# Differential Equations

1st Order Equations# Separable Equations

Problem 1

Problems 2-8

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 \(y:\)

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 \(h\left( y \right):\)

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

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

where \(C\) is an integration constant.

Calculating the integrals, we get the expression

representing the general solution of the separable differential equation.

## Solved Problems

Click on problem description to see 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:

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,

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:

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:

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

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.

Problem 1

Problems 2-8