Internal problem ID [4138]
Internal file name [OUTPUT/3631_Sunday_June_05_2022_09_51_59_AM_94456927/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 31
Problem number: 902.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "dAlembert"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _dAlembert]
\[ \boxed {x^{2} {y^{\prime }}^{2}-x \left (x -2 y\right ) y^{\prime }+y^{2}=0} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} x^{2} p^{2}-x \left (x -2 y \right ) p +y^{2} = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \left (-p +\sqrt {p}\right ) x\tag {1A}\\ y &= \left (-p -\sqrt {p}\right ) x\tag {2A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved. Solving ode 1A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= -p +\sqrt {p}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} 2 p -\sqrt {p} = x \left (-1+\frac {1}{2 \sqrt {p}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} 2 p -\sqrt {p} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=0\\ p&={\frac {1}{4}} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=0\\ y&=\frac {x}{4} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {2 p \left (x \right )-\sqrt {p \left (x \right )}}{x \left (-1+\frac {1}{2 \sqrt {p \left (x \right )}}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {2 \left (2 p -\sqrt {p}\right ) \sqrt {p}}{x \left (2 \sqrt {p}-1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(p)=\frac {\left (2 p -\sqrt {p}\right ) \sqrt {p}}{2 \sqrt {p}-1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (2 p -\sqrt {p}\right ) \sqrt {p}}{2 \sqrt {p}-1}} \,dp &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {\left (2 p -\sqrt {p}\right ) \sqrt {p}}{2 \sqrt {p}-1}} \,dp} &= \int {-\frac {2}{x} \,d x} \\ \ln \left (p \right )&=-2 \ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} p &= {\mathrm e}^{-2 \ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} p &= \frac {c_{2}}{x^{2}} \end {align*}
Which simplifies to \[ p \left (x \right ) = \frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}} \] Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = \left (-\frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}}+\sqrt {\frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}}}\right ) x\\ \end {align*}
Solving ode 2A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= -p -\sqrt {p}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} 2 p +\sqrt {p} = x \left (-1-\frac {1}{2 \sqrt {p}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} 2 p +\sqrt {p} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=0 \end {align*}
Substituting these in (1A) gives \begin {align*} y&=0 \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {2 p \left (x \right )+\sqrt {p \left (x \right )}}{x \left (-1-\frac {1}{2 \sqrt {p \left (x \right )}}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {2 \sqrt {p}\, \left (2 p +\sqrt {p}\right )}{x \left (2 \sqrt {p}+1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(p)=\frac {\sqrt {p}\, \left (2 p +\sqrt {p}\right )}{2 \sqrt {p}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\sqrt {p}\, \left (2 p +\sqrt {p}\right )}{2 \sqrt {p}+1}} \,dp &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {\sqrt {p}\, \left (2 p +\sqrt {p}\right )}{2 \sqrt {p}+1}} \,dp} &= \int {-\frac {2}{x} \,d x} \\ \ln \left (p \right )&=-2 \ln \left (x \right )+c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} p &= {\mathrm e}^{-2 \ln \left (x \right )+c_{3}} \end {align*}
Which simplifies to \begin {align*} p &= \frac {c_{4}}{x^{2}} \end {align*}
Which simplifies to \[ p \left (x \right ) = \frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}} \] Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = \left (-\frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}}-\sqrt {\frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}}}\right ) x\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \tag{2} y &= \frac {x}{4} \\ \tag{3} y &= \left (-\frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}}+\sqrt {\frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}}}\right ) x \\ \tag{4} y &= 0 \\ \tag{5} y &= \left (-\frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}}-\sqrt {\frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}}}\right ) x \\ \end{align*}
Verification of solutions
\[ y = 0 \] Verified OK.
\[ y = \frac {x}{4} \] Verified OK.
\[ y = \left (-\frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}}+\sqrt {\frac {c_{2} {\mathrm e}^{c_{1}}}{x^{2}}}\right ) x \] Verified OK.
\[ y = 0 \] Verified OK.
\[ y = \left (-\frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}}-\sqrt {\frac {c_{4} {\mathrm e}^{c_{3}}}{x^{2}}}\right ) x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} {y^{\prime }}^{2}-x \left (x -2 y\right ) y^{\prime }+y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {\frac {x}{2}-y-\frac {\sqrt {-4 x y+x^{2}}}{2}}{x}, y^{\prime }=\frac {\frac {x}{2}-y+\frac {\sqrt {-4 x y+x^{2}}}{2}}{x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\frac {x}{2}-y-\frac {\sqrt {-4 x y+x^{2}}}{2}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\frac {x}{2}-y+\frac {\sqrt {-4 x y+x^{2}}}{2}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: *** Sublevel 2 *** 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 simple symmetries for implicit equations <- symmetries for implicit equations successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 32
dsolve(x^2*diff(y(x),x)^2-x*(x-2*y(x))*diff(y(x),x)+y(x)^2 = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x}{4} \\ y \left (x \right ) &= \frac {c_{1} \left (-c_{1} +x \right )}{x} \\ y \left (x \right ) &= -\frac {c_{1} \left (c_{1} +x \right )}{x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.217 (sec). Leaf size: 64
DSolve[x^2 (y'[x])^2-x(x-2 y[x])y'[x]+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {e^{-4 c_1}-2 i e^{-2 c_1} x}{4 x} \\ y(x)\to \frac {2 i e^{-2 c_1} x+e^{-4 c_1}}{4 x} \\ y(x)\to 0 \\ \end{align*}