( 4 t ) , y ′ ( 0 ) = − 4 c 1 sin ( 4 × 0 ) + 4 c 2 cos ( 4 × 0 ) 0 = − 4 c 1 sin ( 0 ) + 4 c 2 cos ( 0 ) 0 = − 4 c 1 ( 0 ) + 4 c 2 ( 1 ) 4 c 2 = 0 c 2 = 0 That is, c 1 = 2 and c 2 = 0 . The general solution y ( t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) becomes, y ( t ) = 2 cos ( 4 t ) + ( 0 ) sin ( 4 t ) = 2 cos ( 4 t ) It is known that, − 1 ≤ cos ( 4 t ) ≤ 1 . Thus, y ( t ) = 2 cos ( 4 t ) oscillates between − 2 and 2 . Therefore, the solution to the initial value problem y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 is oscillatory _ as t → + ∞ . (b) The behavior of the solution to the initial value problem y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to zero _ as t → + ∞ . The given initial value problem is y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the positive coefficients are m = 1 , b = 100 and k = 1 . Then, 4 m k will be, 4 ( 1 ) ( 1 ) = 2 . Since 100 > 2 , it is obtained that b > 4 m k . Due to damping, the oscillations will not occur for the solution to y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Here, the damping coefficient b is much greater and so it forces the solution to zero. That is, the solution will tend to zero due to damping. Solve y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 + 100 r + 1 = 0 . Factorize the polynomial r 2 + 100 r + 1 = 0 using the quadratic formula − b ± b 2 − 4 a c 2 a . r = − 100 ± 100 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) = − 100 ± 9996 2 = − 100 ± 2 2499 2 = − 50 ± 2499 Note that, the roots r = − 50 + 2499 and r = − 50 − 2499 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t , y ( 0 ) = c 1 e ( − 50 + 2499 ) × 0 + c 2 e ( − 50 − 2499 ) × 0 1 = c 1 e 0 + c 2 e 0 c 1 + c 2 = 1 c 2 = 1 − c 1 Differentiate y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t with respect to t to obtain y ′ ( t ) = ( − 50 + 2499 ) c 1 e ( − 50 + 2499 ) t + ( − 50 − 2499 ) c 2 e ( − 50 − 2499 ) t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 50 + 2499 ) c 1 e ( − 50 + 2499 ) t + ( − 50 − 2499 ) c 2 e ( − 50 − 2499 ) t , y ′ ( 0 ) = ( − 50 + 2499 ) c 1 e ( − 50 + 2499 ) × 0 + ( − 50 − 2499 ) c 2 e ( − 50 − 2499 ) × 0 0 = ( − 50 + 2499 ) c 1 e 0 + ( − 50 − 2499 ) c 2 e 0 ( − 50 + 2499 ) c 1 = ( 50 + 2499 ) c 2 Substitute c 2 = 1 − c 1 in ( − 50 + 2499 ) c 1 = ( 50 + 2499 ) c 2 , ( − 50 + 2499 ) c 1 = ( 50 + 2499 ) ( 1 − c 1 ) − 50 c 1 + 2499 c 1 = 50 + 2499 − 50 c 1 − 2499 c 1 2 2499 c 1 = 50 + 2499 c 1 = 50 + 2499 2 2499 c 1 = 1 2 + 25 2499 Substitute c 1 = 1 2 + 25 2499 in c 2 = 1 − c 1 , c 2 = 1 − ( 1 2 + 25 2499 ) = 1 2 − 25 2499 That is, c 1 = 1 2 + 25 2499 and c 2 = 1 2 − 25 2499 . The general solution y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t becomes y ( t ) = ( 1 2 + 25 2499 ) e ( − 50 + 2499 ) t + ( 1 2 − 25 2499 ) e ( − 50 − 2499 ) t . Note that, the exponential powers ( − 50 + 2499 ) t ≈ − 0.01 t and ( − 50 − 2499 ) t ≈ − 99.99 t . As t → + ∞ , the powers ( − 50 + 2499 ) t and ( − 50 − 2499 ) t tends to − ∞ . That is, the value of e ( − 50 + 2499 ) t and e ( − 50 − 2499 ) t tends to 0. Thus, the solution tends to 0 as t → + ∞ . Therefore, the solution to the initial value problem y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to zero _ as t → + ∞ . (c) The behavior of the solution to the initial value problem y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . The given initial value problem is y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the coefficients are m = 1 , b = − 6 and k = 8 . Then, 4 m k will be, 4 ( 1 ) ( 8 ) = 4 2 . Since − 6 < 4 2 , it is obtained that b < 4 m k . Here, the damping coefficient b is negative. This will give rise to a friction force which imparts energy to the system instead of draining it. That is, the solution will tend to − ∞ due to negative damping coefficient. Solve y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 − 6 r + 8 = 0 . Factorize the polynomial r 2 − 6 r + 8 = 0 as, r 2 − 2 r − 4 r + 8 = 0 r ( r − 2 ) − 4 ( r − 2 ) = 0 ( r − 2 ) ( r − 4 ) = 0 r = 2 , 4 Note that, the roots r = 2 and r = 4 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e 2 t + c 2 e 4 t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e 2 t + c 2 e 4 t , y ( 0 ) = c 1 e 2 × 0 + c 2 e 4 × 0 1 = c 1 e 0 + c 2 e 0 c 1 + c 2 = 1 c 2 = 1 − c 1 Differentiate y ( t ) = c 1 e 2 t + c 2 e 4 t with respect to t to obtain y ′ ( t ) = 2 c 1 e 2 t + 4 c 2 e 4 t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = 2 c 1 e 2 t + 4 c 2 e 4 t , y ′ ( 0 ) = 2 c 1 e 2 × 0 + 4 c 2 e 4 × 0 0 = 2 c 1 e 0 + 4 c 2 e 0 2 c 1 = − 4 c 2 Substitute c 2 = 1 − c 1 in 2 c 1 = − 4 c 2 , 2 c 1 = − 4 ( 1 − c 1 ) 2 c 1 = − 4 + 4 c 1 2 c 1 = 4 c 1 = 2 Substitute c 1 = 2 in c 2 = 1 − c 1 to get c 2 = − 1 . That is, c 1 = 2 and c 2 = − 1 . The general solution y ( t ) = c 1 e 2 t + c 2 e 4 t becomes y ( t ) = 2 e 2 t − e 4 t . Note that, y ( t ) = 2 e 2 t − ( e 2 t ) 2 and 2 a < a 2 for a > 1 . As t → + ∞ , the terms e 2 t tends to ∞ . That is, the value of y ( t ) = 2 e 2 t − ( e 2 t ) 2 will tend to − ∞ as t → + ∞ . Therefore, the solution to the initial value problem y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . (d) The behavior of the solution to the initial value problem y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . The given initial value problem is y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the coefficients are m = 1 , b = 2 and k = − 3 . Here, the stiffness k is negative. This will give rise to a force that intensifies as displacement increases. That is, the spring must exile the mass to ± ∞ and expect all solutions y ( t ) to approach ± ∞ as t increases. Solve y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 + 2 r − 3 = 0 . Factorize the polynomial r 2 + 2 r − 3 = 0 as, r 2 − r + 3 r − 3 = 0 r ( r − 1 ) + 3 ( r − 1 ) = 0 ( r + 3 ) ( r − 1 ) = 0 r = − 3 , 1 Note that, the roots r = − 3 and r = 1 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e − 3 t + c 2 e t . Apply the condition y ( 0 ) = − 2 in y ( t ) = c 1 e − 3 t + c 2 e t , y ( 0 ) = c 1 e − 3 × 0 + c 2 e 0 − 2 = c 1 e 0 + c 2 e 0 c 1 + c 2 = − 2 c 1 = − 2 − c 2 Differentiate y ( t ) = c 1 e − 3 t + c 2 e t with respect to t to obtain y ′ ( t ) = − 3 c 1 e − 3 t + c 2 e t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 3 c 1 e − 3 t + c 2 e t , y ′ ( 0 ) = − 3 c 1 e − 3 × 0 + c 2 e 0 0 = − 3 c 1 e 0 + c 2 e 0 c 2 = 3 c 1 Substitute c 2 = 3 c 1 in c 1 = − 2 − c 2 , c 1 = − 2 − 3 c 1 4 c 1 = − 2 c 1 = − 1 2 Substitute c 1 = − 1 2 in c 2 = 3 c 1 to get c 2 = − 3 2 . That is, c 1 = − 1 2 and c 2 = − 3 2 . The general solution y ( t ) = c 1 e − 3 t + c 2 e t becomes y ( t ) = − 1 2 e − 3 t − 3 2 e t . As t → + ∞ , the term e − 3 t tends to 0 and e t tends to ∞ . That is, the value of y ( t ) = − 1 2 e − 3 t − 3 2 e t will tend to − ∞ as t → + ∞ . Therefore, the solution to the initial value problem y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . (e) The behavior of the solution to the initial value problem y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 tends