Solution of diff. equa. passing thru...

[kiddopop]

Find a solution of the given differential equation dy/dt=y^2-16 whose path passes through (1/4,1).

 

[galactus]

First, separate variables:

\(\displaystyle \frac{dy}{y^{2}-16}=dt\)

Integrate:

\(\displaystyle \int\frac{1}{y^{2}-16}dy=\int dt\)

 

[kiddopop]

I integrated and got (-ln((y+4)(y-4))/8=t. Then I tried to solve for y and got y=(-4)/(tan(h)*(4t)). This doesn't seem right to me. But I don't know what I did wrong.

 

[mmm4444bot]

I get a different antiderivative for 1/(y^2 - 16):

1/8 * ln[(y - 4)/(y + 4)]

 

[galactus]

That's correct. You are both correct. Note that \(\displaystyle -ln(1/a)=ln(a)\)

 

[kiddopop]

Are you saying that both of our antiderivatives were right or that y=(-4)/(tan(h)*(4t)) is right?

 

[mmm4444bot]

I integrated and got (-ln((y+4)(y-4))/8=t.

You are both correct. Note that \(\displaystyle -ln(1/a)=ln(a)\)

kiddopop typed (y + 4)(y - 4) as the input to the natural-log function (i.e., a product, not a ratio).

Kiddopop is also missing a close parenthesis.

 

[galactus]

Yes, mmm, I reckon I did not pay close enpough attention.

 

[Deleted member 4993]

Are you saying that both of our antiderivatives were right or that y=(-4)/(tan(h)*(4t)) is right?

Actually what you have is y = (-4)/[tanh(4t)] - not what you typed.

\(\displaystyle tanh(x) \ = \ \frac{e^x \ - \ e^{-x}}{e^x \ + \ e^{-x}}\)

And yes after those corrections those answers are equivalent.