Differential Equation - Initial Value Problem

[cse63146]

Homework Statement

Solve the following IVP:

$$\frac{dy}{dt} = \frac{1}{2y+3}$$ y(0) = 1

Homework Equations

The Attempt at a Solution

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

Kinda stuck now. Any hints?

 

[EstimatedEyes]

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.

 

[cse63146]

dont I need the left side to be y(t)?

 

[de_brook]

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

 

[cse63146]

wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?

 

[Omerta6]

wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?

I do believe you are correct

 

[cse63146]

so its:
$$y^2 + 3y = t + 4$$
$$y^2 + 3y - 4 = t$$
$$(y + 4)(y - 1) = t$$

stuck again

 

[de_brook]

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

 

[de_brook]

so its:
$$y^2 + 3y = t + 4$$
$$y^2 + 3y - 4 = t$$
$$(y + 4)(y - 1) = t$$

stuck again

solve $$y^2 + 3y -(t + 4) = 0$$
using the general formula for quadratic equation. you y as a function of t

 

[cse63146]

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.

 

[Mark44]

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.

 

[de_brook]

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

 

[cse63146]

My stupidity knowns no bounds.

After applying the quadratic formula, I got the following:

$$y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}$$ and $$y(t) = \frac{- 3 - \sqrt{9 + 4(t+4)}}{2}$$

at time t = 0

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

Therefore:

$$y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}$$ is the correct answer.

 

[de_brook]

My stupidity knowns no bounds.

After applying the quadratic formula, I got the following:

$$y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}$$ and $$y(t) = \frac{- 3 - \sqrt{9 + 4(t+4)}}{2}$$

at time t = 0

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

Therefore:

$$y(t) = \frac{- 3 + \sqrt{9 + 4(t+4)}}{2}$$ is the correct answer.

Thus, you now know your general solution and the unique solution which satisfies the I.V.P.

 

[cse63146]

Still can't believe it took me that long to figure it out.

Thanks for putting up with me.