Differential Equations

First Order Equations

1st Order Diff Equations Logo

Lagrange and Clairaut Equations

Solved Problems

Example 3.

Find the general and singular solutions of the differential equation \[y = xy' + {\left( {y'} \right)^2}.\]

Solution.

This is a Clairaut equation. By setting \(y' = p,\) we write it in the form

\[y = xp + {p^2}.\]

Differentiating in \(x,\), we have

\[dy = xdp + pdx + 2pdp.\]

Replace \(dy\) with \(pdx\) to obtain:

\[\cancel{pdx} = xdp + \cancel{pdx} + 2pdp,\;\; \Rightarrow dp\left( {x + 2p} \right) = 0.\]

By equating the first factor to zero, we have

\[dp = 0,\;\; \Rightarrow p = C.\]

Now we substitute this into the differential equation:

\[y = Cx + {C^2}.\]

Thus, we obtain the general solution of the Clairaut equation, which is an one-parameter family of straight lines.

By equating the second term to zero we find that

\[x + 2p = 0,\;\; \Rightarrow x = - 2p.\]

This gives us the singular solution of the differential equation in parametric form:

\[\left\{ \begin{array}{l} x = - 2p\\ y = xp + {p^2} \end{array} \right..\]

By eliminating \(p\) from this system, we get the equation of the integral curve:

\[p = - \frac{x}{2},\;\; \Rightarrow y = x\left( { - \frac{x}{2}} \right) + {\left( { - \frac{x}{2}} \right)^2} = - \frac{{{x^2}}}{2} + \frac{{{x^2}}}{4} = - \frac{{{x^2}}}{4}.\]

From geometric point of view, the curve \(y = - \frac{{{x^2}}}{4}\) is the envelope of the family of straight lines defined by the general solution (see Figure \(1\)).

The envelope of a family of straight lines
Figure 1.

Example 4.

Find the general and singular solutions of the differential equation \[y = xy' + \sqrt {{{\left( {y'} \right)}^2} + 1}.\]

Solution.

As it can be seen, this is a Clairaut equation. Introduce the parameter \(y' = p:\)

\[y = xp + \sqrt {{p^2} + 1} .\]

Differentiating both sides with respect to \(x,\) we get:

\[dy = xdp + pdx + \frac{{pdp}}{{\sqrt {{p^2} + 1} }}.\]

Since \(dy = pdx,\) one can write:

\[\cancel{pdx} = xdp + \cancel{pdx} + \frac{{pdp}}{{\sqrt {{p^2} + 1} }},\;\; \Rightarrow \left( {x + \frac{p}{{\sqrt {{p^2} + 1} }}} \right)dp = 0.\]

Consider the case \(dp = 0.\) Then \(p = C.\) Substituting this in the equation, we find the general solution:

\[y = Cx + \sqrt {{C^2} + 1} .\]

Graphically, this solution corresponds to the family of one-parameter straight lines.

The second case is described by the equation \(x = - \frac{p}{{\sqrt {{p^2} + 1} }}.\) Find the corresponding parametric expression for \(y:\)

\[y = xp + \sqrt {{p^2} + 1} = - \frac{{p^2}}{{\sqrt {{p^2} + 1} }} + \sqrt {{p^2} + 1} = \frac{{ - \cancel{p^2} + \cancel{p^2} + 1}}{{\sqrt {{p^2} + 1} }} = \frac{1}{{\sqrt {{p^2} + 1} }}.\]

The parameter \(p\) can be eliminated from the formulas for \(x\) and \(y.\) Squaring and adding these equations, we obtain:

\[{x^2} + {y^2} = {\left( { - \frac{p}{{\sqrt {{p^2} + 1} }}} \right)^2} + {\left( {\frac{1}{{\sqrt {{p^2} + 1} }}} \right)^2} = \frac{\cancel{{p^2} + 1}}{\cancel{{p^2} + 1}} = 1.\]

The last expression is the equation of the circle with radius \(1\) and centered at the origin. Thus, the singular solution is represented by the unit circle on the \(xy\)-plane, which is the envelope of the family of the straight lines (Figure \(2\)).

The singular solution is represented by the unit circle
Figure 2.
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