2.3.21 Problem 21
Internal
problem
ID
[8879]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
3.0
Problem
number
:
21
Date
solved
:
Friday, February 21, 2025 at 08:43:42 PM
CAS
classification
:
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]
Solve
\begin{align*} \left (x^{2}+1\right ) y^{\prime \prime }+1+{y^{\prime }}^{2}&=0 \end{align*}
Solved as second order missing y ode
Time used: 0.680 (sec)
This is second order ode with missing dependent variable \(y\). Let
\begin{align*} u(x) &= y^{\prime } \end{align*}
Then
\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} \left (x^{2}+1\right ) u^{\prime }\left (x \right )+1+u \left (x \right )^{2} = 0 \end{align*}
Which is now solved for \(u(x)\) as first order ode.
The ode
\begin{equation}
u^{\prime }\left (x \right ) = -\frac {u \left (x \right )^{2}+1}{x^{2}+1}
\end{equation}
is separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )^{2}+1}{x^{2}+1}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x^{2}+1}\\ g(u) &= -u^{2}-1 \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\
\int { \frac {1}{-u^{2}-1}\,du} &= \int { \frac {1}{x^{2}+1} \,dx} \\
\end{align*}
\[
-\arctan \left (u \left (x \right )\right )=\arctan \left (x \right )+c_1
\]
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or
\[
-u^{2}-1=0
\]
for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=-i\\ u \left (x \right )&=i \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
Therefore the solutions found are
\begin{align*}
-\arctan \left (u \left (x \right )\right ) &= \arctan \left (x \right )+c_1 \\
u \left (x \right ) &= -i \\
u \left (x \right ) &= i \\
\end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= -i \\
u \left (x \right ) &= i \\
u \left (x \right ) &= -\tan \left (\arctan \left (x \right )+c_1 \right ) \\
\end{align*}
In summary, these are the solution found for \(u(x)\)
\begin{align*}
u \left (x \right ) &= -i \\
u \left (x \right ) &= i \\
u \left (x \right ) &= -\tan \left (\arctan \left (x \right )+c_1 \right ) \\
\end{align*}
For solution \(u \left (x \right ) = -i\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = -i \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-i\, dx}\\ y &= -i x + c_2 \end{align*}
For solution \(u \left (x \right ) = i\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = i \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {i\, dx}\\ y &= i x + c_3 \end{align*}
For solution \(u \left (x \right ) = -\tan \left (\arctan \left (x \right )+c_1 \right )\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = -\tan \left (\arctan \left (x \right )+c_1 \right ) \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-\tan \left (\arctan \left (x \right )+c_1 \right )\, dx}\\ y &= \frac {i {\mathrm e}^{4 i c_1} x}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}}-\frac {i x}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-{\mathrm e}^{2 i c_1}+1\right ) x +i {\mathrm e}^{2 i c_1}+i\right )}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} + c_4 \end{align*}
\begin{align*} y&= \frac {-4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-x +i\right ) {\mathrm e}^{2 i c_1}+i+x \right )-2 \,{\mathrm e}^{2 i c_1} c_4 +\left (i x +c_4 \right ) {\mathrm e}^{4 i c_1}-i x +c_4}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= -i x +c_2 \\
y &= i x +c_3 \\
y &= \frac {-4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-x +i\right ) {\mathrm e}^{2 i c_1}+i+x \right )-2 \,{\mathrm e}^{2 i c_1} c_4 +\left (i x +c_4 \right ) {\mathrm e}^{4 i c_1}-i x +c_4}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= \frac {-4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-x +i\right ) {\mathrm e}^{2 i c_1}+i+x \right )-2 \,{\mathrm e}^{2 i c_1} c_4 +\left (i x +c_4 \right ) {\mathrm e}^{4 i c_1}-i x +c_4}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \\
y &= -i x +c_2 \\
y &= i x +c_3 \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+1+\left (\frac {d}{d x}y \left (x \right )\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d}{d x}u \left (x \right )\right )+1+u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {-1-u \left (x \right )^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{-1-u \left (x \right )^{2}}=\frac {1}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{-1-u \left (x \right )^{2}}d x =\int \frac {1}{x^{2}+1}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (u \left (x \right )\right )=\arctan \left (x \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\tan \left (\arctan \left (x \right )+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int -\tan \left (\arctan \left (x \right )+\mathit {C1} \right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\mathrm {I} \,{\mathrm e}^{4 \,\mathrm {I} \mathit {C1}} x}{\left ({\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}-1\right )^{2}}-\frac {\mathrm {I} x}{\left ({\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} \mathit {C1}} \ln \left (\left (-{\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}+1\right ) x +\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}+\mathrm {I}\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}-1\right )^{2}}+\mathit {C2} \end {array} \]
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) = -(1+_b(_a)^2)/(_a^2+1), _b(_a)` *** Sublevel 2 ***
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
<- separable successful
<- differential order: 2; canonical coordinates successful
<- differential order 2; missing variables successful`
Maple dsolve solution
Solving time : 0.018 (sec)
Leaf size : 33
dsolve((x^2+1)*diff(diff(y(x),x),x)+1+diff(y(x),x)^2 = 0,y(x),singsol=all)
\[
y = \frac {\ln \left (c_{1} x -1\right ) c_{1}^{2}+c_{2} c_{1}^{2}+c_{1} x +\ln \left (c_{1} x -1\right )}{c_{1}^{2}}
\]
✓Mathematica DSolve solution
Solving time : 7.826 (sec)
Leaf size : 33
DSolve[{(1+x^2)*D[y[x],{x,2}]+1+(D[y[x],x])^2==0,{}},y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to -x \cot (c_1)+\csc ^2(c_1) \log (-x \sin (c_1)-\cos (c_1))+c_2
\]
✓Sympy solution
Solving time : 1.689 (sec)
Leaf size : 24
Python version: 3.13.1 (main, Dec 4 2024, 18:05:56) [GCC 14.2.1 20240910]
Sympy version 1.13.3
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x**2 + 1)*Derivative(y(x), (x, 2)) + Derivative(y(x), x)**2 + 1,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
[Eq(y(x), C1 + Integral(tan(C2 - atan(x)), x)), Eq(y(x), C1 +
Integral(tan(C2 - atan(x)), x))]
\[
\left [ y{\left (x \right )} = C_{1} + \int \tan {\left (C_{2} - \operatorname {atan}{\left (x \right )} \right )}\, dx, \ y{\left (x \right )} = C_{1} + \int \tan {\left (C_{2} - \operatorname {atan}{\left (x \right )} \right )}\, dx\right ]
\]