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Chapter 1: Q28 E (page 14)

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

dydx=3x-y-13,y(2)=1

Short Answer

Expert verified

The hypotheses of Theorem 1 are not satisfied.

The initial value problem does not have a unique solution.

Step by step solution

01

Finding the partial derivative of the given relation concerning y

Here,fx,y=3x-y-13and∂f∂y=-23y-113

02

Determining whether Theorem 1 implies the existence of a unique solution or not

From Step 1, we find that∂f∂yis not continuous or even defined when y=1. Consequently, there is no rectangle containing the point 2,1, in which bothfx,y and∂f∂y are continuous. Because the hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the given initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution.

Hence, Theorem 1 implies that the given initial value problem does not have a unique solution.

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Most popular questions from this chapter

In Problems , identify the equation as separable, linear, exact, or having an integrating factor that is a function of either x alone or y alone.

2x+yx-1dx+xy-1dy=0

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.

x2-sin(x+y)=1,dydx=2xsec(x+y)-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.

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).

Use Euler’s method with step size h = 0.1 to approximate the solution to the initial value problem

y'=x-y2, y (1) = 0 at the points x=1.1,1.2,1.3,1.4and1.5.
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