Finding Implicit and Explicit Solutions for Initial-Value Problems

[Saladsamurai]

Homework Statement

Find an implicit and explicit solution for the given initial-value problem

$$\frac{dx}{dt}=4(x^2+1)$$ for $$x(\frac{\pi}{4})=1$$

$$\frac{dx}{dt}=4(x^2+1)$$

$$\Rightarrow \frac{dx}{x^2+1}=4dt$$

$$\Rightarrow \tan^{-1}x+C=4t$$

Now I am a little stuck. Usually I just plug in my values. I am thinking of taking the tan of both sides, eh?

 

[Feldoh]

x = tan(4t-C) looks like a good idea to me :)

 

[HallsofIvy]

Yes, x= tan(4t- C) is sensible. But you don't HAVE to do that to "plug in the values": your equation, as it is, says $tan^{-1} + C= \pi$. What is arctan 1?

 

[Saladsamurai]

Okay, I see now. But I have a new question, or rather just need some clarification.

In Calculus, we always integrated one dependent variable wrt one independent, which is what I am doing above here, except that something new arises in these separation of variables problems.

Since I am integrating TWO functions that are said to be equivalent, I end up with TWO arbitrary constants. Now I have been shown that they could just be moved to one side of the equation, and since one is just as arbitrary as the other, I can just write them as C (just one). I am fine with this.

But in the above example, suppose I had chose to put +C on the RHS instead of the left.
Now solving for this arbitrary constant I would have gotten that $C=-\frac{3}{4}\pi$ instead of $+\frac{3}{4}\pi$

What does this all mean?
Sorry if it is obvious!

Casey

 

[neutrino]

It just means that

$$\tan^{-1}x + \frac{3\pi}{4} = 4t$$

is equivalent to

$$\tan^{-1}x = 4t - \frac{3\pi}{4}$$