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Solving Clairaut and d’Alembert first order differential equations

Nasser M. Abbasi

November 29, 2022   Compiled on November 29, 2022 at 2:58pm

Contents

 1 Introduction
 2 Solved examples
  2.1 Example 1
  2.2 Example 2
  2.3 Example 3
  2.4 Example 4
  2.5 Example 5
  2.6 Example 6
  2.7 Example 7
  2.8 Example 8
  2.9 Example 9
  2.10 Example 10
  2.11 Example 11
  2.12 Example 12
 3 references

1 Introduction

This gives an overview on solving first order ode where \(y^{\prime }\) enters the ode as nonlinear. Examples are \(x\left ( y^{\prime }\right ) ^{2}+yy^{\prime }+x=0\) or \(2y^{\prime }x-y+\ln y^{\prime }=0\) and so on. Four general cases exist and these are summarized in the flow chart at the end of this section. Two of these cases are called the Clairaut ode and the d’Alembert ode. Following the flow chart, a number of examples are solved.

Given the ode \(y^{\prime }=f\left ( x,y,y^{\prime }\right ) \), we start by writing \(y^{\prime }=p\) which results in \[ y^{\prime }=f\left ( x,y,p\right ) \]

  1. If able to solve for \(p\) from \(f(x,y,p)=0\) then solve each generated equation as an ode of degree one. Let roots be \(p_{i}\), then solve each ode \(y^{\prime }=p_{i}=f_{i}(x,y)\) where \(i=1,2,\dots ,n\). This can be done easily now as degree is one.
  2. If above is not successful, then solve for \(y\) from \(f(x,y,p)=0\). Let \begin {equation} y=\phi (x,p) \tag {1} \end {equation}

    Differentiating (1) w.r.t. \(x\) gives \begin {equation} p=\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx} \tag {2} \end {equation}

    Now there are two sub cases. One gives rise to the Clairaut ode and the other to the d’Alembert ode.

    1. if \(\frac {\partial \phi }{\partial x}=p\) then this is now called Clairaut ode. ODE (2) becomes \begin {align*} p & =p+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\\ 0 & =\frac {\partial \phi }{\partial p}\frac {dp}{dx} \end {align*}

      The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p & =c \end {align*}

      Substituting this in (1) gives the general solution \[ y=\phi (x,c) \] The term \(\frac {\partial \phi }{\partial p}=0\) is used to find singular solutions. Solving for \(p\) from \(\frac {\partial \phi }{\partial p}=0\) (algebraic) and substituting the result in (1) gives the singular solution. Let the solution be \(p_{i}\), therefore the singular solution is \[ y=\phi (x,p_{i}) \]

    2. if \(\frac {\partial \phi }{\partial x}\neq p\) then this is now called d’Alembert ode. Expanding ODE (2) results in an ode in \(p\) of the form \[ Q(p)=G(x,p)\frac {dp}{dx}\] The general solution is given by solving for \(p\) from the above ode and substituting the result in (1). This can be done by solving for \(p\) directly from the above first order ode if the ode turns out simple one (linear, separable), if not then the ode is inverted to become \begin {align*} \frac {dp}{dx} & =\frac {Q(p)}{G(x,p)}\\ \frac {dx}{dp} & =\frac {G(x,p}{Q(p)} \end {align*}

      Let the solution be \begin {equation} x=R(p)+c \tag {3} \end {equation} Eliminating \(p\) from (1) and (3) gives the general solution for \(y.\) Singular solutions are found when \(\frac {dp}{dx}=0\). Then the singular solution is found by solving for \(p\) from \(Q(p)=0\). Let the solution be \(p_{i}\). Hence singular solution is given by substituting this in in (1) which gives \[ y=\phi (x,p_{i}) \]

  3. If solving for \(y\) was not possible, then we solve for \(x\) from \(f(x,y,p)=0\). This is similar to the above but instead of solving for \(y\), we try to solve for \(x\). If able to solve for \(x\) from \(f(x,y,p)=0\) then let \[ x=\phi (y,p) \] Differentiating (1) w.r.t. \(y\) gives \begin {align} \frac {dx}{dy} & =\frac {\partial \phi }{\partial y}+\frac {\partial \phi }{\partial p}\frac {dp}{dy}\nonumber \\ \frac {1}{p} & =\frac {\partial \phi }{\partial y}+\frac {\partial \phi }{\partial p}\frac {dp}{dy} \tag {4} \end {align}

    Solving (4) for \(p\) and substituting the result in (1) gives the solution for \(y\)

The above is summarized in the following flow chart.

pict
Figure 1: Algorithm for solving first order ode with nonlinear \(y'\)

To show how these method work, the following ODE’s are now solved.

