Equations not resolved with respect to the derivative

Equation of the form

where F is a continuous function, is called a first-order equation that is not solved for the derivative. If this equation can be solved for y’, then we obtain one or more explicit differential equations of the form

which are solved by methods discussed in other sections.

In what follows, we assume that the differential equation is not reduced to explicit form. The main method for solving such implicit equations is the parameterization method. Below, we show how this method is used to find a general solution for some important special cases of equations that are not solved for the derivative.

Note that a general solution may not cover all possible solutions of the differential equation. In addition to the general solution, a differential equation may also contain so-called singular solutions. This is discussed in more detail on the page Singular Solutions of Differential Equations.

In this case, the variable x is expressed explicitly in terms of the variable y and its derivative y’. We introduce the parameter . We differentiate the equation x = f ( y,y’ ) with respect to the variable y. We obtain:

Since , the last expression can be rewritten as:

We obtain an explicit differential equation, the general solution of which is described by the function.

where C is an arbitrary constant.

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

If we exclude the parameter p from this system, then the general solution can be expressed explicitly as x = f ( y, C ).

Here we encounter a similar case, but now the variable y explicitly depends on x and y’. We introduce a parameter and differentiate the equation y = f (x, y’) with respect to the variable x. The result is:

Solving the last differential equation, we obtain the algebraic equation g ( x, p, C ) = 0. Together with the original equation, it forms the following system of equations:

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

In this case, the differential equation does not contain the variable y. Using the parameter , it is easy to construct a general solution to the equation. Since dy = pdx and

then the following relation is valid:

By integrating the last equation, we obtain the general solution in parametric form:

An equation of this type does not contain the variable x and is solved in a similar manner. Using the parameter , we can write: . From this, it follows that.

By integrating the last expression, we obtain the general solution of the original differential equation in parametric form:

Find the general solution of the equation g ( y’ ) ² − 4 x = 0.

This equation is of the type x = f ( y’ ) (Case 3). We introduce the parameter p = y’ and write the equation as:

      

Let’s take the differentials of both parts of the equation:

      

Since dy = pdx, the last expression can be represented as

      

By integrating, we find the dependence of the variable y on the parameter p :

      

where C is an arbitrary constant.

Thus, we have obtained the general solution of the equation in parametric form:

      

The parameter p can be eliminated from the system of equations. From the second equation, we find:

      

After substituting into the first equation, we obtain a general solution in the form of an explicit function y = f ( x ) :

      

Find the general solution of the differential equation y = ln(25 + ( y’ ) ² ).

This differential equation is case 1 because it contains the variable y and its derivative y’. Using the parameter p, we can rewrite this equation as follows:

      

Let’s take the differentials of both sides:

      

Since dy = pdx, we obtain:

      

Now we can integrate the last expression and find x as a function of p.

      

As a result, we obtain the following parametric representation of the solution of the differential equation:

      

where C is an arbitrary constant.

Solve the differential equation   2 y = 2 x ² + 4 xy’ + ( y’ ) ² .

This equation corresponds to special case 2. Let   y’ = p, so that the equation can be rewritten as:

      

Let’s find the differentials of both sides of the equation, taking into account that dy = p dx. As a result, we obtain:

      

The last equation has two solutions. The first solution is:

      

Hence,

      

Integrating this simple equation, we obtain:

      

where C is an arbitrary constant. To determine the value of C, we substitute the obtained answer into the original differential equation:

      

The constant C must be zero to satisfy the equation. Therefore, the first solution is expressed by the function.

      

Now let us consider the second solution, which is determined by the differential equation.

      

Then

      

At the beginning of the solution, we wrote the differential equation in the form.

      

We substitute the known expression for x (as a function of the parameter p ) to find the dependence of y on p :

Thus, the second solution is described in parametric form by the following system of equations:

      

where C is an arbitrary constant. The final answer looks like this: