A first order differential equation \(y’ = f\left( {x,y} \right)\) is called a separable equation if the function \(f\left( {x,y} \right)\) 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 \(p\left( x \right)\) and \(h\left( y \right)\) are continuous functions.

Considering the derivative \({y’}\) as the ratio of two differentials \({\large\frac{{dy}}{{dx}}\normalsize},\) 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 \({x_0}\) such that \(h\left( {{x_0}} \right) = 0,\) 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 \(q\left( y \right) = {\large\frac{1}{{h\left( y \right)}}\normalsize},\) 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

Click a problem to see the solution.

### Example 1

Solve the differential equation \({\large\frac{{dy}}{{dx}}\normalsize} = y\left( {y + 2} \right).\)### Example 2

Solve the differential equation \(\left( {{x^2} + 4} \right)y’ = 2xy.\)### Example 3

Find all solutions of the differential equation \(y’ = – x{e^y}.\)### Example 4

Find a particular solution of the differential equation \(x\left( {y + 2} \right)y’ =\) \(\ln x + 1\) provided \(y\left( 1 \right) = – 1.\)### Example 5

Solve the differential equation \(y'{\cot ^2}x +\) \(\tan y = 0.\)### Example 6

Find a particular solution of the equation \(\left( {1 + {e^x}} \right)y’ = {e^x}\) satisfying the initial condition \(y\left( 0 \right) = 0.\)### Example 7

Solve the equation \(y\left( {1 + xy} \right)dx =\) \(x\left( {1 – xy} \right)dy.\)### Example 8

Find the general solution of the differential equation \(\left( {x + y + 1} \right)dx +\) \(\left( {4x + 4y + 10} \right)dy \) \(= 0.\)### Example 1.

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

\[{\ln \left| {\frac{y}{{y + 2}}} \right| = 2x + {C_1},\;\;\;}\kern-0.3pt{y = 0,\;\;\;}\kern-0.3pt{y = – 2.}\]

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.