Differential Existence and Uniqueness

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SUMMARY

The discussion centers on verifying the solutions y1(t) = 1 - t and y2(t) = -t²/4 for the initial value problem defined by the differential equation y' = (-t + √(t² + 4y))/2 with the condition y(2) = -1. Both solutions are valid within specific intervals, and the existence of two solutions does not contradict the uniqueness theorem due to the nature of the differential equation. The user is guided to substitute y1 and y2 into the equation to confirm their validity and understand the implications of Theorem 2.4.2 regarding uniqueness.

PREREQUISITES
  • Understanding of first-order differential equations
  • Familiarity with initial value problems
  • Knowledge of the uniqueness theorem for differential equations
  • Basic algebraic manipulation and substitution techniques
NEXT STEPS
  • Study the implications of the existence and uniqueness theorem in differential equations
  • Practice solving initial value problems using various methods
  • Explore the concept of continuous functions and their role in differential equations
  • Investigate the behavior of solutions to nonlinear differential equations
USEFUL FOR

Students studying differential equations, mathematics educators, and anyone seeking to deepen their understanding of initial value problems and solution uniqueness in the context of differential equations.

swooshfactory
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Homework Statement



a) Verify that both y1(t)= 1-t and y2(t)= (-t^2)/4 are solutions of the initial value problem

y-prime = (-t + (t^2 + 4y)^(1/2)) / 2 , for y(2) = -1

Where are these solutions valid?

b) Explain why the existence of two solutions of the given problem does not contradict the uniqueness part of Theorem 2.4.2 (typed in section 2. below) .
C) Show that y=ct+c^2 where c is an arbitrary constant satisfies the differential equation in part (a) for t> or equal to -2c. If c=-1 the initial condition is also satisfied and the solution y=y1(t) is obtained. Show that there is no choice of c that gi es the second solution y=y2(t)

Homework Equations




Let the functions f and df/dy be continiuous in some rectangle alpha < t < beta, gamma < y < theta containing the point (t0, y0). Then in some interval t0 - h < t < t0 + h contained in alpha < t < beta, there is a unique solution y = phi (t) of the initial value problem y-prime = f ( t,y ) , y(t0) = y0

i think this is the relevant equation, at least


The Attempt at a Solution



I am very unsure of my method in attempting to solve this problem. My problem is that I don't understand the concepts I'm employing or why I'm employing them. What I did to solve it was separate the "-t/2" and "[(t^2+4y)^(1/2)]/2]" terms and add "-t+2" to each side. Then, I didn't know how to get the y on its own, so I squared both sides. I don't know if this is right to do. Now I have (d^2)y/(dt^2) + (t/2)^2 = [(t^2+4y)^(1/2)]/2]^2.

I got to (d^2)y/(dt^2) = y(1-t) . I don't know if this is right, what this means, or what to do here. I am very confused. My homework is due Monday, so I would really appreciate it if someone who understands this problem could explain any part of it to me, but especially a) since that is the part I am stuck on.
 
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swooshfactory said:

Homework Statement



a) Verify that both y1(t)= 1-t and y2(t)= (-t^2)/4 are solutions of the initial value problem

y' = (-t + √(t² + 4y))/ 2 , for y(2) = -1

I am very unsure of my method in attempting to solve this problem. My problem is that I don't understand the concepts I'm employing or why I'm employing them. What I did to solve it was separate the "-t/2" and "[(t^2+4y)^(1/2)]/2]" terms and add "-t+2" to each side. Then, I didn't know how to get the y on its own, so I squared both sides. I don't know if this is right to do. Now I have (d^2)y/(dt^2) + (t/2)^2 = [(t^2+4y)^(1/2)]/2]^2.

I got to (d^2)y/(dt^2) = y(1-t) . I don't know if this is right, what this means, or what to do here. I am very confused. My homework is due Monday, so I would really appreciate it if someone who understands this problem could explain any part of it to me, but especially a) since that is the part I am stuck on.

Hi swooshfactory! :smile:

(have a square-root: √ and a squared: ² :smile:)

Nooo … (dy/dt) is not d²y/dt². :cry:

Just substitute y1 or y2 inside the square-root, and prove that the whole RHS is -1 or t2/2, respectively. :smile:
 

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