to ∞ _ as t → + ∞ . The given initial value problem is y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 . Comparing with m y ″ + b y ′ + k y = 0 , the coefficients are m = 1 , b = − 1 and k = − 3 . Here, the stiffness k is negative, which will give rise to a force that intensifies as displacement increases. That is, the spring must exile the mass to ± ∞ and expect all solutions y ( t ) to approach ± ∞ as t increases. Solve y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 as follows: The associated auxiliary equation is r 2 − r − 6 = 0 . Factorize the polynomial r 2 − r − 6 = 0 as, r 2 − 2 r + 3 r − 6 = 0 r ( r − 2 ) + 3 ( r − 2 ) = 0 ( r + 3 ) ( r − 2 ) = 0 r = − 3 , 2 Note that, the roots r = − 3 and r = 2 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e − 3 t + c 2 e 2 t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − 3 t + c 2 e 2 t , y ( 0 ) = c 1 e − 3 × 0 + c 2 e 2 × 0 1 = c 1 e 0 + c 2 e 0 c 1 + c 2 = 1 c 1 = 1 − c 2 Differentiate y ( t ) = c 1 e − 3 t + c 2 e 2 t with respect to t to obtain y ′ ( t ) = − 3 c 1 e − 3 t + 2 c 2 e 2 t . Apply the condition y ′ ( 0 ) = 1 in y ( t ) = c 1 e − 3 t + c 2 e 2 t , y ′ ( 0 ) = − 3 c 1 e − 3 × 0 + 2 c 2 e 2 × 0 0 = − 3 c 1 e 0 + 2 c 2 e 0 2 c 2 = 3 c 1 c 2 = 3 2 c 1 Substitute c 2 = 3 2 c 1 in c 1 = 1 − c 2 , c 1 = 1 − 3 2 c 1 5 2 c 1 = 1 c 1 = 2 5 Substitute c 1 = 2 5 in c 2 = 3 2 c 1 to get c 2 = 3 5 . That is, c 1 = 2 5 and c 2 = 3 5 . The general solution y ( t ) = c 1 e − 3 t + c 2 e 2 t becomes y ( t ) = 2 5 e − 3 t + 3 5 e 2 t . As t → + ∞ , the term e − 3 t tends to 0 and e 2 t tends to ∞ . That is, the value of y ( t ) = 2 5 e − 3 t + 3 5 e 2 t will tend to ∞ as t → + ∞ . Therefore, the solution to the initial value problem y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 tends to ∞ _ as t → + ∞ .

(a) The behavior of the solution to the initial value problem y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 is oscillatory _ as t → + ∞ . Procedure used: The second-order constant-coefficient differential equation a y ″ + b y ′ + c y = 0 can be interpreted as a mass-spring system with mass a , damping coefficient b , spring stiffness c and displacement y , if these a is positive and b and c are nonnegative. With a = m and c = k , the conclusions are: (a) The oscillations will not occur for b > 4 m k this is called over damping phenomenon. (b) Negative damping coefficient b will give rise to a friction force F friction = − b y ′ that imparts energy to the system instead of draining it. The increase in energy over time reveal itself in oscillations of ever-greater amplitude. (c) For negative stiffness k , as the mass is moved away from the spring’s equilibrium point, the spring repels the mass farther with a force F spring = − k y that intensifies as the displacement increases. The spring must exile the mass to ± ∞ and expect all solutions y ( t ) to approach ± ∞ as t increases. Calculation: The given initial value problem is y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the positive coefficients are m = 1 , b = 0 and k = 16 . Then, 4 m k will be, 4 ( 1 ) ( 16 ) = 8 . Since 0 < 8 , it is obtained that b < 4 m k . As there is no damping, the solution to y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 will be oscillatory. Solve y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 + 16 = 0 . Factorize the polynomial r 2 + 16 = 0 as, r 2 = − 16 r = ± − 16 r = ± 4 i ( − 1 = i ) That is, r = 0 ± 4 i . Here, α = 0 and β = 4 . Thus, the general solution will be, y ( t ) = c 1 e 0 cos ( 4 t ) + c 2 e 0 sin ( 4 t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) Apply the condition y ( 0 ) = 2 in y ( t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) , y ( 0 ) = c 1 cos ( 4 × 0 ) + c 2 sin ( 4 × 0 ) 2 = c 1 cos ( 0 ) + c 2 sin ( 0 ) 2 = c 1 ( 1 ) + c 2 ( 0 ) c 1 = 2 Differentiate y ( t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) with respect to t as, y ′ ( t ) = − 4 c 1 sin ( 4 t ) + 4 c 2 cos ( 4 t ) . