# Separable Equations

A first order differential equation y' = f (x, y) is called a separable equation if the function f (x, y) can be factored into the product of two functions of x and y:

$f\left( {x,y} \right) = p\left( x \right)h\left( y \right),$

where p (x) and h (y) are continuous functions.

Considering the derivative $${y'}$$ as the ratio of two differentials $${\frac{{dy}}{{dx}}},$$ we move $$dx$$ to the right side and divide the equation by $$h\left( y \right):$$

$\frac{{dy}}{{dx}} = p\left( x \right)h\left( y \right),\;\; \Rightarrow \frac{{dy}}{{h\left( y \right)}} = p\left( x \right)dx.$

Of course, we need to make sure that $$h\left( y \right) \ne 0.$$ If there's a number $${y_0}$$ such that $$h\left( {{y_0}} \right) = 0,$$ then this number will also be a solution of the differential equation. Division by $$h\left( y \right)$$ causes loss of this solution.

By denoting $$q\left( y \right) = {\frac{1}{{h\left( y \right)}}},$$ we write the equation in the form

$q\left( y \right)dy = p\left( x \right)dx.$

We have separated the variables so now we can integrate this equation:

$\int {q\left( y \right)dy} = \int {p\left( x \right)dx} + C,$

where $$C$$ is an integration constant.

Calculating the integrals, we get the expression

$Q\left( y \right) = P\left( x \right) + C,$

representing the general solution of the separable differential equation.

## Solved Problems

### Example 1.

Solve the differential equation $\frac{{dy}}{{dx}} = y\left( {y + 2} \right).$

Solution.

In the given case $$p\left( x \right) = 1$$ and $$h\left( y \right) = y\left( {y + 2} \right).$$ We divide the equation by $$h\left( y \right)$$ and move $$dx$$ to the right side:

$\frac{{dy}}{{y\left( {y + 2} \right)}} = dx.$

One can notice that after dividing we can lose the solutions $$y = 0$$ and $$y = -2$$ when $$h\left( y \right)$$ becomes zero. In fact, let's see that $$y = 0$$ is a solution of the differential equation. Obviously,

$y = 0,\;\;dy = 0.$

Substituting this into the equation gives $$0 = 0.$$ Hence, $$y = 0$$ is one of the solutions. Similarly, we can check that $$y = -2$$ is also a solution.

Returning to the differential equation, we integrate it:

$\int {\frac{{dy}}{{y\left( {y + 2} \right)}}} = \int {dx} + C.$

We can calculate the left integral using the fractional decomposition of the integrand:

$\frac{1}{{y\left( {y + 2} \right)}} = \frac{A}{y} + \frac{B}{{y + 2}},\;\; \Rightarrow \frac{1}{{y\left( {y + 2} \right)}} = \frac{{A\left( {y + 2} \right) + By}}{{y\left( {y + 2} \right)}},\;\; \Rightarrow 1 \equiv Ay + 2A + By,\;\; \Rightarrow 1 \equiv \left( {A + B} \right)y + 2A,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {A + B = 0}\\ {2A = 1} \end{array}} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {A = \frac{1}{2}}\\ {B = - \frac{1}{2}} \end{array}} \right..$

Thus, we get the following decomposition of the rational integrand:

$\frac{1}{{y\left( {y + 2} \right)}} = \frac{1}{2}\left( {\frac{1}{y} - \frac{1}{{y + 2}}} \right).$

Hence,

$\frac{1}{2}\int {\left( {\frac{1}{y} - \frac{1}{{y + 2}}} \right)dy} = \int {dx} + C,\;\; \Rightarrow \frac{1}{2}\left( {\int {\frac{{dy}}{y}} - \int {\frac{{dy}}{{y + 2}}} } \right) = \int {dx} + C,\;\; \Rightarrow \frac{1}{2}\left( {\ln \left| y \right| - \ln \left| {y + 2} \right|} \right) = x + C,\;\; \Rightarrow \frac{1}{2}\ln \left| {\frac{y}{{y + 2}}} \right| = x + C,\;\; \Rightarrow \ln \left| {\frac{y}{{y + 2}}} \right| = 2x + 2C.$

We can rename the constant: $$2C = {C_1}.$$ Thus, the final solution of the equation is written in the form

$\ln \left| {\frac{y}{{y + 2}}} \right| = 2x + {C_1},\;\; y = 0,\;\; y = - 2.$

Here the general solution is expressed in implicit form. In the given case we can transform the expression to obtain the answer as an explicit function $$y = f\left( {x,{C_1}} \right),$$ where $${C_1}$$ is a constant. However, it is possible to do not for all differential equations.

### Example 2.

Solve the differential equation $\left( {{x^2} + 4} \right)y' = 2xy.$

Solution.

We can rewrite this equation in the following way:

$\left( {{x^2} + 4} \right)dy = 2xydx.$

Divide both sides by $$\left( {{x^2} + 4} \right)y$$ to get

$\frac{{dy}}{y} = \frac{{2xdx}}{{\left( {{x^2} + 4} \right)}}.$

Obviously, that $${{x^2} + 4} \ne 0$$ for all real $$x.$$ Check if $$y = 0$$ is a solution of the equation. Substituting $$y = 0$$ and $$dy = 0$$ into the differential equation, we see that the function $$y = 0$$ is one of the solutions of the equation.

Now we can integrate it:

$\int {\frac{{dy}}{y}} = \int {\frac{{2xdx}}{{\left( {{x^2} + 4} \right)}}} + C,\;\; \Rightarrow \ln \left| y \right| = \int {\frac{{d\left( {{x^2}} \right)}}{{{x^2} + 4}}} + C.$

Notice that $$d\left( {{x^2}} \right) = d\left( {{x^2} + 4} \right).$$ Hence,

$\ln \left| y \right| = \int {\frac{{d\left( {{x^2} + 4} \right)}}{{{x^2} + 4}}} + C,\;\; \Rightarrow \ln \left| y \right| = \ln \left( {{x^2} + 4} \right) + C.$

We can represent the constant $$C$$ as $$\ln {C_1},$$ where $${C_1} \gt 0.$$ Then

$\ln \left| y \right| = \ln \left( {{x^2} + 4} \right) + \ln {C_1},\;\; \Rightarrow \ln \left| y \right| = \ln \left( {{C_1}\left( {{x^2} + 4} \right)} \right),\;\; \Rightarrow \left| y \right| = {C_1}\left( {{x^2} + 4} \right),\;\; \Rightarrow y = \pm {C_1}\left( {{x^2} + 4} \right).$

Thus, the given differential equation has the following solutions:

$y = \pm {C_1}\left( {{x^2} + 4} \right),\;\; y = 0,\;\; \text{where}\;\; {C_1} \gt 0.$

This answer can be simplified. Indeed, if using an arbitrary constant $$C,$$ which takes values from $$-\infty$$ to $$\infty,$$ the solution can be written in the form:

$y = C\left( {{x^2} + 4} \right).$

When $$C = 0$$, it becomes $$y = 0.$$

See more problems on Page 2.