# Show that it satisfies the differential equation

1. Jan 25, 2012

### crybllrd

1. The problem statement, all variables and given/known data

Show that

$$y=tanh(t)$$

satisfies the differential equation

$\frac{dy}{dt}=1-y^{2}$

with initial conditions y(0) = 0

2. Relevant equations

3. The attempt at a solution

I'm not sure how to start, but we have only dealt with one D.E., and we had moved all the variables to a respective side then integrated.

$dy=(1-y^{2})dt$

$\frac{dy}{1-y^{2}}=dt$

Integrate both sides to get

$ln|1-y^{2}|=t$

Not sure where to go from here

2. Jan 25, 2012

### tiny-tim

hi crybllrd!
nooo

d/dy (ln(1 - y2)) = 2y/(1 - y2)

try again, splitting 1/(1 - y2) into partial fractions first

3. Jan 26, 2012

### crybllrd

Oh ok , I got it. I am in Calc 2 and when I read Diff. Eq. it just scared me off.
That was easy enough. It is a seven part question, and the third part looks very similar:

Show that for the arbitrary constants A and B, the function y=Atanh(Bt) satisfies:

$\frac{dy}{dt}=AB-\frac{B}{A}y^{2}$

I will try to get it to match the form:

$\int\frac{1}{A^{2}+x^{2}}=\frac{1}{A}tan^{-1}(\frac{x}{A})+c$
_______________________________________________________

$dy=B(A-\frac{1}{A}y^{2})(dt)$<--- factored out the B and multiplied by dt

$\frac{dy}{A-\frac{1}{A}y^{2}}=Bdt$

It looks like I need to do a little more manipulation/simplification to get the left side where I want it (namely getting rid of the 1/A in the denominator) but I can't get it to the form I want.

Last edited: Jan 26, 2012
4. Jan 26, 2012

### HallsofIvy

Staff Emeritus
No, no, no! Did you not understand what sheriff89 said? You are not asked to solve the differential equation. In Calculus II you wouldn't be expected to do that.
Differentiate y= Atanh(Bt) and put it and $y^2= A^2tanh^2(Bt)$ into the equation and determine if you get a true equation or not.

5. Jan 26, 2012

### crybllrd

Why would I put the y^2 into the equation and square the other side? And what is a "true equation"?
I have a feeling these will be on our exam next week, so I want to understand it well.
Thanks

Last edited: Jan 26, 2012
6. Jan 26, 2012

### Dick

You still aren't getting the point of just substituting into the differential equation to show you have a solution. Go back to the first one. If y=tanh(x) then dy/dx=sech(x)^2. The other side is 1-y^2=1-tanh(x)^2. Aren't they equal? You don't have to solve any ODE's, you just have to show you have a solution.

7. Jan 28, 2012

### crybllrd

Oh ok I get it.
On part A I algebraically rearranged to get dy/(1-y^2 )=dt, then I integrated both sides to get (after simplification) y= tanh⁡t, done. Just did it the long way, not even thinking because all the other problems were integration.

then, when I started integrating part b, people starting getting upset :P

So this is what I have for part b:

y(t)=Atanh(Bt)
y(t)'=ABsech2(Bt)

y(t)'=AB-(B/A)y2

ABsech2(Bt)=AB-(B/A)Atanh2(Bt)

B's cancel out

Asech2(Bt)=A-tanh2(Bt)

I can see this being true if A=1
Is that all I need?
And, where does the y(0)=0 come in to play? I see where that is true, but is that information actually needed to solve it?

8. Jan 28, 2012

### Dick

y(t)^2=A^2*tanh(Bt)^2. If you work it out correctly the A factor will cancel as well. And you don't need y(0)=0 to show that. But given the solution, you can say y(0)=0.

9. Jan 28, 2012

### crybllrd

Cool, thanks a lot.
I think I'm definitely ready for an exam on Monday.

EDIT: I had a math error on my last post when I substituted for y2. When done properly, all the A's cancelled out.

Last edited: Jan 28, 2012
10. Jan 29, 2012

### crybllrd

Ugh, this problem is killing me. On a later part, it asks:
Let v(t) be the velocity of a falling object of mass m that started from rest. For large velocities, air resistance is proportional to the square of velocity v(t)2.
If we choose coordinates so that v(t)>0 for a falling object, then there is a constant k>0 such that
$\frac{dv}{dt}=g-\frac{k}{m}v^{2}$
Solve for v(t) by applying the rsults of part (b) (what I just worked out) with
$A=\sqrt{gm/k}, B=\sqrt{gk/m}$

I am not sure what "result" he is wanting from part B, I just had to show that a function satisfied a differential, and in doing so I cancelled out all of the constants except for the B within the hyperbolic.

11. Jan 29, 2012

### Dick

You know that v(t)=A*tanh(Bt) satisfies $\frac{dv}{dt}=A B - \frac{B}{A}v^{2}$. So to find a solution to $\frac{dv}{dt}=g-\frac{k}{m}v^{2}$ you just set $A B=g$ and $\frac{B}{A}=\frac{k}{m}$ and solve for A and B. But it looks like they already did that for you as well. So maybe they just want to to substitute those values for A and B into v(t)=A*tanh(Bt)? Or maybe they want you to show explicitly that it works?

12. Jan 29, 2012

### crybllrd

Alright I just did both. Thanks again.