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I am trying to solve the following: $y' = (y-5)(y+5)$ if $y(4) = 0$.

So far, I have tried separating the variables and then use partial fractions and have followed these steps:

(1) $A(y+5) + B (y-5) =1$.

(2) For $y = -5$ we have $B = \dfrac{-1}{10}$.

(3) For $ y= 5$, we have $A = \dfrac{1}{10}$.

(4) so we have $\dfrac{1}{10}[ \ln|y^2 - 25|] = x + C$

(5) After a chain of algebraic operations, we have $y = \pm \sqrt{e^{10x + 10C} + 25}$.

However, if I plug in my initial condition, then I get: $0 = e^{40 + 10C} + 25$ which implies $\ln(-25) = 40 + C$ which gets into complex numbers. Any help anyone can provide on where my reasoning went wrong is appreciated.

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  • $\begingroup$ is there anything you don't understand/like about my answer? $\endgroup$ – etothepitimesi May 27 '15 at 12:47
  • $\begingroup$ I think I mostly understand your answer however I do not understand how $\dfrac{1}{(y-5)(y+5)} dy = dx \ \implies \ (\dfrac{1}{10(y-5)} - \dfrac{1}{10(y+5)})dy = dx$. Can you please explain how you derived this? $\endgroup$ – letsmakemuffinstogether May 28 '15 at 20:48
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    $\begingroup$ done! I hope it's clear now. $\endgroup$ – etothepitimesi May 28 '15 at 23:09
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Let's start by decomposing $\frac{1}{(y-5)(y+5)}$ into partial fractions. We want to to find $A$ and $B$ such that $$\frac{1}{(y-5)(y+5)} = \frac{A}{(y-5)} + \frac{B}{(y+5)} = \frac{A(y+5)}{(y-5)(y+5)} + \frac{B(y-5)}{(y-5)(y+5)}.$$ Since we have the same denominator for all terms, the numerators of the fractions must then satisfy $$\tag{0} 1 = A(y+5) + B(y-5)$$ Now, if we take $y=5$ in $(0)$, we get $$1 = A (5+5) + B(5-5) \Rightarrow A = 1/10.$$ On the other hand, by taking $y=-5$ in $(0)$, we get $$1 = A (-5+5) + B(-5-5) \Rightarrow B = -1/10.$$

Therefore,

$$\frac{1}{(y-5)(y+5)} = \frac{A}{(y-5)} + \frac{B}{(y+5)} = \frac{1}{10(y-5)} - \frac{1}{10(y+5)}.$$

We are now able to solve the differential equation, by noting that

$$\frac{dy}{dx} = (y-5)(y+5) \Rightarrow \frac{1}{(y-5)(y+5)}dy = dx \Rightarrow \left(\frac{1}{10(y-5)} - \frac{1}{10(y+5)} \right) dy = dx.$$

Integrating this expression on both sides, we get

$$\tag{1} \frac{1}{10}\left(\ln|y-5| - \ln|y+5|\right) = x + C \Rightarrow \ln{\left|\frac{y-5}{y+5} \right|} = 10x + \hat C.$$

Exponentiating both sides of $(1)$, we arrive at

$$\tag{2} \left|\frac{y-5}{y+5} \right| = e^{10x + \hat C} = e^{10x} \cdot \underbrace{e^{\hat C}}_{=K} = Ke^{10x}.$$

Now, plugging the initial condition, $y(4)=0$, in $(2)$, we finally get

$$\underbrace{\left|\frac{0-5}{0+5} \right|}_{=1} = K \cdot e^{10\times4} \Rightarrow K = e^{-40}.$$


If you really want to solve explicitly for $y$, you need to know the following identity

$$\tag{3} \ln{\left| \frac{1+u}{1-u} \right|} = 2 \, \text{arctanh}(u), \; |u|<1.$$

Then,

$$\frac{1}{10}\left(\ln|y-5| - \ln|y+5|\right) = x + C \Rightarrow \ln{\left( |-5| \left|1-\frac{y}{5} \right| \right)} - \ln{\left( |5| \left|1+\frac{y}{5} \right| \right)} = 10x + \hat C,$$

and, resorting to identity $(3)$, we get

$$\ln{\left| \frac{1-\frac{y}{5}}{1+\frac{y}{5}} \right|} = \ln{\left| \left( \frac{1+\frac{y}{5}}{1-\frac{y}{5}} \right)^{-1} \right|} = -\ln{\left| \frac{1+\frac{y}{5}}{1-\frac{y}{5}} \right|} = - 2 \, \text{arctanh}\left( \frac{y}{5} \right), \; |y|<5.$$

Therefore,

$$\tag{4}\text{arctanh}\left( \frac{y}{5} \right) = \bar C - 5x.$$

Applying $\tanh$ on both sides of $(4)$, we arrive at

$$\frac{y}{5} = \tanh{(\bar C - 5x)} \Rightarrow y = 5\tanh{(\bar C - 5x)}.$$

Now, plugging the initial condition, $y(4)=0$, we finally get

$$0 = \tanh{(\bar C - 5 \cdot 4)} \Rightarrow \bar C - 20 = 0 \Rightarrow \bar C = 20,$$

so

$$y(x) = 5 \tanh{(20 - 5x)}.$$

PS: Another way to solve this problem was to notice that $\frac{d}{du} \text{arctanh}(u) = \frac{1}{1-u^2}$, so all we had to do was to write the inital differential equation in that form.

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  • $\begingroup$ I have done this problem several times in physics, I normally carry forward until I can cast it as $y(x) = -5 \tanh{(5x -20)}$, of course thats fairly more advanced than most people want $\endgroup$ – Triatticus May 20 '15 at 4:36

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