Internal problem ID [8759]
Internal file name [OUTPUT/7693_Sunday_June_05_2022_11_41_36_PM_63945327/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 424.
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`], _dAlembert]
\[ \boxed {x {y^{\prime }}^{2}+a y y^{\prime }=-x b} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} a y p +x \,p^{2} = -x b \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= -\frac {x \left (p^{2}+b \right )}{a p}\tag {1A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved. 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 &= \frac {-p^{2}-b}{a p}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {-p^{2}-b}{a p} = x \left (-\frac {2}{a}-\frac {-p^{2}-b}{a \,p^{2}}\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*} p -\frac {-p^{2}-b}{a p} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=\frac {\sqrt {-\left (a +1\right ) b}}{a +1}\\ p&=-\frac {\sqrt {-\left (a +1\right ) b}}{a +1} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=-\frac {b x}{\sqrt {-a b -b}}\\ y&=\frac {b x}{\sqrt {-a b -b}} \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 {p \left (x \right )-\frac {-p \left (x \right )^{2}-b}{a p \left (x \right )}}{x \left (-\frac {2}{a}-\frac {-p \left (x \right )^{2}-b}{a p \left (x \right )^{2}}\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 {\left (\left (a +1\right ) p^{2}+b \right ) p}{x \left (-p^{2}+b \right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(p)=-\frac {\left (\left (a +1\right ) p^{2}+b \right ) p}{-p^{2}+b}\). Integrating both sides gives \begin{align*} \frac {1}{-\frac {\left (\left (a +1\right ) p^{2}+b \right ) p}{-p^{2}+b}} \,dp &= -\frac {1}{x} \,d x \\ \int { \frac {1}{-\frac {\left (\left (a +1\right ) p^{2}+b \right ) p}{-p^{2}+b}} \,dp} &= \int {-\frac {1}{x} \,d x} \\ \frac {\left (2+a \right ) \ln \left (a \,p^{2}+p^{2}+b \right )}{2+2 a}-\ln \left (p \right )&=-\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\frac {\left (2+a \right ) \ln \left (a \,p^{2}+p^{2}+b \right )}{2+2 a}-\ln \left (p \right )} &= {\mathrm e}^{-\ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {\left (\left (a +1\right ) p^{2}+b \right )^{\frac {2+a}{2+2 a}}}{p} &= \frac {c_{2}}{x} \end {align*}
Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = \frac {x \left (-{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+\textit {\_Z}^{2}+b \right )^{\frac {2+a}{2+2 a}} x +c_{2} \textit {\_Z} \right )}^{2}-b \right )}{a \operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+\textit {\_Z}^{2}+b \right )^{\frac {2+a}{2+2 a}} x +c_{2} \textit {\_Z} \right )}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {b x}{\sqrt {-a b -b}} \\ \tag{2} y &= \frac {b x}{\sqrt {-a b -b}} \\ \tag{3} y &= \frac {x \left (-{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+\textit {\_Z}^{2}+b \right )^{\frac {2+a}{2+2 a}} x +c_{2} \textit {\_Z} \right )}^{2}-b \right )}{a \operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+\textit {\_Z}^{2}+b \right )^{\frac {2+a}{2+2 a}} x +c_{2} \textit {\_Z} \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {b x}{\sqrt {-a b -b}} \] Verified OK.
\[ y = \frac {b x}{\sqrt {-a b -b}} \] Verified OK.
\[ y = \frac {x \left (-{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+\textit {\_Z}^{2}+b \right )^{\frac {2+a}{2+2 a}} x +c_{2} \textit {\_Z} \right )}^{2}-b \right )}{a \operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+\textit {\_Z}^{2}+b \right )^{\frac {2+a}{2+2 a}} x +c_{2} \textit {\_Z} \right )} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x {y^{\prime }}^{2}+a y y^{\prime }=-x b \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {y a -\sqrt {y^{2} a^{2}-4 x^{2} b}}{2 x}, y^{\prime }=-\frac {y a +\sqrt {y^{2} a^{2}-4 x^{2} b}}{2 x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y a -\sqrt {y^{2} a^{2}-4 x^{2} b}}{2 x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y a +\sqrt {y^{2} a^{2}-4 x^{2} b}}{2 x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
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: 0.109 (sec). Leaf size: 217
dsolve(x*diff(y(x),x)^2+a*y(x)*diff(y(x),x)+b*x = 0,y(x), singsol=all)
\begin{align*} \frac {-c_{1} 2^{\frac {a +2}{2 a +2}} \left (a y \left (x \right )-\sqrt {a^{2} y \left (x \right )^{2}-4 b \,x^{2}}\right ) {\left (\frac {\left (-y \left (x \right ) \left (a +1\right ) \sqrt {a^{2} y \left (x \right )^{2}-4 b \,x^{2}}+\left (a^{2}+a \right ) y \left (x \right )^{2}-2 b \,x^{2}\right ) a}{x^{2}}\right )}^{\frac {-a -2}{2 a +2}}+x^{2}}{x} &= 0 \\ \frac {c_{1} 2^{\frac {a +2}{2 a +2}} \left (a y \left (x \right )+\sqrt {a^{2} y \left (x \right )^{2}-4 b \,x^{2}}\right ) {\left (\frac {a \left (y \left (x \right ) \left (a +1\right ) \sqrt {a^{2} y \left (x \right )^{2}-4 b \,x^{2}}+\left (a^{2}+a \right ) y \left (x \right )^{2}-2 b \,x^{2}\right )}{x^{2}}\right )}^{\frac {-a -2}{2 a +2}}+x^{2}}{x} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.076 (sec). Leaf size: 423
DSolve[b*x + a*y[x]*y'[x] + x*y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} \text {Solve}\left [-\frac {i \left (2 \log \left (-i \sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}+\frac {a y(x)}{x}+2 i \sqrt {b}\right )+2 (a+1) \log \left (i \sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}+\frac {a y(x)}{x}-2 i \sqrt {b}\right )-(a+2) \log \left (\frac {i (a+2) y(x) \sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}}{x}+2 \sqrt {b} \left (\sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}-\frac {i (a+2) y(x)}{x}\right )+\frac {a^2 y(x)^2}{x^2}-4 b\right )\right )}{4 (a+1)}&=c_1-\frac {1}{2} i \log (x),y(x)\right ] \\ \text {Solve}\left [\frac {i \left (2 (a+1) \log \left (-i \sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}+\frac {a y(x)}{x}+2 i \sqrt {b}\right )+2 \log \left (i \sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}+\frac {a y(x)}{x}-2 i \sqrt {b}\right )-(a+2) \log \left (-\frac {i (a+2) y(x) \sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}}{x}+2 \sqrt {b} \left (\sqrt {4 b-\frac {a^2 y(x)^2}{x^2}}+\frac {i (a+2) y(x)}{x}\right )+\frac {a^2 y(x)^2}{x^2}-4 b\right )\right )}{4 (a+1)}&=\frac {1}{2} i \log (x)+c_1,y(x)\right ] \\ \end{align*}