Internal problem ID [9948]
Internal file name [OUTPUT/8895_Monday_June_06_2022_05_47_21_AM_63181951/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1626 (6.36).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_y_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }+2 y^{\prime } y+f \left (x \right ) y^{\prime }+y f^{\prime }\left (x \right )=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+\left (2 y+f \left (x \right )\right ) y^{\prime }+y f^{\prime }\left (x \right )\right )d x &= 0 \\ f \left (x \right ) y+y^{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -f \left (x \right ) y -y^{2}+c_{1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -f \left (x \right ) y -y^{2}+c_{1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c_{1}\), \(f_1(x)=-f \left (x \right )\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=f \left (x \right )\\ f_2^2 f_0 &=c_{1} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-f \left (x \right ) u^{\prime }\left (x \right )+c_{1} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{2}\) gives, after renaming the constant \(\frac {c_{3}}{c_{2}}=c_{4}\) the following solution
\[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
Writing the ode as \[ y^{\prime \prime }+\left (2 y+f \left (x \right )\right ) y^{\prime }+y f^{\prime }\left (x \right ) = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+\left (2 y+f \left (x \right )\right ) y^{\prime }+y f^{\prime }\left (x \right )\right )d x &= 0 \\ f \left (x \right ) y+y^{2}+y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -f \left (x \right ) y -y^{2}+c_{1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -f \left (x \right ) y -y^{2}+c_{1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c_{1}\), \(f_1(x)=-f \left (x \right )\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=f \left (x \right )\\ f_2^2 f_0 &=c_{1} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-f \left (x \right ) u^{\prime }\left (x \right )+c_{1} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{2}\) gives, after renaming the constant \(\frac {c_{3}}{c_{2}}=c_{4}\) the following solution
\[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= 1\\ a_1 &= 2 y+f \left (x \right )\\ a_0 &= y f^{\prime }\left (x \right ) \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {1\,d y'} + \int {2 y+f \left (x \right )\,d y} + \int {y f^{\prime }\left (x \right )\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} y^{\prime }+y \left (f \left (x \right )+y\right )+f \left (x \right ) y = c_{1} \end {align*}
Which is now solved In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -2 f \left (x \right ) y -y^{2}+c_{1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -2 f \left (x \right ) y -y^{2}+c_{1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c_{1}\), \(f_1(x)=-2 f \left (x \right )\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=2 f \left (x \right )\\ f_2^2 f_0 &=c_{1} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-2 f \left (x \right ) u^{\prime }\left (x \right )+c_{1} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{2}\) gives, after renaming the constant \(\frac {c_{3}}{c_{2}}=c_{4}\) the following solution
\[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{-c_{1} \textit {\_Y} \left (x \right )+2 f \left (x \right ) \textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
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 -> trying 2nd order, dynamical_symmetries, fully reducible to Abel through one integrating factor of the form G(x,y)/(1+H(x,y)*y)^2 --- trying a change of variables {x -> y(x), y(x) -> x} and re-entering methods for dynamical symmetries --- -> trying 2nd order, dynamical_symmetries, fully reducible to Abel through one integrating factor of the form G(x,y)/(1+H(x,y)*y) trying 2nd order, integrating factors of the form mu(x,y)/(y)^n, only the singular cases trying symmetries linear in x and y(x) trying differential order: 2; exact nonlinear -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -f(_a)*_b(_a)-_b(_a)^2-c__1, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(diff(y(x), x))*f(x)-c__1*y(x), y(x)` *** Sublevel 3 *** Methods for second order ODEs: -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-(-y(x)^2+y(x)-f(x)*y(x)*x-c__1*x^2)/x, y(x), explicit` *** Sublevel 3 Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 6`
✗ Solution by Maple
dsolve(diff(diff(y(x),x),x)+2*y(x)*diff(y(x),x)+f(x)*diff(y(x),x)+diff(f(x),x)*y(x)=0,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y[x]*f'[x] + f[x]*y'[x] + 2*y[x]*y'[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
Not solved