Explanation Result used: The general solution of ordinary differential equation: “Suppose that, the linear ordinary differential equation of the form y ″ + a y ′ + b y = 0 . The auxiliary equation is m 2 + a m + b = 0 . Then, then the general solution of the equation is of the form: Case 1: If the roots are real distinct m 1 and m 2 , then the general solution is, y h ( x ) = c 1 e m 1 + c 2 e m 2 . Case 2: If the roots are real double roots m 1 and m 1 , then the general solution is, y h ( x ) = ( c 1 + c 2 x ) e m 1 . Case 2: If the roots are complex a ± i b , then the general solution is, y h ( x ) = e a x ( A cos ( b x ) + B sin ( b x ) ) . Formula used: The particular solution of nonhomogeneous ordinary differential equation of the form y ″ + ω 0 y = F 0 m sin ω 0 t is given by, y p ( t ) = − F 0 2 m ω 0 t cos ω 0 t . Calculation: The given differential equation is y ″ + 4 y = sin t + 1 3 sin 3 t + 1 5 sin 5 t . First obtain the solution of the homogeneous part as follows. The auxiliary equation of y ″ + 4 y = 0 is m 2 + 4 = 0 . Obtain the roots of the equation m 2 + 4 = 0 as shown below. m 2 + 4 = 0 m 2 = − 4 m = ± 2 i Here the roots are complex conjugate. Therefore, the solution y h is y h = A cos 2 t + B sin 2 t . By sum rule, the particular solution y p = y p 1 + y p 2 + y p 3 . Compare the equation y ″ + 4 y = sin t with that of m y ″ + c y ′ + k y = F 0 sin ω t and obtain the values of m , c , k , F 0 and ω as m = 1 , c = 0 , k = 4 , F 0 = 1 and ω = 1 . First compute the value of ω 0 as shown below. ω 0 = k m = 4 1 = 4 = 2 Now obtain the values of a and b as follows. a = − F 0 ω c m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = − ( 1 ) ( 2 ) ( 0 ) ( 1 ) 2 ( ( 2 ) 2 − ( 1 ) 2 ) 2 + ( 1 ) 2 ( 0 ) 2 = 0 Compute the value of b as shown below. b = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = ( 1 ) ( ( 2 ) 2 − ( 1 ) 2 ) ( 1 ) 2 ( ( 2 ) 2 − ( 1 ) 2 ) 2 + ( 1 ) 2 ( 0 ) 2 = ( 1 ) ( 4 − 1 ) ( 1 ) 2 ( 4 − 1 ) 2 + ( 1 ) ( 0 ) = 3 ( 3 ) 2 + 0 = 1 3 Therefore, the particular solution y p 1 is y p 1 ( t ) = a cos ω t + b sin ω t = 1 3 sin t . Now compare y ″ + 4 y = 1 3 sin 3 t with that of m y ″ + c y ′ + k y = F 0 sin ω t and obtain the values of m , c , k , F 0 and ω as m = 1 , c = 0 , k = 4 , F 0 = 1 3 and ω = 3 . Obtain the values of a and b as follows. a = − F 0 ω c m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = − ( 1 3 ) ( 3 ) ( 0 ) ( 1 ) 2 ( ( 2 ) 2 − ( 3 ) 2 ) 2 + ( 3 ) 2 ( 0 ) 2 = 0 Compute the value of b as shown below. b = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = ( 1 3 ) ( ( 2 ) 2 − ( 3 ) 2 ) ( 1 ) 2 ( ( 2 ) 2 − ( 3 ) 2 ) 2 + ( 3 ) 2 ( 0 ) 2 = ( 1 ) ( 4 − 9 ) ( 1 ) 2 ( 4 − 9 ) 2 + ( 9 ) ( 0 ) = ( − 5 3 ) ( − 5 ) 2 + 0 = − 1 15 Therefore, the particular solution y p 2 is y p 2 ( t ) = a cos ω t + b sin ω t = − 1 15 sin 3 t

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Videos

Question

Chapter 2.8, Problem 17P

To determine

The motion of the mass-spring system modeled by the ordinary differential equation (D2+4I)y=sint+13sin3t+15sin5t and the initial conditions y(0)=0, y′(0)=335 and sketch the solution.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 2.8, Problem 16P

Chapter 2.8, Problem 18P

Students have asked these similar questions

For the Big-M tableau (of a maximization LP and row0 at bottom and M=1000), Z Ꮖ 1 x2 x3 81 82 83 e4 a4 RHS 0 7 0 0 1 0 4 3 -3 20 0 -4.5 0 0 0 1 -8 -2.5 2.5 6 0 7 0 1 0 0 8 3 -3 4 0 -1 50 1 0 0 0-2 -1 1 4 0000 0 30 970 200 If the original value of c₁ is increased by 60, what is the updated value of c₁ (meaning keeping the same set for BV. -10? Having made that change, what is the new optimal value for ž?

Here is the optimal tableau for a standard Max problem. zx1 x2 x3 24 81 82 83 rhs 1 0 5 3 0 6 0 1 .3 7.5 0 - .1 .2 0 0 28 360 0 -8 522 0 2700 0 6 12 1 60 0 0 -1/15-3 1 1/15 -1/10 0 2 Using that the dual solution y = CBy B-1 and finding B = (B-¹)-¹ we find the original CBV and rhs b. The allowable increase for b₂ is If b₂ is increased by 3 then, using Dual Theorem, the new value for * is If c₂ is increased by 10, then the new value for optimal > is i.e. if no change to BV, then just a change to profit on selling product 2. The original coefficients c₁ = =☐ a and c4 = 5 If c4 is changed to 512, then (first adjusting other columns of row0 by adding Delta times row belonging to x4 or using B-matrix method to update row0) the new optimal value, after doing more simplex algorithm, for > is

Please show in mathematical form.

Chapter 2 Solutions

Advanced Engineering Mathematics

Knowledge Booster

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

The motion of the mass-spring system modeled by the ordinary differential equation ( D 2 + 4 I ) y = sin t + 1 3 sin 3 t + 1 5 sin 5 t and the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 3 35 and sketch the solution.

Videos

The motion of the mass-spring system modeled by the ordinary differential equation (D2+4I)y=sint+13sin3t+15sin5t and the initial conditions y(0)=0, y′(0)=335 and sketch the solution.

Want to see the full answer?

Chapter 2 Solutions