Internal
problem
ID
[9084]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
13
Date
solved
:
Wednesday, March 05, 2025 at 07:19:21 AM
CAS
classification
:
[[_2nd_order, _missing_x]]
Solve
Time used: 1.829 (sec)
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
Then
Hence the ode becomes
Which is now solved as first order ode for \(p(y)\).
Factoring the ode gives these factors
Now each of the above equations is solved in turn.
Solving equation (1)
Solving for \(p\) from
Solving gives \(p = 0\)
Solving equation (2)
Solving for the derivative gives these ODE’s to solve
Now each of the above is solved separately.
Solving Eq. (1)
Integrating gives
Solving Eq. (2)
Integrating gives
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
Let \(p=y^{\prime }\) the ode becomes
Solving for \(y\) from the above results in
This has the form
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
Comparing the form \(y=x f + g\) to (1A) shows that
Hence (2) becomes
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
Solving the above for \(p\) results in
Substituting these in (1A) and keeping singular solution that verifies the ode gives
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
This ODE is now solved for \(p \left (x \right )\). No inversion is needed.
Integrating gives
Singular solutions are found by solving
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.
Substituing the above solution for \(p\) in (2A) gives
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
Let \(p=y^{\prime }\) the ode becomes
Solving for \(y\) from the above results in
This has the form
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
Comparing the form \(y=x f + g\) to (1A) shows that
Hence (2) becomes
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
Solving the above for \(p\) results in
Substituting these in (1A) and keeping singular solution that verifies the ode gives
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
This ODE is now solved for \(p \left (x \right )\). No inversion is needed.
Integrating gives
Singular solutions are found by solving
for \(p \left (x \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.
Substituing the above solution for \(p\) in (2A) gives
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
Will add steps showing solving for IC soon.
The solution
was found not to satisfy the ode or the IC. Hence it is removed.
Summary of solutions found
Time used: 0.405 (sec)
This is second order ode with missing dependent variable \(y\). Let
Then
Hence the ode becomes
Which is now solved for \(u(x)\) as first order ode.
Let \(p=u^{\prime }\left (x \right )\) the ode becomes
Solving for \(u \left (x \right )\) from the above results in
This has the form
Where \(f,g\) are functions of \(p=u'(x)\). The above ode is dAlembert ode which is now solved.
Taking derivative of (*) w.r.t. \(x\) gives
Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that
Hence (2) becomes
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives
Solving the above for \(p\) results in
Substituting these in (1A) and keeping singular solution that verifies the ode gives
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in
This ODE is now solved for \(p \left (x \right )\). No inversion is needed.
Since the ode has the form \(p^{\prime }\left (x \right )=f(x)\), then we only need to integrate \(f(x)\).
Substituing the above solution for \(p\) in (2A) gives
In summary, these are the solution found for \(u(x)\)
For solution \(u \left (x \right ) = 0\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
For solution \(u \left (x \right ) = -\frac {\left (-x +2 c_1 \right )^{2}}{4}\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
In summary, these are the solution found for \((y)\)
Will add steps showing solving for IC soon.
Summary of solutions found
ode:=diff(diff(y(x),x),x)^2+diff(y(x),x) = 0; dsolve(ode,y(x), singsol=all);
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** Methods for second order ODEs: Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 3 *** 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 -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)+1/2, y(x)` *** Sublevel 4 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful <- 2nd order ODE linearizable_by_differentiation successful ------------------- * Tackling next ODE. *** Sublevel 3 *** 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 <- 2nd order ODE linearizable_by_differentiation successful`
ode=(D[y[x],{x,2}])^2+D[y[x],x]==0; ic={}; DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
from sympy import * x = symbols("x") y = Function("y") ode = Eq(Derivative(y(x), x) + Derivative(y(x), (x, 2))**2,0) ics = {} dsolve(ode,func=y(x),ics=ics)