2 Solved examples

All the ode’s below have the form \(y=\phi \left ( x,y,p\right ) \). Notice that an ode is Clairaut only when \(\frac {\partial \phi }{\partial x}=p\) and d’Alembert otherwise.

number original ode \(y=\phi \left ( x,y,p\right ) \) \(\frac {\partial \phi }{\partial x}\) type
\(1\) \(x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1\) \(y=xp+\frac {1}{p}\) \(p\) Clairaut
\(2\) \(y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}\) \(y=xp-p^{2}\) \(p\) Clairaut
\(3\) \(y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}\) \(y=xp-\frac {1}{4}p^{2}\) \(p\) Clairaut
\(4\) \(y=x\left ( y^{\prime }\right ) ^{2}\) \(y=xp^{2}\) \(p^{2}\) d’Alembert
\(5\) \(y=x+\left ( y^{\prime }\right ) ^{2}\) \(y=x+p^{2}\) \(1\) d’Alembert
\(6\) \(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) \(y=-x+\left ( p^{2}-1\right ) \) \(-1\) d’Alembert
\(7\) \(yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x\) \(y=\frac {1}{p}x+p\) \(\frac {1}{p}\) d’Alembert
\(8\) \(y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}\) \(y=xp^{2}+p^{2}\) \(p^{2}\) d’Alembert
\(9\) \(y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}\) \(y=\frac {x}{a}p+\frac {b}{a}p^{-1}\) \(\frac {p}{a}\) d’Alembert
\(10\) \(y=x\left ( y^{\prime }+a\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\right ) \) \(y=x\left ( p+a\sqrt {1+p^{2}}\right ) \) \(p+a\sqrt {1+p^{2}}\) d’Alembert
\(11\) \(y=x+\left ( y^{\prime }\right ) ^{2}\left ( 1-\frac {2}{3}y^{\prime }\right ) \) \(y=x+p^{2}\left ( 1-\frac {2}{3}p\right ) \) \(1\) d’Alembert

2.1 Example 1

\(x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1\), is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp+\frac {1}{p}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align*} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\\ p & =p+\left ( x-\frac {1}{p^{2}}\right ) \frac {dp}{dx}\\ 0 & =\left ( x-\frac {1}{p^{2}}\right ) \frac {dp}{dx} \end {align*}

Since \(\frac {\partial \phi }{\partial x}=p\) then this is Clairaut ode. The general solution is found from \(\frac {dp}{dx}=0\). This implies \(p=c_{1}\). Substituting this into (1) gives\begin {equation} y\left ( x\right ) =xc_{1}+\frac {1}{c_{1}} \tag {2} \end {equation} The singular solution is found from \(\left ( x-\frac {1}{p^{2}}\right ) =0\) Hence \(p^{2}=\frac {1}{x}\) or \(p=\pm \sqrt {\frac {1}{x}}\). Substituting these back in (1) gives\begin {align} y_{1}\left ( x\right ) & =x\sqrt {\frac {1}{x}}+\sqrt {x}\nonumber \\ & =2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}\nonumber \\ & =-2\sqrt {x} \tag {4} \end {align}

Eq. (2) is the general solution and (3,4) are the singular solutions.

2.2 Example 2

\(y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp-p^{2}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align*} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\\ p & =p+\left ( x-2p\right ) \frac {dp}{dx}\\ 0 & =\left ( x-2p\right ) \frac {dp}{dx} \end {align*}

Since \(\frac {\partial \phi }{\partial x}=p\) then this is Clairaut ode. The general solution is found from \(\frac {dp}{dx}=0\). This implies \(p=c_{1}\). Substituting this into (1) gives\begin {equation} y\left ( x\right ) =xc_{1}-c_{1}^{2} \tag {2} \end {equation} The singular solution is found from \(x-2p=0\). Hence \(p=\frac {x}{2}\). Substituting this back in (1) gives\begin {align} y\left ( x\right ) & =\frac {x^{2}}{2}-\frac {x^{2}}{4}\nonumber \\ & =\frac {x^{2}}{4} \tag {3} \end {align}

Eq. (2) is the general solution and (3) is the singular solution.

