Show that it satisfies the differential equation

[crybllrd]

Homework Statement

Show that

$$y=tanh(t)$$

satisfies the differential equation

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

with initial conditions y(0) = 0

Homework Equations

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

 

[tiny-tim]

hi crybllrd!

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

Integrate both sides to get

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

nooo …

d/dy (ln(1 - y²)) = 2y/(1 - y²)

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

 

[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.

 

[HallsofIvy]

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.

 

[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

 

[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.

 

[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 into play? I see where that is true, but is that information actually needed to solve it?

 

[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.

 

[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 canceled out.

 

[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 canceled out all of the constants except for the B within the hyperbolic.

 

[Dick]

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 canceled out all of the constants except for the B within the hyperbolic.

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?

 

[crybllrd]

Alright I just did both. Thanks again.