Differential Equation - Initial Value Problem

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  • #1
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Homework Statement



Solve the following IVP:

[tex]\frac{dy}{dt} = \frac{1}{2y+3}[/tex] y(0) = 1

Homework Equations





The Attempt at a Solution



[tex]\frac{dy}{dt} = \frac{1}{2y+3}[/tex]
[tex]\Rightarrow \int 2y + 3 dy= \int dt[/tex]
[tex]\Rightarrow y^2 + 3y= t +c[/tex] where C is a constant

Kinda stuck now. Any hints?
 

Answers and Replies

  • #2
If I am understanding the problem correctly, all you have to do is plug in the values for y and t given by the problem text and solve for C.
 
  • #3
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dont I need the left side to be y(t)?
 
  • #4
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to get your constant c, put in the value of y at t = 0 i.e y(0) = 1,
we have 1 + 3(0) = 0 + c implies c = 1. Thus you get a quadratic equation, then you can solve for y from there
 
  • #5
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wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?
 
  • #6
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wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?
I do believe you are correct
 
  • #7
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so its:
[tex]y^2 + 3y = t + 4[/tex]
[tex]y^2 + 3y - 4 = t[/tex]
[tex](y + 4)(y - 1) = t[/tex]

stuck again
 
  • #8
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wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?
Oh, sorry for the mistake. But i know you got what i mean't
 
  • #9
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so its:
[tex]y^2 + 3y = t + 4[/tex]
[tex]y^2 + 3y - 4 = t[/tex]
[tex](y + 4)(y - 1) = t[/tex]

stuck again
solve [tex]y^2 + 3y -(t + 4) = 0[/tex]
using the general formula for quadratic equation. you y as a function of t
 
  • #10
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So would the general solution be y^2 + 3y = t + c and the unique solution be y^2 + 3y = t + 4?

I don't think that's it.
 
  • #11
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Does (0, 1) satisfy your unique solution?
Does your relation satisfy the differential equation? (You'll need to differentiate implicitly if you use y^2 + 3y = t + 4.)

If the answers to both questions are "yes" you have the unique solution.
 
  • #12
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So would the general solution be y^2 + 3y = t + c and the unique solution be y^2 + 3y = t + 4?

I don't think that's it.
the general solution is y^2 + 3y = t + c because simply it satisfies the differential equation.(check)
It will be unique if it satisfies the following
1. dy/dx is continuous in the given interval which contains the initial point

2. if Lipschitz condition is satisfied.

y has two roots when you solve the quadratic equation provided the discriminant is not zero. thus the solution of the differential equation is not unique
 
Last edited:
  • #13
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My stupidity knowns no bounds.

After applying the quadratic formula, I got the following:

[tex]y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}[/tex] and [tex]y(t) = \frac{- 3 - \sqrt{9 + 4(t+4)}}{2}[/tex]

at time t = 0

[tex]y(t) = \frac{- 3 + \sqrt{9 + 4(0+4)}}{2} = 1[/tex] which satisfies the IVP, while [tex]y(t) = \frac{- 3 - \sqrt{9 + 4(0+4)}}{2} = -4[/tex] which doesn't satisfy the IVP.

Therefore:

[tex]y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}[/tex] is the correct answer.
 
  • #14
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My stupidity knowns no bounds.

After applying the quadratic formula, I got the following:

[tex]y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}[/tex] and [tex]y(t) = \frac{- 3 - \sqrt{9 + 4(t+4)}}{2}[/tex]

at time t = 0

[tex]y(t) = \frac{- 3 + \sqrt{9 + 4(0+4)}}{2} = 1[/tex] which satisfies the IVP, while [tex]y(t) = \frac{- 3 - \sqrt{9 + 4(0+4)}}{2} = -4[/tex] which doesn't satisfy the IVP.

Therefore:

[tex]y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}[/tex] is the correct answer.
Thus, you now know your general solution and the unique solution which satisfies the I.V.P.
 
  • #15
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Still can't believe it took me that long to figure it out.

Thanks for putting up with me.
 

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