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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 186
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

A mass–spring system is driven by a sinusoidal external force g(t)=5sint. The mass equals 1, the spring constant equals 3, and the damping coefficient equals 4. If the mass is initially located at y(0)=12and at rest, i.e., y'(0)=0, find its equation of motion.

The equation of motion is y=2e-t-12e-3t-cost+12sint.

See the step by step solution

Step by Step Solution

Step 1: Use the given information for writing the differential equation.

Given that,

The mass equals m=1,

The spring constant equals c=3,

And the damping coefficient equals b=4.

The differential equation is,

my''+by'+cy=g(t)y''+4y'+3y=5sint                      ......(1)

Write the homogeneous differential equation of the equation (1),


Step 2: Now find the complementary solution of the given equation.

The auxiliary equation for the above equation,


Solve the auxiliary equation,


The roots of auxiliary equation are,

m1=-1,      m2=-3

The complimentary solution of the given equation is,


Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

yp(t)=Acost+Bsint                   ......(2)

Now find the first and second derivative of above equation,


Substitute the value of yp(t),  yp'(t) and yp''(t) in the equation (1),


Comparing the all coefficients of the above equation,

2B-4A=5                     .......(3)4B+2A=0                     ......(4)

Solve the above equations,

2(2B-4A)=5×24B-8A=104B+2A=0              A=-1

Substitute the value of A in the equation (4),

role="math" localid="1655115721256" 4B+2(-1)=0B=12

Therefore, the particular solution of equation (1),


Step 4: Find the general solution and use the given initial condition, 

Therefore, the general solution is,

y=yc(t)+yp(t)y=c1e-t+c2e-3t-cost+12sint                      ......(5)

Given initial condition,

y(0)=12,  y'(0)=0

Substitute the value of y=12 and t = 0 in the equation (3),

12=c1e-(0)+c2e-3(0)-cos(0)+12sin(0)12=c1+c2-1c1+c2=32                                .....(6)

Now find the derivative of above equation,


Substitute the value of y’ = 0 and t = 0 in the above equation,

0=-c1e-(0)-3c2e-3(0)+sin(0)+12cos(0)0=-c1-3c2+12c1+3c2=12                                        ......(7)

Solve the (6) and (7) equations,

c1+c2=32c1+3c2=12             c2=-12

Substitute the value of c2 in the equation (6),

role="math" localid="1655116289545" c1+c2=32c1+-12=32c1=2

Substitute the value of c1 and c2 in the equation (5),

role="math" localid="1655116374807" y=c1e-t+c2e-3t-cost+12sinty=2e-t-12e-3t-cost+12sint

Thus, the equation of motion is:


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