4.3 Separable equations  (Page 17/8)

Follow the five-step method of separation of variables.

1. In this example, $f\left(x\right)=x^2-4$ and $g\left(y\right)=3y+2.$ Setting $g(y)=0$ gives $y=-\frac{2}{3}$ as a constant solution.
2. Rewrite the differential equation in the form
   
   $\frac{dy}{3y+2}=(x^2-4)dx.$
3. Integrate both sides of the equation:
   
   ${\displaystyle \int \frac{dy}{3y+2}}={\displaystyle \int \left(x^2-4\right)dx}.$
   
   Let $u=3y+2.$ Then $du=3\frac{dy}{dx}dx,$ so the equation becomes
   
   $\begin{array}{ccc}\hfill \frac{1}{3}{\displaystyle \int \frac{1}{u}du}& =\hfill & \frac{1}{3}{x}^3-4x+C\hfill \\ \hfill \frac{1}{3}\text{ln}\left|u\right|& =\hfill & \frac{1}{3}{x}^3-4x+C\hfill \\ \hfill \frac{1}{3}\text{ln}\left|3y+2\right|& =\hfill & \frac{1}{3}{x}^3-4x+C.\hfill \end{array}$
4. To solve this equation for $y,$ first multiply both sides of the equation by $3.$
   
   $\text{ln}\left|3y+2\right|=x^3-12x+3C$
   
   Now we use some logic in dealing with the constant $C.$ Since $C$ represents an arbitrary constant, $3C$ also represents an arbitrary constant. If we call the second arbitrary constant $C_1,$ the equation becomes
   
   $\text{ln}\left|3y+2\right|=x^3-12x+C_1.$
   
   Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base $e).$
   
   $\begin{array}{ccc}\hfill e^{\text{ln}\left|3y+2\right|}& =\hfill & e^{x^3-12x+C_1}\hfill \\ \hfill \left|3y+2\right|& =\hfill & e^{C_1}{e}^{x^3-12x}\hfill \end{array}$
   
   Again define a new constant $C_2=e^{c_1}$ (note that $C_2>0)\text{:}$
   
   $\left|3y+2\right|=C_2{e}^{x^3-12x}.$
   
   This corresponds to two separate equations: $3y+2=C_2{e}^{x^3-12x}$ and $3y+2=\text{−}{C}_2{e}^{x^3-12x}.$
   The solution to either equation can be written in the form $y=\frac{\mathrm{-2}\pm C_2{e}^{x^3-12x}}{3}.$
   Since $C_2>0,$ it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant $C$ is entirely arbitrary, and can be dropped. Therefore the solution can be written as
   
   $y=\frac{\mathrm{-2}+C{e}^{x^3-12x}}{3}.$
5. No initial condition is imposed, so we are finished.