2.3 Example 3

\(y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives\begin {align} y & =xp-\frac {1}{4}p^{2}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align*} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\\ p & =p+\left ( x-\frac {1}{2}p\right ) \frac {dp}{dx}\\ 0 & =\left ( x-\frac {1}{2}p\right ) \frac {dp}{dx} \end {align*}

Since \(\frac {\partial \phi }{\partial x}=p\) then this is Clairaut ode. The general solution is found from \(\frac {dp}{dx}=0\). This implies \(p=c_{1}\). Substituting this into (1) gives\begin {equation} y\left ( x\right ) =xc_{1}-\frac {1}{4}c_{1}^{2} \tag {2} \end {equation} The singular solution is found from \(x-\frac {1}{2}p=0\). Hence \(p=2x\). Substituting this back in (1) gives\begin {align} y\left ( x\right ) & =2x^{2}-x^{2}\nonumber \\ & =x^{2} \tag {3} \end {align}

Eq. (2) is the general solution and (3) is the singular solution.

2.4 Example 4

\(y=x\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives

\begin {align} y & =xp^{2}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =p^{2}+2xp\frac {dp}{dx}\nonumber \\ \frac {p-p^{2}}{p} & =2x\frac {dp}{dx}\nonumber \\ \frac {p}{p-p^{2}}\frac {dp}{dx} & =\frac {1}{2x} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=p^{2}\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p\left ( p-1\right ) =0\). Hence \(p=0\) or \(p=1\).  Substituting these values of \(p\) in (1) gives singular solutions\begin {align} y & =0\tag {3}\\ y & =x \tag {4} \end {align}

general solution is found when \(\frac {dp}{dx}\neq 0\,\). We need to eliminate \(p\) between (2,1). Now we could either solve ode (2) directly as it is for \(p\left ( x\right ) \), and do an inversion and solve for \(x\left ( p\right ) \). Since (2) is separable as is, no need to do an inversion. Solution of (2) is\begin {align*} p & =0\\ p & =1+\frac {c_{1}}{\sqrt {x}} \end {align*}

For each \(p\), there is a general solution. Substituting each of the above in (1) gives\begin {align*} y\left ( x\right ) & =0\\ y\left ( x\right ) & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \end {align*}

Hence the final solutions are\begin {align*} y & =x\qquad \text {(singular)}\\ y & =0\\ y & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \end {align*}

But \(y=x\) can be obtained from the general solution when \(c_{1}=0\). Hence it is removed. Therefore the final solutions are\begin {align} y & =0\tag {5}\\ y & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \tag {6} \end {align}

What will happen if we had done an inversion to \(x\left ( p\right ) \)? Lets find out. ode(2) now becomes\begin {align*} \frac {p-p^{2}}{p}\frac {dx}{dp} & =2x\\ \frac {dx}{2x} & =\frac {p}{p-p^{2}}dp \end {align*}

Solving this for \(x\) gives\[ x=\frac {c_{1}}{\left ( p-1\right ) ^{2}}\] Solving for \(p\) from the above gives\begin {align*} p & =\frac {x+\sqrt {xc_{1}}}{x}\\ p & =\frac {x-\sqrt {xc_{1}}}{x} \end {align*}

Substituting each of the above in (1) gives\begin {align*} y & =x\left ( \frac {x+\sqrt {xc_{1}}}{x}\right ) ^{2}=\frac {\left ( x+\sqrt {xc_{1}}\right ) }{x}^{2}\\ y & =x\left ( \frac {x-\sqrt {xc_{1}}}{x}\right ) ^{2}=\frac {\left ( x-\sqrt {xc_{1}}\right ) }{x}^{2} \end {align*}

Now we see that singular solution \(y=x\) can be obtained from the above general solutions from \(c_{1}=0\). But \(y=0\) can not. Hence the final solutions are\begin {align} y & =0\qquad \text {(singular)}\tag {7}\\ y & =\frac {\left ( x+\sqrt {xc_{1}}\right ) }{x}^{2}\tag {8}\\ y & =\frac {\left ( x-\sqrt {xc_{1}}\right ) }{x}^{2} \tag {9} \end {align}

All solutions (5,6,7,8,9) are correct and verify. Maple gives the solutions given in (7,8,9) and not those in (5,6).

