Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
Question
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Chapter 12, Problem 12.1P

(a)

To determine

The solution for x1(t) by separating variables.

(a)

Expert Solution
Check Mark

Answer to Problem 12.1P

The solution for x1(t) by separating variables is x1(t)=(t+k)2+1 for any constant k.

Explanation of Solution

Given nonlinear first-order equation,

x˙=2x1dxdt=2x1

Separating same variables in one side of the equation

dxx1=2dt

Integrate the above equation and solving

dxx1=2dt(x1)12+112+1=2t+C

Here, C is the integration constant.

(x1)1212=2t+C2x1=2t+Cx1=t+C2

Squaring on both sides

x1=(t+C2)2

Let, k=C2 another constant.

x1=(t+k)2x=(t+k)2+1

Conclusion:

Therefore, the solution for x1(t) is (t+k)2+1 for any constant k.

(b)

To determine

Show that there is another solution for x˙=2x1 is x2(t)=1.

(b)

Expert Solution
Check Mark

Answer to Problem 12.1P

The another solution for x˙=2x1 is x2(t)=1.

Explanation of Solution

Assume x2(t)=1 is the another solution of the nonlinear first order differential equation, then it will satisfy the nonlinear first order differential equation.

Substitute 1 for x in x˙=2x1 and solve

x˙=2x1=211=0

Since, x2(t)=1 satisfies the nonlinear first order differential equation x˙=2x1. It is true that the another solution for x˙=2x1 is x2(t)=1.

Conclusion:

The another solution for x˙=2x1 is x2(t)=1.

(c)

To determine

Show that although x1(t) and x2(t) are solutions for x˙=2x1, neither Ax1(t), nor Bx2(t), nor x1(t)+x2(t) are solutions.

(c)

Expert Solution
Check Mark

Answer to Problem 12.1P

It is shown that Ax1(t), Bx2(t), and x1(t)+x2(t) are not the solutions for x˙=2x1.

Explanation of Solution

From part (a), the solution for x1(t) is (t+k)2+1.

Now, consider x=Ax1(t).

Substitute (t+k)2+1 for x1(t) in x=Ax1(t)

x=Ax1(t)x=A[(t+k)2+1]

Differentiate the above equation with respect to t

x˙=ddtA[(t+k)2+1]=A[2(t+k)]=2A[(t+k)]

Now, substitute A[(t+k)2+1] for x in x˙=2x1

x˙=2A[(t+k)2+1]1=2A(t+k)2+A1

Substitute the above equation in x˙=2A[(t+k)]

2A(t+k)2+A1=2A(t+k)A(t+k)2+A1=A(t+k)

Thus, Ax1(t) does not satisfy the nonlinear first order equation x˙=2x1. Ax1(t) is not the solution of the equation x˙=2x1.

From part (b), the solution for x2(t) is 1.

Now, consider x=Bx2(t).

Substitute 1 for x2(t) in x=Bx2(t)

x=B(1)x=B

Differentiate the above equation with respect to t

x˙=dBdt=0

Now, substitute B for x in x˙=2x1

x˙=2B1

Substitute the above equation in x˙=0

2B1=0B=1

Thus, Bx2(t) does not satisfy the nonlinear first order equation x˙=2x1. Bx2(t) is not the solution of the equation x˙=2x1.

Now, consider x=x1(t)+x2(t).

Substitute (t+k)2+1 for x1(t) and 1 for x2(t) in x=x1(t)+x2(t)

x=(t+k)2+1+1=(t+k)2+2

Differentiate the above equation with respect to t

dxdt=ddt((t+k)2+2)x˙=2(t+k)

Now, substitute (t+k)2+2 for x in x˙=2x1

x˙=2(t+k)2+21=2(t+k)2+1

Substitute the above equation in x˙=2(t+k)

2(t+k)2+1=2(t+k)(t+k)2+1=(t+k)

Thus, x1(t)+x2(t) does not satisfy the nonlinear first order equation x˙=2x1. x1(t)+x2(t) is not the solution of the equation x˙=2x1.

Conclusion:

It is shown that Ax1(t), Bx2(t), and x1(t)+x2(t) are not the solutions for x˙=2x1.

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