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 4 c 1 sin ( 4 t ) + 4 c 2 cos ( 4 t ) , y ′ ( 0 ) = − 4 c 1 sin ( 4 × 0 ) + 4 c 2 cos ( 4 × 0 ) 0 = − 4 c 1 sin ( 0 ) + 4 c 2 cos ( 0 ) 0 = − 4 c 1 ( 0 ) + 4 c 2 ( 1 ) 4 c 2 = 0 c 2 = 0 That is, c 1 = 2 and c 2 = 0 . The general solution y ( t ) = c 1 cos ( 4 t ) + c 2 sin ( 4 t ) becomes, y ( t ) = 2 cos ( 4 t ) + ( 0 ) sin ( 4 t ) = 2 cos ( 4 t ) It is known that, − 1 ≤ cos ( 4 t ) ≤ 1 . Thus, y ( t ) = 2 cos ( 4 t ) oscillates between − 2 and 2 . Therefore, the solution to the initial value problem y ″ + 16 y = 0 , y ( 0 ) = 2 , y ′ ( 0 ) = 0 is oscillatory _ as t → + ∞ . (b) The behavior of the solution to the initial value problem y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to zero _ as t → + ∞ . The given initial value problem is y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the positive coefficients are m = 1 , b = 100 and k = 1 . Then, 4 m k will be, 4 ( 1 ) ( 1 ) = 2 . Since 100 > 2 , it is obtained that b > 4 m k . Due to damping, the oscillations will not occur for the solution to y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Here, the damping coefficient b is much greater and so it forces the solution to zero. That is, the solution will tend to zero due to damping. Solve y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 + 100 r + 1 = 0 . Factorize the polynomial r 2 + 100 r + 1 = 0 using the quadratic formula − b ± b 2 − 4 a c 2 a . r = − 100 ± 100 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) = − 100 ± 9996 2 = − 100 ± 2 2499 2 = − 50 ± 2499 Note that, the roots r = − 50 + 2499 and r = − 50 − 2499 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t , y ( 0 ) = c 1 e ( − 50 + 2499 ) × 0 + c 2 e ( − 50 − 2499 ) × 0 1 = c 1 e 0 + c 2 e 0 c 1 + c 2 = 1 c 2 = 1 − c 1 Differentiate y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t with respect to t to obtain y ′ ( t ) = ( − 50 + 2499 ) c 1 e ( − 50 + 2499 ) t + ( − 50 − 2499 ) c 2 e ( − 50 − 2499 ) t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 50 + 2499 ) c 1 e ( − 50 + 2499 ) t + ( − 50 − 2499 ) c 2 e ( − 50 − 2499 ) t , y ′ ( 0 ) = ( − 50 + 2499 ) c 1 e ( − 50 + 2499 ) × 0 + ( − 50 − 2499 ) c 2 e ( − 50 − 2499 ) × 0 0 = ( − 50 + 2499 ) c 1 e 0 + ( − 50 − 2499 ) c 2 e 0 ( − 50 + 2499 ) c 1 = ( 50 + 2499 ) c 2 Substitute c 2 = 1 − c 1 in ( − 50 + 2499 ) c 1 = ( 50 + 2499 ) c 2 , ( − 50 + 2499 ) c 1 = ( 50 + 2499 ) ( 1 − c 1 ) − 50 c 1 + 2499 c 1 = 50 + 2499 − 50 c 1 − 2499 c 1 2 2499 c 1 = 50 + 2499 c 1 = 50 + 2499 2 2499 c 1 = 1 2 + 25 2499 Substitute c 1 = 1 2 + 25 2499 in c 2 = 1 − c 1 , c 2 = 1 − ( 1 2 + 25 2499 ) = 1 2 − 25 2499 That is, c 1 = 1 2 + 25 2499 and c 2 = 1 2 − 25 2499 . The general solution y ( t ) = c 1 e ( − 50 + 2499 ) t + c 2 e ( − 50 − 2499 ) t becomes y ( t ) = ( 1 2 + 25 2499 ) e ( − 50 + 2499 ) t + ( 1 2 − 25 2499 ) e ( − 50 − 2499 ) t . Note that, the exponential powers ( − 50 + 2499 ) t ≈ − 0.01 t and ( − 50 − 2499 ) t ≈ − 99.99 t . As t → + ∞ , the powers ( − 50 + 2499 ) t and ( − 50 − 2499 ) t tends to − ∞ . That is, the