2.5 Example 5

\(y=x+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =x+p^{2}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =1+2p\frac {dp}{dx}\nonumber \\ \frac {p-1}{2p} & =\frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=1\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=1\). Substituting this in (1) gives singular solution as\begin {equation} y=x+1 \tag {3} \end {equation} The general solution is found by finding \(p\) from (3). No need here to do the inversion as it is separable already. Solving (2) gives\[ p=\operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1 \] Substituting this in (1) gives the general solution\begin {equation} y\left ( x\right ) =x+\left ( \operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1\right ) ^{2} \tag {4} \end {equation} Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =x+1\). Hence (3) is a particular solution and not a singular solution. (4) is the only solution.

2.6 Example 6

\(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =-x+\left ( p^{2}-1\right ) \tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =-1+2p\frac {dp}{dx}\nonumber \\ \frac {p+1}{2p} & =\frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=-1\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives singular solution as\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} The general solution is found by finding \(p\) from (2). No need here to do the inversion as (2) is separable already. Solving (2) gives\begin {align*} p & =-\operatorname *{LambertW}\left ( -e^{-\frac {x}{2}-1+\frac {c_{2}}{2}}\right ) -1\\ & =-\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1 \end {align*}

Substituting the above in (1) gives the general solution\begin {align} y\left ( x\right ) & =-x+\left ( p^{2}-1\right ) \nonumber \\ y\left ( x\right ) & =-x+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1\right ) ^{2}-1 \tag {4} \end {align}

Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =-x\). Hence (3) is a particular solution and not a singular solution. Solution (4) is therefore the only solution.

2.7 Example 7

\(yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

\begin {align} y & =\frac {1}{p}x+p\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =\frac {1}{p}+\left ( -\frac {x}{p^{2}}+1\right ) \frac {dp}{dx}\nonumber \\ p-\frac {1}{p} & =\left ( 1-\frac {x}{p^{2}}\right ) \frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=\frac {1}{p}\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p^{2}=1\). Solving for this gives \(p=\pm 1\). Substituting these values in (1) gives the solutions\begin {align} y_{1}\left ( x\right ) & =x+1\tag {3}\\ y_{2}\left ( x\right ) & =-\left ( x+1\right ) \tag {4} \end {align}

The general solution is found by finding \(p\) from (d3). Since (d3) is not linear in \(p\), then inversion is needed. Writing (2) as\[ \frac {\frac {p^{2}-1}{p}}{1-\frac {x}{p^{2}}}=\frac {dp}{dx}\] Since this is not linear or separable in \(p\), it is inverted.\begin {align*} \frac {dx}{dp} & =\frac {\frac {p^{2}-x}{p^{2}}}{\frac {p^{2}-1}{p}}\\ \frac {dx}{dp} & =\frac {p^{2}-x}{p\left ( p^{2}-1\right ) }\\ \frac {dx}{dp}+x\frac {1}{p\left ( p^{2}-1\right ) } & =\frac {p}{\left ( p^{2}-1\right ) } \end {align*}

This is now linear ODE in \(x\left ( p\right ) \). The solution is\begin {align} x & =\frac {p\sqrt {\left ( p-1\right ) \left ( 1+p\right ) }\ln \left ( p+\sqrt {p^{2}-1}\right ) }{\left ( 1+p\right ) \left ( p-1\right ) }+c_{1}\frac {p}{\sqrt {\left ( 1+p\right ) \left ( p-1\right ) }}\nonumber \\ & =\frac {p\sqrt {p^{2}-1}\ln \left ( p+\sqrt {p^{2}-1}\right ) }{p^{2}-1}+c_{1}\frac {p}{\sqrt {p^{2}-1}} \tag {5} \end {align}

