2.3.21 Problem 21

Internal problem ID [8879]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 21
Date solved : Wednesday, March 05, 2025 at 07:00:30 AM
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: 0.723 (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*}

\[ -\arctan \left (u \left (x \right )\right )=\arctan \left (x \right )+c_1 \]

In summary, these are the solution found for \(u(x)\)

\begin{align*} -\arctan \left (u \left (x \right )\right ) &= \arctan \left (x \right )+c_1 \\ \end{align*}

For solution \(-\arctan \left (u \left (x \right )\right ) = \arctan \left (x \right )+c_1\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} -\arctan \left (y^{\prime }\right ) = \arctan \left (x \right )+c_1 \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_2 \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_2 +\left (i x +c_2 \right ) {\mathrm e}^{4 i c_1}-i x +c_2}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \end{align*}

In summary, these are the solution found for \((y)\)

\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_2 +\left (i x +c_2 \right ) {\mathrm e}^{4 i c_1}-i x +c_2}{\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_2 +\left (i x +c_2 \right ) {\mathrm e}^{4 i c_1}-i x +c_2}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \\ \end{align*}

✓ Maple. Time used: 0.018 (sec). Leaf size: 33

ode:=(x^2+1)*diff(diff(y(x),x),x)+1+diff(y(x),x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {\ln \left (c_{1} x -1\right ) c_{1}^{2}+c_{2} c_{1}^{2}+c_{1} x +\ln \left (c_{1} x -1\right )}{c_{1}^{2}} \]

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 step by step

\[ \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} \]

✓ Mathematica. Time used: 7.826 (sec). Leaf size: 33

ode=(1+x^2)*D[y[x],{x,2}]+1+(D[y[x],x])^2==0; 
ic={}; 
DSolve[{ode,ic},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. Time used: 1.689 (sec). Leaf size: 24

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)
 

\[ \left [ y{\left (x \right )} = C_{1} + \int \tan {\left (C_{2} - \operatorname {atan}{\left (x \right )} \right )}\, dx, \ y{\left (x \right )} = C_{1} + \int \tan {\left (C_{2} - \operatorname {atan}{\left (x \right )} \right )}\, dx\right ] \]