Internal problem ID [10080]
Internal file name [OUTPUT/9027_Monday_June_06_2022_06_14_39_AM_68881698/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1759 (book 6.168).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {\left (a y+b \right ) y^{\prime \prime }+c {y^{\prime }}^{2}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} \left (y a +b \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+c p \left (y \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {c p}{y a +b} \end {align*}
Where \(f(y)=-\frac {c}{y a +b}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -\frac {c}{y a +b} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {-\frac {c}{y a +b} \,d y}\\ \ln \left (p \right )&=-\frac {c \ln \left (y a +b \right )}{a}+c_{1}\\ p&={\mathrm e}^{-\frac {c \ln \left (y a +b \right )}{a}+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {c \ln \left (y a +b \right )}{a}} \end {align*}
Which simplifies to \[ p \left (y \right ) = c_{1} \left (y a +b \right )^{-\frac {c}{a}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = c_{1} \left (a y+b \right )^{-\frac {c}{a}} \end {align*}
Integrating both sides gives \begin{align*} \int \frac {\left (y a +b \right )^{\frac {c}{a}}}{c_{1}}d y &= \int d x \\ \frac {\left (a y+b \right )^{1+\frac {c}{a}}}{c_{1} \left (a +c \right )}&=x +c_{2} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-b +{\mathrm e}^{\frac {\ln \left (a c_{1} c_{2} +a c_{1} x +c c_{1} c_{2} +c c_{1} x \right ) a}{a +c}}}{a} \\ \end{align*}
Verification of solutions
\[ y = \frac {-b +{\mathrm e}^{\frac {\ln \left (a c_{1} c_{2} +a c_{1} x +c c_{1} c_{2} +c c_{1} x \right ) a}{a +c}}}{a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a y+b \right ) y^{\prime \prime }+c {y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (y a +b \right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+c u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {c u \left (y \right )}{y a +b} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=-\frac {c}{y a +b} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int -\frac {c}{y a +b}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=-\frac {c \ln \left (y a +b \right )}{a}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{-\frac {c \ln \left (y a +b \right )-c_{1} a}{a}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{-\frac {c \ln \left (y a +b \right )-c_{1} a}{a}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{-\frac {c \ln \left (a y+b \right )-c_{1} a}{a}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{-\frac {c \ln \left (a y+b \right )-c_{1} a}{a}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\mathrm e}^{-\frac {c \ln \left (a y+b \right )-c_{1} a}{a}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\mathrm e}^{-\frac {c \ln \left (a y+b \right )-c_{1} a}{a}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {a^{2} y+a b}{{\mathrm e}^{-\frac {c \ln \left (a y+b \right )-c_{1} a}{a}} \left (a +c \right ) a}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {c_{2} {\mathrm e}^{\frac {c_{1} a +\frac {c \left (\ln \left (\frac {1}{\left (a +c \right ) \left (x +c_{2} \right )}\right )-c_{1} \right )}{1+\frac {c}{a}}}{a}} a +c_{2} {\mathrm e}^{\frac {c_{1} a +\frac {c \left (\ln \left (\frac {1}{\left (a +c \right ) \left (x +c_{2} \right )}\right )-c_{1} \right )}{1+\frac {c}{a}}}{a}} c +x \,{\mathrm e}^{\frac {c_{1} a +\frac {c \left (\ln \left (\frac {1}{\left (a +c \right ) \left (x +c_{2} \right )}\right )-c_{1} \right )}{1+\frac {c}{a}}}{a}} a +x \,{\mathrm e}^{\frac {c_{1} a +\frac {c \left (\ln \left (\frac {1}{\left (a +c \right ) \left (x +c_{2} \right )}\right )-c_{1} \right )}{1+\frac {c}{a}}}{a}} c -b}{a} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 73
dsolve((a*y(x)+b)*diff(diff(y(x),x),x)+c*diff(y(x),x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {b}{a} \\ y \left (x \right ) &= \frac {\left (-\left (\frac {1}{\left (a +c \right ) \left (c_{1} x +c_{2} \right )}\right )^{-\frac {c}{a +c}} b +\left (a +c \right ) \left (c_{1} x +c_{2} \right )\right ) \left (\frac {1}{\left (a +c \right ) \left (c_{1} x +c_{2} \right )}\right )^{\frac {c}{a +c}}}{a} \\ \end{align*}
✓ Solution by Mathematica
Time used: 16.845 (sec). Leaf size: 31
DSolve[c*y'[x]^2 + (b + a*y[x])*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {-b+(c_1 (a+c) (x+c_2)){}^{\frac {a}{a+c}}}{a} \]