From (1) since \(y=\frac {1}{p}x+p\) then solving for \(p\) gives\begin {align*} p_{1} & =\frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\\ p_{2} & =\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x} \end {align*}

Substituting each \(p_{i}\) in (5) gives the general solution (implicit) in \(y\left ( x\right ) \). First solution is\[ x=\frac {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}\] And second solution is\[ x=\frac {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}\]

2.8 Example 8

\(y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

\begin {align} y & =xp^{2}+p^{2}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =p^{2}+\left ( 2px+2p\right ) \frac {dp}{dx}\nonumber \\ \frac {p-p^{2}}{p} & =\left ( 2x+2\right ) \frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=p^{2}\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=0\) or \(p=1\). Substituting these values in (1) gives the singular solutions\begin {align} y_{1}\left ( x\right ) & =0\tag {3}\\ y_{2}\left ( x\right ) & =x+1 \tag {4} \end {align}

The general solution is found by finding \(p\) from (d3). Since (d3) is not linear in \(p\), then inversion is needed. Writing (d3) as\[ \frac {p\left ( 1-p\right ) }{2p\left ( x+1\right ) }=\frac {dp}{dx}\] Inverting gives\begin {align*} \frac {dx}{dp} & =\frac {2\left ( x+1\right ) }{\left ( 1-p\right ) }\\ \frac {dx}{dp}-x\frac {2}{\left ( 1-p\right ) } & =\frac {2}{\left ( 1-p\right ) } \end {align*}

This is now linear \(x\left ( p\right ) \). The solution is\[ x=\frac {C^{2}}{\left ( p-1\right ) ^{2}}-1 \] Solving for \(p\) gives\begin {align*} \frac {C^{2}}{\left ( p-1\right ) ^{2}} & =x+1\\ \left ( p-1\right ) ^{2} & =\frac {C^{2}}{x+1}\\ \left ( p-1\right ) & =\pm \frac {C}{\sqrt {x+1}}\\ p & =1\pm \frac {C}{\sqrt {x+1}} \end {align*}

Substituting the above in (1) gives the general solutions\[ y=\left ( x+1\right ) p^{2}\] Therefore\begin {align*} y\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2} \end {align*}

The solution \(y_{1}\left ( x\right ) =0\) found earlier can not be obtained from the above general solution hence it is singular solution. But \(y_{2}\left ( x\right ) =x+1\) can be obtained from the general solution when \(C=0\). Hence there are only three solutions, they are\begin {align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y_{3}\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2} \end {align*}

2.9 Example 9

\(y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives

\begin {align} y & =\frac {x}{a}p+\frac {b}{a}p^{-1}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =\frac {p}{a}+\left ( \frac {x}{a}-\frac {b}{a}\frac {1}{p^{2}}\right ) \frac {dp}{dx}\nonumber \\ p-\frac {p}{a} & =\left ( \frac {x}{a}-\frac {b}{a}\frac {1}{p^{2}}\right ) \frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=\frac {p}{a}\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=0\). Substituting this in (1) does not generate any solutions due to division by zero. Hence no singular solution exist.

The general solution is found by finding \(p\) from (2). Since (2) is not linear in \(p\), then inversion is needed. Writing (2) as\[ \frac {p\left ( 1-\frac {1}{a}\right ) }{\frac {x}{a}-\frac {b}{a}p^{-2}}=\frac {dp}{dx}\] Since this is nonlinear, then inversion is needed\begin {align*} \frac {dx}{dp} & =\frac {\frac {x}{a}-\frac {b}{a}p^{-2}}{p\left ( 1-\frac {1}{a}\right ) }\\ \frac {dx}{dp}-x\frac {1}{p\left ( a-1\right ) } & =-\frac {b}{a}\frac {1}{p^{3}\left ( 1-\frac {1}{a}\right ) } \end {align*}

This is now linear ode in \(x\left ( p\right ) \). The solution is\begin {equation} x=\frac {b}{(2a-1)p^{2}}+C_{1}p^{\frac {1}{a-1}} \tag {3} \end {equation} There are now two choices to take. The first is by solving for \(p\) from the above in terms of \(x\) and then substituting the result in (1) to obtain explicit solution for \(y\left ( x\right ) \), and the second choice is by solving for \(p\) algebraically from (1) and substituting the result in (3). The second choice is easier in this case but gives an implicit solution. Solving for \(p\) from (1) gives

