Follow the five-step method of separation of variables.

1. In this example, [latex]f\left(x\right)={x}^{2}-4[/latex] and [latex]g\left(y\right)=3y+2[/latex]. Setting [latex]g\left(y\right)=0[/latex] gives [latex]y=-\frac{2}{3}[/latex] as a constant solution.
2. Rewrite the differential equation in the form
   
   [latex]\frac{dy}{3y+2}=\left({x}^{2}-4\right)dx[/latex].

    

3. Integrate both sides of the equation:
   
   [latex]\displaystyle\int \frac{dy}{3y+2}=\displaystyle\int \left({x}^{2}-4\right)dx[/latex].

   
   
   Let [latex]u=3y+2[/latex]. Then [latex]du=3\frac{dy}{dx}dx[/latex], so the equation becomes
   

   [latex]\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}|u|& =\hfill & \frac{1}{3}{x}^{3}-4x+C\hfill \\ \hfill \frac{1}{3}\text{ln}|3y+2|& =\hfill & \frac{1}{3}{x}^{3}-4x+C.\hfill \end{array}[/latex]

    

4. To solve this equation for [latex]y[/latex], first multiply both sides of the equation by [latex]3[/latex].
   
   [latex]\text{ln}|3y+2|={x}^{3}-12x+3C[/latex]

   
   
   Now we use some logic in dealing with the constant [latex]C[/latex]. Since [latex]C[/latex] represents an arbitrary constant, [latex]3C[/latex] also represents an arbitrary constant. If we call the second arbitrary constant [latex]{C}_{1}[/latex], the equation becomes
   

   [latex]\text{ln}|3y+2|={x}^{3}-12x+{C}_{1}[/latex].

   
   
   Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base [latex]e[/latex]).
   

   [latex]\begin{array}{ccc}\hfill {e}^{\text{ln}|3y+2|}& =\hfill & {e}^{{x}^{3}-12x+{C}_{1}}\hfill \\ \hfill |3y+2|& =\hfill & {e}^{{C}_{1}}{e}^{{x}^{3}-12x}\hfill \end{array}[/latex]

   
   
   Again define a new constant [latex]{C}_{2}={e}^{{c}_{1}}[/latex] (note that [latex]{C}_{2}>0[/latex]):
   

   [latex]|3y+2|={C}_{2}{e}^{{x}^{3}-12x}[/latex].

   
   
   This corresponds to two separate equations: [latex]3y+2={C}_{2}{e}^{{x}^{3}-12x}[/latex] and [latex]3y+2=\text{-}{C}_{2}{e}^{{x}^{3}-12x}[/latex].
   
   The solution to either equation can be written in the form [latex]y=\frac{-2\pm {C}_{2}{e}^{{x}^{3}-12x}}{3}[/latex].
   
   Since [latex]{C}_{2}>0[/latex], it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant [latex]C[/latex] is entirely arbitrary, and can be dropped. Therefore the solution can be written as
   

   [latex]y=\frac{-2+C{e}^{{x}^{3}-12x}}{3}[/latex].

    

5. No initial condition is imposed, so we are finished.

Follow the five-step method of separation of variables.

1. In this example, [latex]f\left(x\right)=2x+3[/latex] and [latex]g\left(y\right)={y}^{2}-4[/latex]. Setting [latex]g\left(y\right)=0[/latex] gives [latex]y=\pm 2[/latex] as constant solutions.
2. Divide both sides of the equation by [latex]{y}^{2}-4[/latex] and multiply by [latex]dx[/latex]. This gives the equation
   
   [latex]\frac{dy}{{y}^{2}-4}=\left(2x+3\right)dx[/latex].

    

3. Next integrate both sides:
   
   [latex]\displaystyle\int \frac{1}{{y}^{2}-4}dy=\displaystyle\int \left(2x+3\right)dx[/latex].

   
   
   To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
   

   [latex]\frac{1}{{y}^{2}-4}=\frac{1}{4}\left(\frac{1}{y - 2}-\frac{1}{y+2}\right)[/latex].

   
   
   Then integration becomes
   

   [latex]\begin{array}{ccc}\hfill \frac{1}{4}{\displaystyle\int \left(\frac{1}{y - 2}-\frac{1}{y+2}\right)}dy& =\hfill & {\displaystyle\int \left(2x+3\right)dx}\hfill \\ \hfill \frac{1}{4}\left(\text{ln}|y - 2|-\text{ln}|y+2|\right)& =\hfill & {x}^{2}+3x+C.\hfill \end{array}[/latex]

   
   
   Multiplying both sides of this equation by [latex]4[/latex] and replacing [latex]4C[/latex] with [latex]{C}_{1}[/latex] gives
   

   [latex]\begin{array}{ccc}\hfill \text{ln}|y - 2|-\text{ln}|y+2|& =\hfill & 4{x}^{2}+12x+{C}_{1}\hfill \\ \hfill \text{ln}|\frac{y - 2}{y+2}|& =\hfill & 4{x}^{2}+12x+{C}_{1}.\hfill \end{array}[/latex]

    

4. It is possible to solve this equation for y. First exponentiate both sides of the equation and define [latex]{C}_{2}={e}^{{C}_{1}}\text{:}[/latex]
   
   [latex]|\frac{y - 2}{y+2}|={C}_{2}{e}^{4{x}^{2}+12x}[/latex].

   
   
   Next we can remove the absolute value and let [latex]{C}_{2}[/latex] be either positive or negative. Then multiply both sides by [latex]y+2[/latex].
   

   [latex]\begin{array}{}\\ \\ y - 2={C}_{2}\left(y+2\right){e}^{4{x}^{2}+12x}\hfill \\ y - 2={C}_{2}y{e}^{{}^{4{x}^{2}+12x}}+2{C}_{2}{e}^{{}^{4{x}^{2}+12x}}.\hfill \end{array}[/latex]

   
   
   Now collect all terms involving y on one side of the equation, and solve for [latex]y\text{:}[/latex]
   

   [latex]\begin{array}{ccc}\hfill y-{C}_{2}y{e}^{4{x}^{2}+12x}& =\hfill & 2+2{C}_{2}{e}^{4{x}^{2}+12x}\hfill \\ \hfill y\left(1-{C}_{2}{e}^{4{x}^{2}+12x}\right)& =\hfill & 2+2{C}_{2}{e}^{4{x}^{2}+12x}\hfill \\ \hfill y& =\hfill & \frac{2+2{C}_{2}{e}^{4{x}^{2}+12x}}{1-{C}_{2}{e}^{4{x}^{2}+12x}}.\hfill \end{array}[/latex]

    

5. To determine the value of [latex]{C}_{2}[/latex], substitute [latex]x=0[/latex] and [latex]y=-1[/latex] into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation [latex]\frac{y - 2}{y+2}={C}_{2}{e}^{4{x}^{2}+12}[/latex]. This is much easier to solve for [latex]{C}_{2}\text{:}[/latex]
   
   [latex]\begin{array}{ccc}\hfill \frac{y - 2}{y+2}& =\hfill & {C}_{2}{e}^{4{x}^{2}+12x}\hfill \\ \hfill \frac{-1 - 2}{-1+2}& =\hfill & {C}_{2}{e}^{4{\left(0\right)}^{2}+12\left(0\right)}\hfill \\ \hfill {C}_{2}& =\hfill & -3.\hfill \end{array}[/latex]

   
   
   Therefore the solution to the initial-value problem is
   

   [latex]y=\frac{2 - 6{e}^{4{x}^{2}+12x}}{1+3{e}^{4{x}^{2}+12x}}[/latex].

   
   
   A graph of this solution appears in Figure 1.