Differential Equations

1st Order Equations

Implicit Differential Equations

Page 1
Theory
Page 2
Problems 1-3

Definition and Methods of Solution

An equation of type

\[F\left( {x,y,y’} \right) = 0,\]

where \(F\) is a continuous function, is called the first order implicit differental equation.

If this equation can be solved for \(y’,\) we get one or several explicit differental equations of type

\[y’ = f\left( {x,y} \right),\]

that can be solved by methods covered in other sections.

Further we suppose that the differential equation can not be solved in the explicit form, so we should use different methods. The main techniques for solving an implicit differential equation is the method of introducing a parameter. Below we show how this method works to find the general solution for some most important particular cases of implicit differential equations.

Here we note that the general solution may not cover all possible solutions of a differential equation. Besides the general solution, the differential equation may also have so-called singular solutions.

We consider this in more detail on the page Singular Solutions of Differential Equations.

Case \(1.\) Implicit Differential Equation of Type \(x = f\left( {y,y’} \right).\)

In this case, the variable \(x\) is expressed explicitly in terms of \(y\) and the derivative \(y’.\) We introduce the parameter \(p = y’ \) \(= {\large\frac{{dy}}{{dx}}\normalsize}.\) Differentiate the equation \(x = f\left( {y,y’} \right)\) with respect to \(y.\) This produces:

\[{\frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {f\left( {y,p} \right)} \right] }={ \frac{{\partial f}}{{\partial y}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dy}}.}\]

As \({\large\frac{{dx}}{{dy}}\normalsize} = {\large\frac{1}{p}\normalsize},\) the last expression can be written as follows:

\[\frac{1}{p} = \frac{{\partial f}}{{\partial y}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dy}}.\]

We obtain an explicit differential equation such that its general solution is given by the function

\[g\left( {y,p,C} \right) = 0,\]

where \(C\) is a constant.

Thus, the general solution of the original implicit differential equation is defined in the parametric form by the system of two algebraic equations:

\[\left\{ \begin{array}{l}
g\left( {y,p,C} \right) = 0\\
x = f\left( {y,p} \right)
\end{array} \right..\]

If the parameter \(p\) can be eliminated from the system, the general solution is given in the explicit form \(x = f\left( {y,C} \right).\)

Case \(2.\) Implicit Differential Equation of Type \(y = f\left( {x,y’} \right).\)

Here we consider a similar case, when the variable \(y\) is an explicit function of \(x\) and \(y’.\) Introduce the parameter \(p = y’ \) \(= {\large\frac{{dy}}{{dx}}\normalsize}\) and differentiate the equation \(y = f\left( {x,y’} \right)\) with respect to \(x.\) As a result, we have:

\[
{{\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {f\left( {x,p} \right)} \right] }={ \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dx}}\;\;}}\kern-0.3pt
{\text{or}\;\;p = \frac{{\partial f}}{{\partial x}} + \frac{{\partial f}}{{\partial p}}\frac{{dp}}{{dx}}.}
\]

Solving the last differential equation, we get the algebraic equation \(g\left( {x,p,C} \right)\) \( = 0.\) Together with the original equation, they form the following system of equations:

\[\left\{ \begin{array}{l}
g\left( {x,p,C} \right) = 0\\
y = f\left( {x,p} \right)
\end{array} \right.,\]

which is the general solution of the given differential equation in the parametric form. In some cases, when the parameter \(p\) can be eliminated from the system, the general solution can be written in the explicit form \(y = f\left( {x,C} \right).\)

Case \(3.\) Implicit Differential Equation of Type \(x = f\left( {y’} \right).\)

Here the differential equation does not contain the variable \(y.\) Using the parameter \(p = y’ \) \(= {\large\frac{{dy}}{{dx}}\normalsize},\) it’s easy to construct the general solution of the equation. As \(dy = pdx\) and

\[{dx = d\left[ {f\left( p \right)} \right] }={ \frac{{df}}{{dp}}dp,}\]

then the following relationship holds:

\[dy = p\frac{{df}}{{dp}}dp.\]

Integrating the last equation gives the general solution in the parametric form:

\[\left\{ \begin{array}{l}
y = \int {p\frac{{df}}{{dp}}dp} + C\\
x = f\left( p \right)
\end{array} \right..\]

Case \(4.\) Implicit Differential Equation of Type \(y = f\left( {y’} \right).\)

The equation of this kind does not contain the variable \(x\) and can be solved the similar way. Using the parameter \(p = y’ \) \(= {\large\frac{{dy}}{{dx}}\normalsize},\) we can write \(dx = \large\frac{{dy}}{p}\normalsize.\) Then it follows from the equation that

\[{dx = \frac{{dy}}{p} }={ \frac{1}{p}\frac{{df}}{{dp}}dp.}\]

Integrating the last expression gives the general solution of the original implicit equation in parametric form:

\[\left\{ \begin{array}{l}
x = \int {\frac{1}{p}\frac{{df}}{{dp}}dp} \\
y = f\left( p \right)
\end{array} \right..\]

Solved Problems

Click on problem description to see solution.

 Example 1

Find the general solution of the equation \(9{\left( {y’} \right)^2} – 4x = 0.\)

 Example 2

Find the general solution of the differential equation \(y = \ln \left[ {25 + {{\left( {y’} \right)}^2}} \right].\)

 Example 3

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

Page 1
Theory
Page 2
Problems 1-3