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Chapter 1: Q11RP (page 30)

The initial value problem \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}{\bf{,}}\;{\bf{x(0) = 1}}\] has a unique solution in some open interval around t = 0.

Short Answer

Expert verified

The given statement is true.

Step by step solution

01

Finding partial derivatives

Since \[\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = 3}}{{\bf{x}}^{\frac{{\bf{2}}}{{\bf{3}}}}}\]

Then\[\frac{{\partial {\bf{f}}}}{{\partial {\bf{x}}}}{\bf{ = }}\frac{{\bf{2}}}{{\sqrt[{\bf{3}}]{{\bf{x}}}}}\]

02

Checking the final result

Apply the initial conditions\[x\left( 0 \right) = 1\]

\[\frac{{\partial f}}{{\partial x}}{\bf{ = }}\frac{{\bf{2}}}{{\bf{0}}}\]

The result is infinite.

The given function is discontinuous a x = 0. So the function is not continuous in a rectangle containing the point (0,1).

Therefore, the statement is True.

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Most popular questions from this chapter

Find a general solution for the given differential equation.

(a)y(4)+2y'''-4y''-2y'+3y=0

(b)y'''+3y''-5y'+y=0

(c)y(5)-y(4)+2y'''-2y''+y'-y=0

(d)y'''-2y''-y'+2y=ex+x

Implicit Function Theorem. Let G(x,y)have continuous first partial derivatives in the rectangleR={x,y:a<x<b,c<y<d}containing the pointlocalid="1664009358887" (x0,y0). IfG(x0,y0)=0 and the partial derivativeGy(x0,y0)≠0, then there exists a differentiable function y=ϕ(x), defined in some intervalI=(x0-δ,y0+δ),that satisfies G for allforG(x,ϕx)all x∈I.

The implicit function theorem gives conditions under which the relationshipG(x,y)=0 implicitly defines yas a function of x. Use the implicit function theorem to show that the relationshipx+y+exy=0 given in Example 4, defines y implicitly as a function of x near the point(0,-1).

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

Nonlinear Spring.The Duffing equationy''+y+ry3=0 where ris a constant is a model for the vibrations of amass attached to a nonlinearspring. For this model, does the period of vibration vary as the parameter ris varied?

Does the period vary as the initial conditions are varied? [Hint:Use the vectorized Runge–Kutta algorithm with h= 0.1 to approximate the solutions for r= 1 and 2,

with initial conditionsy(0)=a,y'(0)=0 for a = 1, 2, and 3.]

The direction field for dydx=2x+yas shown in figure 1.13.

  1. Sketch the solution curve that passes through (0, -2). From this sketch, write the equation for the solution.

b. Sketch the solution curve that passes through (-1, 3).

c. What can you say about the solution in part (b) as x→+∞? How about x→-∞?
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