\begin {align*} p_{1} & =\frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\\ p_{1} & =\frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x} \end {align*}

Substituting each one of these solutions back in (3) gives two implicit solutions\begin {align*} x & =\frac {b}{(2a-1)\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}}\\ x & =\frac {b}{(2a-1)\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}} \end {align*}

2.10 Example 10

\(y=xy^{\prime }+ax\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =x\left ( p+a\sqrt {1+p^{2}}\right ) \tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =\left ( p+a\sqrt {1+p^{2}}\right ) +x\left ( 1+\frac {ap}{\sqrt {1+p^{2}}}\right ) \frac {dp}{dx}\nonumber \\ -a\sqrt {1+p^{2}} & =x\left ( 1+\frac {ap}{\sqrt {1+p^{2}}}\right ) \frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=p+a\sqrt {1+p^{2}}\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(-a\sqrt {1+p^{2}}=0\). This gives no real solution for \(p.\) Hence no singular solution exist.

The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). Since (2) is nonlinear, inversion is needed.\begin {align*} \frac {-a\sqrt {1+p^{2}}}{x+\frac {1}{2}x\frac {2ap}{\sqrt {1+p^{2}}}} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {x\left ( 1+\frac {1}{2}\frac {2ap}{\sqrt {1+p^{2}}}\right ) }{-a\sqrt {1+p^{2}}}\\ \frac {dx}{x} & =\frac {1+\frac {1}{2}\frac {2ap}{\sqrt {1+p^{2}}}}{-a\sqrt {1+p^{2}}}dp\\ \frac {dx}{x} & =\frac {\sqrt {1+p^{2}}+\frac {1}{2}2ap}{-a\left ( 1+p^{2}\right ) }dp\\ \frac {dx}{x} & =\left ( -\frac {1}{a\sqrt {1+p^{2}}}-\frac {p}{\left ( 1+p^{2}\right ) }\right ) dp \end {align*}

Integrating gives\[ \ln x\left ( p\right ) =-\frac {1}{2}\ln \left ( p^{2}+1\right ) -\frac {1}{a}\operatorname {arcsinh}\left ( p\right ) \] Therefore\begin {equation} x=c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( p\right ) \right ) }}{\sqrt {p^{2}+1}}\tag {3} \end {equation} There are now two choices to take. The first is by solving for \(p\) from the above in terms of \(x\) and substituting the result in (1) to obtain explicit solution for \(y\left ( x\right ) \), and the second choice is by solving for \(p\) algebraically from (1) and substituting the result in (3). The second choice is easier in this case but gives an implicit solution. Solving for \(p\) from (1) gives\begin {align*} p_{1} & =-\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\\ p_{2} & =\frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1} \end {align*}

Substituting each one of these solutions back in (3) gives two implicit solutions\begin {align*} x & =c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( -\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt {\left ( -\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}\\ x & =c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( \frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt {\left ( \frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}} \end {align*}

2.11 Example 11

\begin {align*} y & =x+\left ( y^{\prime }\right ) ^{2}\left ( 1-\frac {2}{3}y^{\prime }\right ) \\ & =\phi \left ( x,y^{\prime }\right ) \end {align*}

Let \(y^{\prime }=p\)\begin {align} y & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \tag {1}\\ & =\phi \left ( x,p\right ) \nonumber \end {align}

Differentiating w.r.t. \(x\) gives\begin {align} y^{\prime } & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =1+\left ( 2p-2p^{2}\right ) \frac {dp}{dx}\nonumber \\ \frac {dp}{dx} & =\frac {p-1}{2\left ( p-p^{2}\right ) }\tag {2} \end {align}

The singular solution is when \(\frac {dp}{dx}=0\) which results in \(p=1\). Substituting this in (1) gives\begin {align*} y & =x-\left ( 1-\frac {2}{3}\right ) \\ & =x+\frac {1}{3} \end {align*}