value of e ( − 50 + 2499 ) t and e ( − 50 − 2499 ) t tends to 0. Thus, the solution tends to 0 as t → + ∞ . Therefore, the solution to the initial value problem y ″ + 100 y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to zero _ as t → + ∞ . (c) The behavior of the solution to the initial value problem y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . The given initial value problem is y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the coefficients are m = 1 , b = − 6 and k = 8 . Then, 4 m k will be, 4 ( 1 ) ( 8 ) = 4 2 . Since − 6 < 4 2 , it is obtained that b < 4 m k . Here, the damping coefficient b is negative. This will give rise to a friction force which imparts energy to the system instead of draining it. That is, the solution will tend to − ∞ due to negative damping coefficient. Solve y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 − 6 r + 8 = 0 . Factorize the polynomial r 2 − 6 r + 8 = 0 as, r 2 − 2 r − 4 r + 8 = 0 r ( r − 2 ) − 4 ( r − 2 ) = 0 ( r − 2 ) ( r − 4 ) = 0 r = 2 , 4 Note that, the roots r = 2 and r = 4 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e 2 t + c 2 e 4 t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e 2 t + c 2 e 4 t , y ( 0 ) = c 1 e 2 × 0 + c 2 e 4 × 0 1 = c 1 e 0 + c 2 e 0 c 1 + c 2 = 1 c 2 = 1 − c 1 Differentiate y ( t ) = c 1 e 2 t + c 2 e 4 t with respect to t to obtain y ′ ( t ) = 2 c 1 e 2 t + 4 c 2 e 4 t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = 2 c 1 e 2 t + 4 c 2 e 4 t , y ′ ( 0 ) = 2 c 1 e 2 × 0 + 4 c 2 e 4 × 0 0 = 2 c 1 e 0 + 4 c 2 e 0 2 c 1 = − 4 c 2 Substitute c 2 = 1 − c 1 in 2 c 1 = − 4 c 2 , 2 c 1 = − 4 ( 1 − c 1 ) 2 c 1 = − 4 + 4 c 1 2 c 1 = 4 c 1 = 2 Substitute c 1 = 2 in c 2 = 1 − c 1 to get c 2 = − 1 . That is, c 1 = 2 and c 2 = − 1 . The general solution y ( t ) = c 1 e 2 t + c 2 e 4 t becomes y ( t ) = 2 e 2 t − e 4 t . Note that, y ( t ) = 2 e 2 t − ( e 2 t ) 2 and 2 a < a 2 for a > 1 . As t → + ∞ , the terms e 2 t tends to ∞ . That is, the value of y ( t ) = 2 e 2 t − ( e 2 t ) 2 will tend to − ∞ as t → + ∞ . Therefore, the solution to the initial value problem y ″ − 6 y ′ + 8 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . (d) The behavior of the solution to the initial value problem y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . The given initial value problem is y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 . Comparing with m y ″ + b y ′ + k y = 0 , the coefficients are m = 1 , b = 2 and k = − 3 . Here, the stiffness k is negative. This will give rise to a force that intensifies as displacement increases. That is, the spring must exile the mass to ± ∞ and expect all solutions y ( t ) to approach ± ∞ as t increases. Solve y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 as follows: The associated auxiliary equation is r 2 + 2 r − 3 = 0 . Factorize the polynomial r 2 + 2 r − 3 = 0 as, r 2 − r + 3 r − 3 = 0 r ( r − 1 ) + 3 ( r − 1 ) = 0 ( r + 3 ) ( r − 1 ) = 0 r = − 3 , 1 Note that, the roots r = − 3 and r = 1 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e − 3 t + c 2 e t . Apply the condition y ( 0 ) = − 2 in y ( t ) = c 1 e − 3 t + c 2 e t , y ( 0 ) = c 1 e − 3 × 0 + c 2 e 0 − 2 = c 1 e 0 + c 2 e 0 c 1 + c 2 = − 2 c 1 = − 2 − c 2 Differentiate y ( t ) = c 1 e − 3 t + c 2 e t with respect to t to obtain y ′ ( t ) = − 3 c 1 e − 3 t + c 2 e t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 3 c 1 e − 3 t + c 2 e t , y ′ ( 0 ) = − 3 c 1 e − 3 × 0 + c 2 e 0 0 = − 3 c 1 e 0 + c 2 e 0 c 2 = 3 c 1 Substitute c 2 = 3 