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Chapter 5: Q4E (page 259)

In Problems 1鈥�7, convert the given initial value problem into an initial value problem for a system in normal form.

y6(t)=y'(t)3-sin(y(t))+e2t;y(0)=y'(0)=......=y(5)(0)=0

Short Answer

Expert verified

x'6(t)=x2(t)3-sin(x1(t))+e2t

Step by step solution

01

Express the equation in form of x

Here giveny6(t)=y'(t)3-sin(y(t))+e2t

Denote,

x1(t)=y(t)x2(t)=y'(t)x3(t)=y''(t)x4(t)=y3(t)x5(t)=y4(t)x6(t)=y5(t)

The equation transforms as;

x'1(t)=x2(t)x'2(t)=y''(t)=x3(t)x'3(t)=y'''(t)=x4(t)x'4(t)=x5(t)x'5(t)=x6(t)x'6(t)=x2(t)3-sin(x1(t))+e2t

02

The initial conditions

The given initial conditions arey(0)=y'(0)=......=y(5)(0)=0.

Initial conditions after transformationsx1(0)=x2(0)=x3(0)=x4(0)=x5(0)=x6(0)=0.

This is the required result.

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Most popular questions from this chapter

Use the Runge鈥揔utta algorithm for systems with h= 0.1 to approximate the solution to the initial value problem.

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At t=1.

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