The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2) is now separable. Solving for \(p\) gives\begin {align*} p & =-\sqrt {c_{1}-x}\\ p & =\sqrt {c_{1}-x} \end {align*}

Substituting each one of the above solutions of \(p\) in (1) gives \begin {align*} y_{1} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( -\sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( -\sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}

And\begin {align*} y_{2} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( \sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( \sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}

Therefore the solutions are\begin {align*} y & =x+\frac {1}{3}\\ y & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\\ y & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}} \end {align*}

2.12 Example 12

This ode is an example where \(y\) does not appear explicitly in the ode so not possible to directly solve for \(y\). It is given here to show possible problems with this method.\begin {equation} y^{\prime }=\sqrt {1+x+y} \tag {1A} \end {equation} This ode is squared to first solve for \(y\) which gives\begin {equation} \left ( y^{\prime }\right ) ^{2}=1+x+y \tag {2A} \end {equation} However, here care is needed. To get back to original ode (1A) then (2A) means two possible equations\[ y^{\prime }=\pm \sqrt {1+x+y}\] Hence the solutions obtained using (2A) can be the solution to one of these\begin {align} y^{\prime } & =+\sqrt {1+x+y}\tag {B1}\\ y^{\prime } & =-\sqrt {1+x+y} \tag {B2} \end {align}

Therefore the solution obtained by squaring both sides of (1A), which is done in order to solve for \(y\), must be checked to see if it satisfies the original ode, else it will be extraneous solution resulting from squaring both sides of the ode.

Starting from (2A), in normal form (by replacing \(y^{\prime }\) with \(p\)) it becomes\begin {align} y & =-x-1+p^{2}\tag {1}\\ & =\phi \left ( x,p\left ( x\right ) \right ) \nonumber \end {align}

Taking derivative w.r.t. \(x\) gives \begin {align} \frac {dy}{dx} & =\frac {d}{dx}\phi \left ( x,p\left ( x\right ) \right ) \nonumber \\ p & =\frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial p}\frac {dp}{dx}\nonumber \\ p & =-1+2p\frac {dp}{dx}\nonumber \\ p+1 & =2p\frac {dp}{dx} \tag {2} \end {align}

Since \(\frac {\partial \phi }{\partial x}=-1\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=-1\). Substituting this in (1) gives the singular solution\begin {equation} y\left ( x\right ) =-x \tag {3} \end {equation} But this solution does not satisfy the ode, hence it is extraneous. The general solution is found by finding \(p\) from (2). Since (2) is nonlinear, then it is inverted which gives\begin {align*} \frac {p+1}{2p} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {2p}{p+1} \end {align*}

Which is linear in \(x\). Solving gives\begin {equation} x=2p-2\ln \left ( p+1\right ) +c_{1} \tag {4} \end {equation} Instead of inverting this to find \(p\) in terms of \(x\), \(p\) is found from (1) which gives\begin {align*} y+x+1 & =p^{2}\\ p & =\pm \sqrt {y+x+1} \end {align*}

Substituting these solutions in (4) gives implicit solutions as\begin {align*} x & =2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\\ x & =-2\sqrt {y+x+1}-2\ln \left ( 1-\sqrt {y+x+1}\right ) +c_{1} \end {align*}

But only the first one above satisfies the ode. The second is extraneous. Therefore the final solution is\[ \fbox {$x=2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_1$}\] And no singular solutions exist. If instead of doing the above, \(p\) was found from (4) using inversion, then it will be\[ p=-\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1 \] Substituting this in (1) gives\[ y=-x-1+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1\right ) ^{2}\] But this general solution does not satisfy the original ode.  In general, it is best to avoid squaring both side of the ode in order to solve for \(y\) as this can generate extraneous solutions. Only use this method if the original ode is already given in the form where \(y\) shows explicitly.

3 references

  1. An elementary treatise on differential equations. By Abraham Cohen. 1906.
  2. Applied differential equations, N Curle. 1972
  3. Ordinary differential equations, LB Jones. 1976.
  4. Elementary differential equations, William Martin, Eric Reissner. second edition. 1961.
  5. Differentialgleichungen, by E. Kamke, page 30.
  6. Differential and integral calculus by N. Piskunov, Vol II