4.3 Separable equations  (Page 2/8)

Follow the five-step method of separation of variables.

1. In this example, $f(x)=2x+3$ and $g(y)=y^2-4.$ Setting $g(y)=0$ gives $y=\pm 2$ as constant solutions.
2. Divide both sides of the equation by $y^2-4$ and multiply by $dx.$ This gives the equation
   
   $\frac{dy}{y^2-4}=\left(2x+3\right)dx.$
3. Next integrate both sides:
   
   ${\displaystyle \int \frac{1}{y^2-4}dy}={\displaystyle \int \left(2x+3\right)dx}.$
   
   To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
   
   $\frac{1}{y^2-4}=\frac{1}{4}\left(\frac{1}{y-2}-\frac{1}{y+2}\right).$
   
   Then [link] becomes
   
   $\begin{array}{ccc}\hfill \frac{1}{4}{\displaystyle \int \left(\frac{1}{y-2}-\frac{1}{y+2}\right)dy}& =\hfill & {\displaystyle \int \left(2x+3\right)dx}\hfill \\ \hfill \frac{1}{4}\left(\text{ln}\left|y-2\right|-\text{ln}\left|y+2\right|\right)& =\hfill & x^2+3x+C.\hfill \end{array}$
   
   Multiplying both sides of this equation by $4$ and replacing $4C$ with $C_1$ gives
   
   $\begin{array}{ccc}\hfill \text{ln}\left|y-2\right|-\text{ln}\left|y+2\right|& =\hfill & 4x^2+12x+C_1\hfill \\ \hfill \text{ln}\left|\frac{y-2}{y+2}\right|& =\hfill & 4x^2+12x+C_1.\hfill \end{array}$
4. It is possible to solve this equation for y . First exponentiate both sides of the equation and define $C_2=e^{C_1}\text{:}$
   
   $\left|\frac{y-2}{y+2}\right|=C_2{e}^{4x^2+12x}.$
   
   Next we can remove the absolute value and let $C_2$ be either positive or negative. Then multiply both sides by $y+2.$
   
   $\begin{array}{}\\ \\ y-2=C_2\left(y+2\right)e^{4x^2+12x}\hfill \\ y-2=C_{2}y{e}^{{}^{4x^2+12x}}+2C_2{e}^{{}^{4x^2+12x}}.\hfill \end{array}$
   
   Now collect all terms involving y on one side of the equation, and solve for $y\text{:}$
   
   $\begin{array}{ccc}\hfill y-C_{2}y{e}^{4x^2+12x}& =\hfill & 2+2C_2{e}^{4x^2+12x}\hfill \\ \hfill y(1-C_2{e}^{4x^2+12x})& =\hfill & 2+2C_2{e}^{4x^2+12x}\hfill \\ \hfill y& =\hfill & \frac{2+2C_2{e}^{4x^2+12x}}{1-C_2{e}^{4x^2+12x}}.\hfill \end{array}$
5. To determine the value of $C_2,$ substitute $x=0$ and $y=\mathrm{-1}$ into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation $\frac{y-2}{y+2}=C_2{e}^{4x^2+12}.$ This is much easier to solve for $C_2\text{:}$
   
   $\begin{array}{ccc}\hfill \frac{y-2}{y+2}& =\hfill & C_2{e}^{4x^2+12x}\hfill \\ \hfill \frac{\mathrm{-1}-2}{\mathrm{-1}+2}& =\hfill & C_2{e}^{4{\left(0\right)}^2+12\left(0\right)}\hfill \\ \hfill C_2& =\hfill & \mathrm{-3}.\hfill \end{array}$
   
   Therefore the solution to the initial-value problem is
   
   $y=\frac{2-6e^{4x^2+12x}}{1+3e^{4x^2+12x}}.$
   
   A graph of this solution appears in [link] .