c 1 in c 1 = − 2 − c 2 , c 1 = − 2 − 3 c 1 4 c 1 = − 2 c 1 = − 1 2 Substitute c 1 = − 1 2 in c 2 = 3 c 1 to get c 2 = − 3 2 . That is, c 1 = − 1 2 and c 2 = − 3 2 . The general solution y ( t ) = c 1 e − 3 t + c 2 e t becomes y ( t ) = − 1 2 e − 3 t − 3 2 e t . As t → + ∞ , the term e − 3 t tends to 0 and e t tends to ∞ . That is, the value of y ( t ) = − 1 2 e − 3 t − 3 2 e t will tend to − ∞ as t → + ∞ . Therefore, the solution to the initial value problem y ″ + 2 y ′ − 3 y = 0 , y ( 0 ) = − 2 , y ′ ( 0 ) = 0 tends to − ∞ _ as t → + ∞ . (e) The behavior of the solution to the initial value problem y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 as t → + ∞ and confirm the predictions by actually solving the problem. The solution to the initial value problem y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 tends to ∞ _ as t → + ∞ . The given initial value problem is y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 . Comparing with m y ″ + b y ′ + k y = 0 , the coefficients are m = 1 , b = − 1 and k = − 3 . Here, the stiffness k is negative, which will give rise to a force that intensifies as displacement increases. That is, the spring must exile the mass to ± ∞ and expect all solutions y ( t ) to approach ± ∞ as t increases. Solve y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 as follows: The associated auxiliary equation is r 2 − r − 6 = 0 . Factorize the polynomial r 2 − r − 6 = 0 as, r 2 − 2 r + 3 r − 6 = 0 r ( r − 2 ) + 3 ( r − 2 ) = 0 ( r + 3 ) ( r − 2 ) = 0 r = − 3 , 2 Note that, the roots r = − 3 and r = 2 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e − 3 t + c 2 e 2 t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − 3 t + c 2 e 2 t , y ( 0 ) = c 1 e − 3 × 0 + c 2 e 2 × 0 1 = c 1 e 0 + c 2 e 0 c 1 + c 2 = 1 c 1 = 1 − c 2 Differentiate y ( t ) = c 1 e − 3 t + c 2 e 2 t with respect to t to obtain y ′ ( t ) = − 3 c 1 e − 3 t + 2 c 2 e 2 t . Apply the condition y ′ ( 0 ) = 1 in y ( t ) = c 1 e − 3 t + c 2 e 2 t , y ′ ( 0 ) = − 3 c 1 e − 3 × 0 + 2 c 2 e 2 × 0 0 = − 3 c 1 e 0 + 2 c 2 e 0 2 c 2 = 3 c 1 c 2 = 3 2 c 1 Substitute c 2 = 3 2 c 1 in c 1 = 1 − c 2 , c 1 = 1 − 3 2 c 1 5 2 c 1 = 1 c 1 = 2 5 Substitute c 1 = 2 5 in c 2 = 3 2 c 1 to get c 2 = 3 5 . That is, c 1 = 2 5 and c 2 = 3 5 . The general solution y ( t ) = c 1 e − 3 t + c 2 e 2 t becomes y ( t ) = 2 5 e − 3 t + 3 5 e 2 t . As t → + ∞ , the term e − 3 t tends to 0 and e 2 t tends to ∞ . That is, the value of y ( t ) = 2 5 e − 3 t + 3 5 e 2 t will tend to ∞ as t → + ∞ . Therefore, the solution to the initial value problem y ″ − y ′ − 6 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 1 tends to ∞ _ as t → + ∞ .

Solution Summary: The author explains the behavior of the solution to the initial value problem y′′+16y=0 and confirms the predictions by solving the problem.

Student's Solutions Manual for Fundamentals of Differential Equations and Fundamentals of Differential Equations and Boundary Value Problems

Student's Solutions Manual for Fundamentals of Differential Equations and Fundamentals of Differential Equations and Boundary Value Problems

9th Edition

ISBN: 9780321977212

Author: Nagle, R. Kent; Saff, Edward B.; Snider, Arthur David

Publisher: PEARSON

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Student's Solutions Manual for Fundamentals of Differential Equations and Fundamentals of Differential Equations and Boundary Value Problems

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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