Internal problem ID [4322]
Internal file name [OUTPUT/3815_Sunday_June_05_2022_11_08_03_AM_38415039/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 37
Problem number: 1123.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "clairaut"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries], _Clairaut]
\[ \boxed {\sqrt {a^{2}+b^{2} {y^{\prime }}^{2}}+y^{\prime } x -y=0} \]
This is Clairaut ODE. It has the form \[ y=y^{\prime } x+g\left (y^{\prime }\right ) \] Where \(g\) is function of \(y'(x)\). Let \(p=y^{\prime }\) the ode becomes \begin {align*} \sqrt {b^{2} p^{2}+a^{2}}+p x -y = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= p x +\sqrt {b^{2} p^{2}+a^{2}}\tag {1A} \end {align*}
The above ode is a Clairaut ode which is now solved. We start by replacing \(y^{\prime }\) by \(p\) which gives \begin {align*} y&=p x +\sqrt {b^{2} p^{2}+a^{2}}\\ &=p x +\sqrt {b^{2} p^{2}+a^{2}} \end {align*}
Writing the ode as \begin {align*} y&= p x +g \left (p \right ) \end {align*}
We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes \begin {align*} y = p x +g\tag {1} \end {align*}
Then we see that \begin {align*} g&=\sqrt {b^{2} p^{2}+a^{2}} \end {align*}
Taking derivative of (1) w.r.t. \(x\) gives \begin {align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \end {align*}
Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dx} & =0\\ p &=c_{1} \end {align*}
Substituting this in (1) gives the general solution as \begin {align*} y = c_{1} x +\sqrt {b^{2} c_{1}^{2}+a^{2}} \end {align*}
The singular solution is found from solving for \(p\) from \begin {align*} x+g'\left ( p\right ) &=0 \end {align*}
And substituting the result back in (1). Since we found above that \(g=\sqrt {b^{2} p^{2}+a^{2}}\), then the above equation becomes \begin {align*} x+g'\left ( p\right ) &= x +\frac {b^{2} p}{\sqrt {b^{2} p^{2}+a^{2}}}\\ &= 0 \end {align*}
Solving the above for \(p\) results in \begin {align*} p_{1} &=\frac {x a}{\sqrt {b^{2}-x^{2}}\, b}\\ p_{2} &=-\frac {x a}{\sqrt {b^{2}-x^{2}}\, b} \end {align*}
Substituting the above back in (1) results in \begin {align*} y_{1} &=\frac {\sqrt {\frac {a^{2} b^{2}}{b^{2}-x^{2}}}\, \sqrt {b^{2}-x^{2}}\, b +a \,x^{2}}{\sqrt {b^{2}-x^{2}}\, b}\\ y_{2} &=-\frac {a \,x^{2}-\sqrt {\frac {a^{2} b^{2}}{b^{2}-x^{2}}}\, \sqrt {b^{2}-x^{2}}\, b}{\sqrt {b^{2}-x^{2}}\, b} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +\sqrt {b^{2} c_{1}^{2}+a^{2}} \\ \tag{2} y &= \frac {\sqrt {\frac {a^{2} b^{2}}{b^{2}-x^{2}}}\, \sqrt {b^{2}-x^{2}}\, b +a \,x^{2}}{\sqrt {b^{2}-x^{2}}\, b} \\ \tag{3} y &= -\frac {a \,x^{2}-\sqrt {\frac {a^{2} b^{2}}{b^{2}-x^{2}}}\, \sqrt {b^{2}-x^{2}}\, b}{\sqrt {b^{2}-x^{2}}\, b} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +\sqrt {b^{2} c_{1}^{2}+a^{2}} \] Verified OK.
\[ y = \frac {\sqrt {\frac {a^{2} b^{2}}{b^{2}-x^{2}}}\, \sqrt {b^{2}-x^{2}}\, b +a \,x^{2}}{\sqrt {b^{2}-x^{2}}\, b} \] Verified OK.
\[ y = -\frac {a \,x^{2}-\sqrt {\frac {a^{2} b^{2}}{b^{2}-x^{2}}}\, \sqrt {b^{2}-x^{2}}\, b}{\sqrt {b^{2}-x^{2}}\, b} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sqrt {a^{2}+b^{2} {y^{\prime }}^{2}}+y^{\prime } x -y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {y x -\sqrt {b^{2} y^{2}-a^{2} b^{2}+a^{2} x^{2}}}{b^{2}-x^{2}}, y^{\prime }=-\frac {y x +\sqrt {b^{2} y^{2}-a^{2} b^{2}+a^{2} x^{2}}}{b^{2}-x^{2}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y x -\sqrt {b^{2} y^{2}-a^{2} b^{2}+a^{2} x^{2}}}{b^{2}-x^{2}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y x +\sqrt {b^{2} y^{2}-a^{2} b^{2}+a^{2} x^{2}}}{b^{2}-x^{2}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying dAlembert <- dAlembert successful`
✓ Solution by Maple
Time used: 0.312 (sec). Leaf size: 21
dsolve(sqrt(a^2+b^2*diff(y(x),x)^2)+x*diff(y(x),x)-y(x) = 0,y(x), singsol=all)
\[ y \left (x \right ) = \sqrt {b^{2} c_{1}^{2}+a^{2}}+c_{1} x \]
✓ Solution by Mathematica
Time used: 0.383 (sec). Leaf size: 37
DSolve[Sqrt[a^2+b^2 (y'[x])^2] +x y'[x] -y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \sqrt {a^2+b^2 c_1{}^2}+c_1 x \\ y(x)\to \sqrt {a^2} \\ \end{align*}