# Differential Equations

## First Order Equations # Separable Equations

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 $${y_0}$$ 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 $$q\left( y \right) = {\large\frac{1}{{h\left( y \right)}}\normalsize},$$ 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 or tap 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.

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Problem 1
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Problems 2-8