# Differential Equation - Initial Value Problem

## Homework Statement

Solve the following IVP:

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

## 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?

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.

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

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

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

wouldnt it be 1^2 + 3(1) = 0 + c => c = 4?
I do believe you are correct

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

stuck again

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

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

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
Mentor
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.

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:
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.

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.

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

Thanks for putting up with me.