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Chapter 5: Q7E (page 259)

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

x'''-y=t;x(0)=x'(0)=x''(0)=42x''+5y''-2y=1;y(0)=y'(0)=1

Short Answer

Expert verified

x'1(t)=x2(t)x'2(t)=x3(t)x'3(t)=x4(t)+tx'4(t)=x5(t)x'5(t)=2x4(t)-2x3(t)+15x1(0)=x2(0)=x3(0)=4,x4(0)=x5(0)=1

Step by step solution

01

Express equation in form of x

Here givenx'''-y=t and2x''+5y''-2y=1.

Rewrite the equationsx'''=y+t andx''=-5y''+2y+12

Denote,

x1(t)=x(t)x2(t)=x'(t)x3(t)=x''(t)x4(t)=y(t)x5(t)=y'(t)x'1(t)=x2(t)

The equation transforms as:

x'1(t)=x2(t)x'2(t)=x3(t)x'3(t)=x4(t)+tx'4(t)=x5(t)x'5(t)=2x4(t)-2x3(t)+15

02

The initial conditions

The given initial conditions are x(0)=x'(0)=x''(0)=4andy(0)=y'(0)=1.

Initial conditions after transformationsx1(0)=x2(0)=x3(0)=4,x4(0)=x5(0)=1.

This is the required result.

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Most popular questions from this chapter

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