Analytic Solution for Differential Equation with Initial Condition y(0)=1

[Kawakaze]

Homework Statement

$$\sqrt{t^2 + 9}\frac{dy}{dt}=y^2$$
when $$(0 < y)$$

What method would you use to find an analytic solution and why,
Find the general solution
Solve initial condition problem y(0)=1

The Attempt at a Solution

i don't know where to begin here, :/

 

[tiny-tim]

Hi Kawakaze!

What method would you use to find an analytic solution and why, …

What methods do you know?

 

[Kawakaze]

I don't know any method at the minute other than direct integration, I am aware of them, but they are far from 2nd nature as I know they should be :)

Im guessing i need to separate the variables and get the y on one side? Thats as far as I get.

 

[tiny-tim]

Im guessing i need to separate the variables and get the y on one side? Thats as far as I get.

That's the one!

ok: use separation of variables, to get something dy = something dt, and integrate both sides …

what do you get? ​

 

[Kawakaze]

I really have no idea how to separate these. may as well give it a go :)

$$\frac{dy}{dt} = y^2(\frac{1}{\sqrt{t^2 + 9}})$$

 

[tiny-tim]

Hi Kawakaze!

I really have no idea how to separate these …

Wow! you really do have no idea!

"separate the variables" means you put all the y and dy on one side, and all the t and dt on the other side …

dy/y² = dt/√(t² + 9) …

carry on from there ​

 

[Kawakaze]

I am really coming unstuck with this leibniz(sic!) notation. I thought dy/dx is one symbol and cannot be altered? for example if dy/dx = 0, you can't multiply by dx and get dy = dx?

i have no idea how to handle these. if i were to integrate the formula you just wrote, what would happen to the dy and dx terms?

 

[HallsofIvy]

Stictly speaking "dy/dx" is a single symbol. However, it can be treated like a fraction because it is a limit of the difference quotient. You can go back before the limit, prove whatever fraction property you want for the difference quotient, then take the limit again, showing that the derivative has that fraction property. To make use of that fact, we define "differentials" by dy= f'(x)dx. (I intentionally wrote f'(x) there so that I wouldn't be writing "dy= (dy/dx)dx" where, again, it looks like you are canceling the "dx"s. That isn't what is happening although the effect is the same.)

As for "what happens to the dx and dy", surely you know that
$$\int f(x)dx= F(x)+ C$$
where F is an anti-derivative of f and C is a constant. There is no "dx" on the right. That disappears with the integral.

 

[tiny-tim]

Hi Kawakaze!

I am really coming unstuck with this leibniz(sic!) notation. I thought dy/dx is one symbol and cannot be altered? for example if dy/dx = 0, you can't multiply by dx and get dy = dx?

Take our word for it … you can do this, it's perfectly mathematically rigorous.

It happens in the chain rule for derivatives, and it happens here.

(Here, it basically has to do with the fundamental theorem of calculus, with integration being the opposite of differentiation … the dy and dx become separated, and each then becomes part of a ∫ dy or ∫ dx integral.)

Check with your professor if you don't trust us!

I'm sure that when you get used to using it, you'll wonder what you ever worried about.

i have no idea how to handle these. if i were to integrate the formula you just wrote, what would happen to the dy and dx terms?

We "absorb" them by putting an ∫ in front of both sides of the equation (yes, we are allowed to do that!) …

∫ dy/y² = ∫ dt/√(t² + 9) …

and then solve (remembering of course to put in a constant of integration).

Have a go! ​

 

[Char. Limit]

I am really coming unstuck with this leibniz(sic!) notation. I thought dy/dx is one symbol and cannot be altered? for example if dy/dx = 0, you can't multiply by dx and get dy = dx?

Also, no need for the (sic). You spelled it right.

Well, if dy/dx = 0, then multiplying by dx would give dy=0 dx. I think you mean if dy/dx = 1, you can't multiply by dx and get dy=dx...

but you can. And if you then integrate them, you'll get y=x+C, which function fits the condition dy/dx = 1.

 

[Kawakaze]

Ok its becoming a little clearer now, still a long way off though :)

$$\frac{1}{y^2}dy=\frac{1}{\sqrt{t^2+9}}dt$$

Integrate this getting

$$-\frac{1}{y}+C=?$$

Wolphram alpha gives the RHS as = sinh^-1(t/3) (+D)

Seeing as we haven't covered the hyperbolic functions, my prof might get a little upset :) is this going further than needed or something?

 

[Char. Limit]

I believe you can make a similar trig substitution with s=3tan(x), and that should work.

 

[dextercioby]

The +D is not necessary, because you already have the integration constant in the LHS.

 

[HallsofIvy]

I am really coming unstuck with this leibniz(sic!) notation.

Also, no need for the (sic). You spelled it right.

Well, since it is a person's name it really should be "Leibniz", with a capital L.

 

[Char. Limit]

Well, since it is a person's name it really should be "Leibniz", with a capital L.

But that's not spelling. Properly, that's a capitalization error. Perhaps a (cic) is in order?

 

[Kawakaze]

But that's not spelling. Properly, that's a capitalization error. Perhaps a (cic) is in order?

Surely you mean (cnic) :)

 

[Kawakaze]

I believe you can make a similar trig substitution with s=3tan(x), and that should work.

s? where is the s coming from?

 

[Char. Limit]

s? where is the s coming from?

s is my dummy variable. u would also work, so would theta, or anything really.

 

[Kawakaze]

dummy variable? I've not seen these before. would this be a way to integrate in parts and then use on of them rules to put it all back together again?

 

[Char. Limit]

No, actually this is a trig substitution. However, I did have it backwards. It should be x=3tan(s), not s=3tan(x). From there, you get dx = 3 sec^2(s) ds, and x^2 = 9 tan^2(s). Substitute those two values in for dx and x^2 and you should get a solvable integral in s.

EDIT: Removed wording that I considered condescending.

 

[Kawakaze]

Ive not seen this before, I am going to google it when i have time. Strings to the bow and so on! :) I did find this in my handbook, i think this is sufficient for my level. Any thoughts?

Standard Integral
$$\int{\frac{1}{\sqrt{x^2+a^2}}dx=ln(x+\sqrt{x^2+a^2})$$

 

[Char. Limit]

Ive not seen this before, I am going to google it when i have time. Strings to the bow and so on! :) I did find this in my handbook, i think this is sufficient for my level. Any thoughts?

Standard Integral
$$\int{\frac{1}{\sqrt{x^2+a^2}}dx=ln(x+\sqrt{x^2+a^2})$$

Yep, that'll do it. Don't forget the +C at the end.

 

[Kawakaze]

Yep, that'll do it. Don't forget the +C at the end.

So

$$-\frac{1}{y}=ln(x+\sqrt{x^2+3^2})+C$$
$$y=-\frac{1}{ln(x+\sqrt{x^2+3^2})}+C$$

Does this simplify any more than this?

 

[Char. Limit]

So

$$-\frac{1}{y}=ln(x+\sqrt{x^2+3^2})+C$$
$$y=-\frac{1}{ln(x+\sqrt{x^2+3^2})}+C$$

Does this simplify any more than this?

That C needs to stick with the ln, like this:

$$y=-\frac{1}{ln(x+\sqrt{x^2+9})+C}$$

Other than that I think you're done.

 

[Kawakaze]

Thanks again, as always, you guys are great!

 

[Char. Limit]

Thanks again, as always, you guys are great!

As are you. Have a great day!