# Math Help - help w/ difficult IVP

1. ## help w/ difficult IVP

Use separation of variables to obtain the solution to the following IVP:

I can't seem to finish this one.

$\int dy/(y^2-4) = t + C$

$ln(y-2) - ln(y+2) = 4(t+C)$

$(y-2)/(y+2)=e^(4t+4C))$

solving for y gave me:

$y= (2e^(4t+4C)))/(1-e^(4t+4C)))$

so yeah its not turning out well for me

2. Originally Posted by cdlegendary
Use separation of variables to obtain the solution to the following IVP:

I can't seem to finish this one.

$\int dy/(y^2-4) = t + C$

$ln(y-2) - ln(y+2) = 4(t+C)$

$(y-2)/(y+2)=e^(4t+4C))$

solving for y gave me:

$y= (2e^(4t+4C)))/(1-e^(4t+4C)))$

so yeah its not turning out well for me
Unless someone else can see an easy method then this is just the way I would have done it.

Except you're missing a bit in you y = ... part.

I think it should be...

$y = \frac{2(e^{4t+4C} + 1)}{1-e^{4t+4C}}$

Then for the initial condition you get...

$\frac{2(e^{4C} + 1)}{1-e^{4C}} = 0$

Solving this gives... $C = \frac{1}{4}i \pi$.

Hence we get $y = \frac{2(1 - e^{4t})}{1+e^{4t}}$

3. ah yeah I mistyped part of my equation, but this method sems to still give the wrong answer

4. Originally Posted by cdlegendary
Use separation of variables to obtain the solution to the following IVP:

I can't seem to finish this one.

$\int dy/(y^2-4) = t + C$

$ln(y-2) - ln(y+2) = 4(t+C)$

$(y-2)/(y+2)=e^(4t+4C))$

solving for y gave me:

$y= (2e^(4t+4C)))/(1-e^(4t+4C)))$

so yeah its not turning out well for me
$\frac{dy}{dx} = y^2 - 4$

$\frac{dx}{dy} = \frac{1}{y^2 - 4}$

$\frac{dx}{dy} = \frac{1}{(y - 2)(y + 2)}$.

Now using partial fractions:

$\frac{A}{y - 2} + \frac{B}{y + 2} = \frac{1}{(y - 2)(y + 2)}$

$\frac{A(y + 2) + B(y - 2)}{(y - 2)(y + 2)} = \frac{1}{(y - 2)(y + 2)}$.

So $A(y + 2) + B(y - 2) = 1$

$Ay + 2A + By - 2B = 1$

$(A + B)y + 2A - 2B = 0y + 1$.

So $A + B = 0$ and $2A - 2B = 1$.

From the first equation, we can see that $B = -A$.

Therefore $2A - 2(-A) = 1$

$4A = 1$

$A = \frac{1}{4}$.

So $B = -\frac{1}{4}$.

Therefore:

$\frac{1}{(y - 2)(y + 2)} = \frac{1}{4(y - 2)} - \frac{1}{4(y + 2)}$.

So $\frac{dx}{dy} = \frac{1}{4(y - 2)} - \frac{1}{4(y + 2)}$

$x = \frac{1}{4}\int{\frac{1}{y - 2}\,dy} - \frac{1}{4}\int{\frac{1}{y + 2}\,dy}$

$x = \frac{1}{4}\ln{|y - 2|} - \frac{1}{4}\ln{|y + 2|} + C$

$x = \frac{1}{4}\ln{\left|\frac{y - 2}{y + 2}\right|} + C$

$x - C = \frac{1}{4}\ln{\left|1 - \frac{4}{y + 2}\right|}$

$4x - 4C = \ln{\left|1 - \frac{4}{y + 2}\right|}$

$e^{4x - 4C} = \left|1 - \frac{4}{y + 2}\right|$

$e^{-4C}e^{4x} = \left|1 - \frac{4}{y + 2}\right|$

$Ae^{4x} = 1 - \frac{4}{y + 2}$ where $A = \pm e^{-4C}$

$\frac{4}{y + 2} = 1 - Ae^{4x}$

$\frac{y + 2}{4} = \frac{1}{1 - Ae^{4x}}$

$y + 2 = \frac{4}{1 - Ae^{4x}}$

$y = \frac{4}{1 - Ae^{4x}} - 2$.

Since $y(0) = 0$

$0 = \frac{4}{1 - Ae^{4(0)}} - 2$

$2 = \frac{4}{1 - A}$

$\frac{1}{2} = \frac{1 - A}{4}$

$2 = 1 - A$

$A = -1$.

Therefore $y = \frac{4}{1 + e^{4x}} - 2$

$y = \frac{4}{1 + e^{4x}} - \frac{2(1 + e^{4x})}{1 + e^{4x}}$

$y = \frac{4 - 2 - 2e^{4x}}{1 + e^{4x}}$

$y = \frac{2 - 2e^{4x}}{1 + e^{4x}}$.