Internal problem ID [9962]
Internal file name [OUTPUT/8909_Monday_June_06_2022_05_50_12_AM_67846755/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1640 (6.50).
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], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime }+a y {y^{\prime }}^{2}+b y=0} \]
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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+a y p \left (y \right )^{2}+b y = 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 {y \left (a \,p^{2}+b \right )}{p} \end {align*}
Where \(f(y)=-y\) and \(g(p)=\frac {a \,p^{2}+b}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {a \,p^{2}+b}{p}} \,dp &= -y \,d y \\ \int { \frac {1}{\frac {a \,p^{2}+b}{p}} \,dp} &= \int {-y \,d y} \\ \frac {\ln \left (a \,p^{2}+b \right )}{2 a}&=-\frac {y^{2}}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\frac {\ln \left (a \,p^{2}+b \right )}{2 a}} &= {\mathrm e}^{-\frac {y^{2}}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \left (a \,p^{2}+b \right )^{\frac {1}{2 a}} &= c_{2} {\mathrm e}^{-\frac {y^{2}}{2}} \end {align*}
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 } = \operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+b \right )^{\frac {1}{2 a}}+c_{2} {\mathrm e}^{-\frac {y^{2}}{2}+c_{1}}\right ) \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+b \right )^{\frac {1}{2 a}}+c_{2} {\mathrm e}^{-\frac {y^{2}}{2}+c_{1}}\right )}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+b \right )^{\frac {1}{2 a}}+c_{2} {\mathrm e}^{-\frac {\textit {\_a}^{2}}{2}+c_{1}}\right )}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+b \right )^{\frac {1}{2 a}}+c_{2} {\mathrm e}^{-\frac {\textit {\_a}^{2}}{2}+c_{1}}\right )}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (-\left (a \,\textit {\_Z}^{2}+b \right )^{\frac {1}{2 a}}+c_{2} {\mathrm e}^{-\frac {\textit {\_a}^{2}}{2}+c_{1}}\right )}d \textit {\_a} = x +c_{3} \] Warning, solution could not be verified
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+a*_a*_b(_a)^2+_a*b = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 70
dsolve(diff(diff(y(x),x),x)+a*y(x)*diff(y(x),x)^2+b*y(x)=0,y(x), singsol=all)
\begin{align*} a \left (\int _{}^{y \left (x \right )}\frac {1}{\sqrt {a \left ({\mathrm e}^{-a \,\textit {\_a}^{2}} c_{1} a -b \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -a \left (\int _{}^{y \left (x \right )}\frac {1}{\sqrt {a \left ({\mathrm e}^{-a \,\textit {\_a}^{2}} c_{1} a -b \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.909 (sec). Leaf size: 290
DSolve[b*y[x] + a*y[x]*y'[x]^2 + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\sqrt {a}}{\sqrt {e^{2 a c_1-a K[1]^2}-b}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\sqrt {a}}{\sqrt {e^{2 a c_1-a K[2]^2}-b}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\sqrt {a}}{\sqrt {e^{2 a (-c_1)-a K[1]^2}-b}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\sqrt {a}}{\sqrt {e^{2 a c_1-a K[1]^2}-b}}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\sqrt {a}}{\sqrt {e^{2 a (-c_1)-a K[2]^2}-b}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\sqrt {a}}{\sqrt {e^{2 a c_1-a K[2]^2}-b}}dK[2]\&\right ][x+c_2] \\ \end{align*}