I need help with Differential equations.

Question

What is the Implicit answer for this differential equation?

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

Help me i'm a newbie in differential equations.

Answer 1

Here is how you solve equations like yours. You are trying to solve $\frac{dy}{dx}=f(y)$. Rewrite it as $\frac{dy}{f(y)}=dx$, integrate on both sides $\int\frac{1}{f(y)}dy=\int dx=x+c$ 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.

Applying this to your example. You have $\int\frac{1}{y^2-4}dy=x+c$. The integral is $\frac{1}{4}\log{\left(\frac{2-y}{2+y}\right)}$ so that $\frac{2-y}{2+y}=\exp{(4x+4c)}$. From here solving for $y$ is straightforward.

Answer 2

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

Then integrate each side. You can easily use partial fractions on the left hand side.:

$$\int\frac{dy}{y^2-4} = \int\frac {dy}{(y-2)(y+2)} = \int\frac A{y-2} +\frac B{y+ 2}$$

$Ay + 2A + By-2B = 1\iff (A+B)y+ 2(A-B)= 1 \iff (A+B =0, \;2(A-B) = 1) \iff (A=-B \land -4B = 1).$

So $A=\frac 14, B=-\frac 14$.

That gives us $$\frac 12\int \frac 1{y-2} -\frac 1{y+2} dy = \int dx +c$$

$$\iff \frac 14 \ln\left(\frac{y-2}{y+2}\right) = x + C\tag 1$$

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Given your last comment, we can continue as follows from $(1)$:

$$\ln\left(\frac{y-2}{y+2}\right) = 4x + C^*\tag 2$$

Raising each side as a power of $e$ gives us: $$\ln\left(\frac{y-2}{y+2}\right) = 4x + C \iff e^{\large \ln\left(\frac{y-2}{y+2}\right)} = e^{\large 4x + C}$$ $$\iff \frac{y-2}{y+2} = e^{4x}\underbrace{e^C}_{C'}$$

So $$\frac {y-2}{y+2}= C'e^{4x}\tag 3$$