# What is the Implicit answer for this differential equation? (dy)/(dx)=y^2-4

I need help with Differential equations.
What is the Implicit answer for this differential equation?
$\frac{dy}{dx}={y}^{2}-4$
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Leon Webster
Step 1
Here is how you solve equations like yours. You are trying to solve $\frac{dy}{dx}=f\left(y\right)$. Rewrite it as $\frac{dy}{f\left(y\right)}=dx$, integrate on both sides $\int \frac{1}{f\left(y\right)}dy=\int dx=x+c$ and invert to get expression for y(x). If you have conditions on the value of y(x) as some particular x, like $x=0$, then you can determine c as well.
Step 2
Applying this to your example. You have $\int \frac{1}{{y}^{2}-4}dy=x+c$. The integral is $\frac{1}{4}\mathrm{log}\left(\frac{2-y}{2+y}\right)$ so that $\frac{2-y}{2+y}=\mathrm{exp}\left(4x+4c\right)$. From here solving for y is straightforward.
###### Did you like this example?
Modelfino0g
Step 1
$\frac{dy}{dx}={y}^{2}-4\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{dy}{{y}^{2}-4}=dx$
Then integrate each side. You can easily use partial fractions on the left hand side.:
$\int \frac{dy}{{y}^{2}-4}=\int \frac{dy}{\left(y-2\right)\left(y+2\right)}=\int \frac{A}{y-2}+\frac{B}{y+2}$
$Ay+2A+By-2B=1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(A+B\right)y+2\left(A-B\right)=1\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(A+B=0,\phantom{\rule{thickmathspace}{0ex}}2\left(A-B\right)=1\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left(A=-B\wedge -4B=1\right).$
So $A=\frac{1}{4},B=-\frac{1}{4}$
That gives us
$\frac{1}{2}\int \frac{1}{y-2}-\frac{1}{y+2}dy=\int dx+c$
$\begin{array}{}\text{(1)}& \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{1}{4}\mathrm{ln}\left(\frac{y-2}{y+2}\right)=x+C\end{array}$
Step 2
Given your last comment, we can continue as follows from (1):
$\begin{array}{}\text{(2)}& \mathrm{ln}\left(\frac{y-2}{y+2}\right)=4x+{C}^{\ast }\end{array}$
Raising each side as a power of e gives us:
$\mathrm{ln}\left(\frac{y-2}{y+2}\right)=4x+C\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}{e}^{\mathrm{ln}\left(\frac{y-2}{y+2}\right)}={e}^{4x+C}$
$\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\frac{y-2}{y+2}={e}^{4x}\underset{{C}^{\prime }}{\underset{⏟}{{e}^{C}}}$
So $\begin{array}{}\text{(3)}& \frac{y-2}{y+2}={C}^{\prime }{e}^{4x}\end{array}$