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Chapter 1: Q25 E (page 14)

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

3xdxdt+4t=0,x(2)=-蟺

Short Answer

Expert verified

The hypotheses of Theorem 1 are satisfied.

The theorem shows that the given initial value problem has a unique solution.

Step by step solution

01

Finding the partial derivative of the given relation concerning y

Here, ft,x=-4t3xand

鈭�f鈭�x=-12t脳-1脳x-2鈭�f鈭�x=12tx2
02

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

Now from Step 1, we find that both of the functions ft,xand 鈭�f鈭�xare continuous in any rectangle containing the point2,-蟺, so the hypotheses of Theorem 1 are
satisfied. It then follows from the theorem that the given initial value problem has a unique solution in an interval about t=2of the form 2-未,2+未, where 未is some positive number.

Hence, Theorem 1 implies that the given initial value problem has a unique solution.

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Most popular questions from this chapter

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

dydt-ty=sin2t,y(蟺)=5

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 ( 鈮�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=渭-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).

(a) For the initial value problem (12) of Example 9. Show that 蠒1(x)=0 and蠒2(x)=(x-2)3are solutions. Hence, this initial value problem has multiple solutions. (See also Project G in Chapter 2.)

(b) Does the initial value problemy'=3y23,y(0)=10-7have a unique solution in a neighbourhood ofx=0?

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

Question: Show that,28.鈭�n=0鈭�anxn+1+鈭�n=1鈭�nbnxn-1=b1+鈭�n-1鈭�[2an-1+(n+1)bn+1]xn
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