Chapter 5: Q4E (page 259)

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.
$y^{6} (t) = {[y' (t)]}^{3} - \sin (y (t)) + e^{2 t}; y (0) = y' (0) = . . . . . . = y^{(5)} (0) = 0$

Short Answer

Expert verified

$x'_{6} (t) = {[x_{2} (t)]}^{3} - \sin (x_{1} (t)) + e^{2 t}$

Step by step solution

Express the equation in form of x

Here given $y^{6} (t) = {[y' (t)]}^{3} - \sin (y (t)) + e^{2 t}$

Denote,

$x_{1} (t) = y (t) x_{2} (t) = y' (t) x_{3} (t) = y'' (t) x_{4} (t) = y^{3} (t) x_{5} (t) = y^{4} (t) x_{6} (t) = y^{5} (t)$

The equation transforms as;

$x'_{1} (t) = x_{2} (t) x'_{2} (t) = y'' (t) = x_{3} (t) x'_{3} (t) = y''' (t) = x_{4} (t) x'_{4} (t) = x_{5} (t) x'_{5} (t) = x_{6} (t) x'_{6} (t) = {[x_{2} (t)]}^{3} - \sin (x_{1} (t)) + e^{2 t}$

The initial conditions

The given initial conditions are $y (0) = y' (0) = . . . . . . = y^{(5)} (0) = 0 .$

Initial conditions after transformations $x_{1} (0) = x_{2} (0) = x_{3} (0) = x_{4} (0) = x_{5} (0) = x_{6} (0) = 0 .$

This is the required result.

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In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.
$y^{6} (t) = {[y' (t)]}^{3} - \sin (y (t)) + e^{2 t}; y (0) = y' (0) = . . . . . . = y^{(5)} (0) = 0$

Short Answer

Step by step solution

Express the equation in form of x

The initial conditions

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In Problems 1–7, 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

Step by step solution

Express the equation in form of x

The initial conditions

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Decision Maths

Logic and Functions

Geometry

Discrete Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.
$y^{6} (t) = {[y' (t)]}^{3} - \sin (y (t)) + e^{2 t}; y (0) = y' (0) = . . . . . . = y^{(5)} (0) = 0$