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Chapter 1: Q5.3-15E (page 1)

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm, approximate the solution to the initial value problem y''=t2+y2;y(0)=1,y'(0)=0at t=1. Starting with h=1, continue halving the step size until two successive approximations of both y1andy'1differ by at most 0.1.

Short Answer

Expert verified

The solution is y1=1.69 and y'1=1.82.

Step by step solution

01

Transform the equation

Write the equation asy''=t2+y2

The equations can be written as:

x1t=ytx2t=y't=x'1

The transformation of the equation is:

localid="1664091743800" x'1(t)=x2(t)x'2(t)=t2+x21

The initial conditions are:

localid="1664091773344" x11=y11=1x21=y'1=0

02

Apply Runge –Kutta method

For the solution, apply the Runge-Kutta method in MATLAB, and the solution isy1=1.69 and y'1=1.82.

03

Find that y1 and y'(1) differ by at most 0.01.

Subtracting the values ofu1 andv1 then

y'1-y1=1.82-1.69=0.13

Therefore,y1 andy'1 differ by at most 0.1.

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Most popular questions from this chapter

In Problems 13-16, write a differential equation that fits the physical description. The rate of change in the temperature T of coffee at time t is proportional to the difference between the temperature M of the air at time t and the temperature of the coffee at time t.

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

ym+3yn+28y'+26y=0

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

dydx=y4-x4,y(0)=7

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 Problem 19, solve the given initial value problem

y'''−y''−4y'+4y=0y(0)=−4y'(0)=−1y''(0)=−19
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