Differential Equations

First Order Equations

1st Order Differential Equations Logo

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 \({x_0}\) 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.

    Page 1
    Problem 1
    Page 2
    Problems 2-8