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Chapter 1: Q5.3-22E (page 1)

Oscillations and Nonlinear Equations. For the initial value problem x''+(0.1)(1-x2)x'+x=0;x(0)=xo,x'(0)=0using the vectorized Runge–Kutta algorithm with h = 0.02 to illustrate that as t increases from 0 to 20, the solution x exhibits damped oscillations when xo=1, whereas exhibits expanding oscillations when xo=2.1,.

Short Answer

Expert verified

The result can get by the Runge-Kutta method.

Step by step solution

01

Transform the equation

Here, the equationx''+(0.1)(1-x2)x'+x=0.

The system can be written as:

x1=x(t)x2=x'=x'1

The transform equation is:

role="math" localid="1664101777238" x'1=x2x'2=-x1-0.1(1-x21)x2

The initial conditions are,

x1(0)=x(0)=xo=1,2,1x2(0)=x'(0)=0

02

Apply the Runge-Kutta method

Apply Matlab to find the results. And some results are;

T

For x0=1

For xo=2.1

0

1

2.1

0.02

0.9998

2.09957

0.04

0.9992

2.0983

0.06

0.9982

2.096

0.1

0.9950

2.0893

1

0.5441

1.0600

2

-0.36227

-0.9737

Applying the same procedure gets the result.

This is the required result.

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Most popular questions from this chapter

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

(2xy+3)dx+(x2-1)dy=0

Find a general solution for the differential equation with x as the independent variable:

y'''+3y''−4y'−6y=0

Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).

(a) Show that the general solution to (20)dydx+P(x)y=Q(x) has the formy(x)=Cyh(x)+yp(x),whereyh ( ≠0is a solution to equation (20) when Q(x)=0,

C is a constant, andyp(x)=v(x)yh(x) for a suitable function v(x). [Hint: Show that we can takeyh=μ-1(x) and then use equation (8).] We can in fact determine the unknown function yhby solving a separable equation. Then direct substitution of vyh in the original equation will give a simple equation that can be solved for v.

Use this procedure to find the general solution to (21) localid="1663920708127" dydx+3xy=x2, x > 0 by completing the following steps:

(b) Find a nontrivial solutionyh to the separable equation (22) localid="1663920724944" dydx+3xy=0, localid="1663920736626" x>0.

(c) Assuming (21) has a solution of the formlocalid="1663920777078" yp(x)=v(x)yh(x) , substitute this into equation (21), and simplify to obtain localid="1663920789271" v'(x)=x2yh(x).

d) Now integrate to getlocalid="1663920800433" vx

(e) Verify thatlocalid="1663920811828" y(x)=Cyh(x)+v(x)yh(x) is a general solution to (21).

In Problems 13-16, write a differential equation that fits the physical description. The rate of change of the mass A of salt at time t is proportional to the square of the mass of salt present at time t.

Let ϕ(x)denote the solution to the initial value problem

dydx=x-y,y(0)=1

⦁ Show that ϕ(x)=1-ϕ'(x)=1-x+ϕ(x)

⦁ Argue that the graph of ϕ is decreasing for x near zero and that as x increases from zero, ϕ(x)decreases until it crosses the line y = x, where its derivative is zero.

⦁ Let x* be the abscissa of the point where the solution curve y=ϕ(x) crosses the line y=x.Consider the sign of ϕ(x*) and argue that ϕ has a relative minimum at x*.

⦁ What can you say about the graph of y=ϕ(x) for x > x*?

⦁ Verify that y = x – 1 is a solution to dydx=x-y and explain why the graph of y=ϕ(x) always stays above the line y=x-1.

⦁ Sketch the direction field for dydx=x-y by using the method of isoclines or a computer software package.

⦁ Sketch the solution y=ϕ(x) using the direction field in part (f).
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