Internal problem ID [7212]
Internal file name [OUTPUT/6198_Sunday_June_05_2022_04_27_49_PM_82611728/index.tex]
Book: Own collection of miscellaneous problems
Section: section 3.0
Problem number: 22.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }+{y^{\prime }}^{2}=x -1} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (x^{2}+1\right ) p^{\prime }\left (x \right )+1+p \left (x \right )^{2}-x = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= -\frac {p^{2}-x +1}{x^{2}+1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ p' = -\frac {p^{2}}{x^{2}+1}+\frac {x}{x^{2}+1}-\frac {1}{x^{2}+1} \] With Riccati ODE standard form \[ p' = f_0(x)+ f_1(x)p+f_2(x)p^{2} \] Shows that \(f_0(x)=-\frac {1-x}{x^{2}+1}\), \(f_1(x)=0\) and \(f_2(x)=-\frac {1}{x^{2}+1}\). Let \begin {align*} p &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{x^{2}+1}} \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' &=\frac {2 x}{\left (x^{2}+1\right )^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-\frac {1-x}{\left (x^{2}+1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {u^{\prime \prime }\left (x \right )}{x^{2}+1}-\frac {2 x u^{\prime }\left (x \right )}{\left (x^{2}+1\right )^{2}}-\frac {\left (1-x \right ) u \left (x \right )}{\left (x^{2}+1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{1} +\operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{\frac {\sqrt {1+i}}{2}} c_{2} \right ) \left (x -i\right )^{\frac {\sqrt {1-i}}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (x -i\right )^{-1+\frac {\sqrt {1-i}}{2}} \left (\left (i \sqrt {2}+i \sqrt {1-i}+i \sqrt {1+i}+1+i\right ) \sqrt {-2+2 \sqrt {2}}\, \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{1} \left (x^{2}+1\right ) \operatorname {hypergeom}\left (\left [1-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+2-\frac {\sqrt {1+i}}{2}\right ], \left [2-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )-4 \left (\left (-x +i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{1} \operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )-4 \left (x +i\right )^{\frac {\sqrt {1+i}}{2}} c_{2} \left (-\frac {\left (x^{2}+1\right ) \left (-1+\sqrt {1+i}\right ) \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}+1, \frac {\sqrt {2+2 \sqrt {2}}}{2}+2\right ], \left [2+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )}{2}+\left (\left (x -i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )\right )\right )}{8 x +8 i} \] Using the above in (1) gives the solution \[ p \left (x \right ) = -\frac {\left (x -i\right )^{-1+\frac {\sqrt {1-i}}{2}} \left (\left (i \sqrt {2}+i \sqrt {1-i}+i \sqrt {1+i}+1+i\right ) \sqrt {-2+2 \sqrt {2}}\, \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{1} \left (x^{2}+1\right ) \operatorname {hypergeom}\left (\left [1-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+2-\frac {\sqrt {1+i}}{2}\right ], \left [2-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )-4 \left (\left (-x +i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{1} \operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )-4 \left (x +i\right )^{\frac {\sqrt {1+i}}{2}} c_{2} \left (-\frac {\left (x^{2}+1\right ) \left (-1+\sqrt {1+i}\right ) \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}+1, \frac {\sqrt {2+2 \sqrt {2}}}{2}+2\right ], \left [2+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )}{2}+\left (\left (x -i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )\right )\right ) \left (x^{2}+1\right ) \left (x -i\right )^{-\frac {\sqrt {1-i}}{2}}}{\left (8 x +8 i\right ) \left (\operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{1} +\operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{\frac {\sqrt {1+i}}{2}} c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ p \left (x \right ) = \frac {-\left (i \sqrt {2}+i \sqrt {1-i}+i \sqrt {1+i}+1+i\right ) \sqrt {-2+2 \sqrt {2}}\, \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{3} \left (x^{2}+1\right ) \operatorname {hypergeom}\left (\left [1-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+2-\frac {\sqrt {1+i}}{2}\right ], \left [2-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )+4 \left (\left (-x +i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{3} \operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )+4 \left (x +i\right )^{\frac {\sqrt {1+i}}{2}} \left (-\frac {\left (x^{2}+1\right ) \left (-1+\sqrt {1+i}\right ) \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}+1, \frac {\sqrt {2+2 \sqrt {2}}}{2}+2\right ], \left [2+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )}{2}+\left (\left (x -i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )\right )}{8 \operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{3} +8 \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{\frac {\sqrt {1+i}}{2}}} \] Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {-\left (i \sqrt {2}+i \sqrt {1-i}+i \sqrt {1+i}+1+i\right ) \sqrt {-2+2 \sqrt {2}}\, \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{3} \left (x^{2}+1\right ) \operatorname {hypergeom}\left (\left [1-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+2-\frac {\sqrt {1+i}}{2}\right ], \left [2-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )+4 \left (\left (-x +i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{3} \operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )+4 \left (x +i\right )^{\frac {\sqrt {1+i}}{2}} \left (-\frac {\left (x^{2}+1\right ) \left (-1+\sqrt {1+i}\right ) \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}+1, \frac {\sqrt {2+2 \sqrt {2}}}{2}+2\right ], \left [2+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )}{2}+\left (\left (x -i\right ) \sqrt {1+i}+\left (x +i\right ) \sqrt {1-i}\right ) \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right )\right )}{8 \operatorname {hypergeom}\left (\left [-\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {\sqrt {1-i}}{2}+1-\frac {\sqrt {1+i}}{2}\right ], \left [1-\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{-\frac {\sqrt {1+i}}{2}} c_{3} +8 \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [1+\sqrt {1+i}\right ], \frac {1}{2}-\frac {i x}{2}\right ) \left (x +i\right )^{\frac {\sqrt {1+i}}{2}}} \end {align*}
Integrating both sides gives \begin {align*} \text {Expression too large to display} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \text {Expression too large to display} \\ \end{align*}
Verification of solutions
\[ \text {Expression too large to display} \] 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)^2-_a+1)/(_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 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: <- Abel AIR successful: ODE belongs to the 2F1 2-parameter class <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 460
dsolve((1+x^2)*diff(y(x),x$2)+1+diff(y(x),x)^2=x,y(x), singsol=all)
\[ y \left (x \right ) = -\left (\int \frac {-\left (\frac {1}{2}-\frac {i x}{2}\right )^{\frac {i \sqrt {-2+2 \sqrt {2}}}{2}} \left (x +i\right ) \left (\frac {1}{2}+\frac {i x}{2}\right )^{i \sqrt {-1+i}} \sqrt {-1+i}\, \operatorname {hypergeom}\left (\left [\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {i \sqrt {-1+i}}{2}+\frac {\sqrt {1+i}}{2}+1\right ], \left [i \sqrt {-1+i}+1\right ], \frac {1}{2}+\frac {i x}{2}\right )-4 \left (\frac {1}{2}-\frac {i x}{2}\right )^{\frac {\sqrt {2+2 \sqrt {2}}}{2}} \left (-\frac {1}{2}+\frac {i x}{2}\right )^{i \sqrt {-1+i}} \left (x +i\right ) \sqrt {-1+i}\, c_{1} \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [-i \sqrt {-1+i}+1\right ], \frac {1}{2}+\frac {i x}{2}\right )-8 \left (\operatorname {HeunCPrime}\left (0, -i \sqrt {-1+i}, -1, 0, \frac {1}{2}-\frac {i}{2}, \frac {x -i}{x +i}\right ) c_{1} \left (-\frac {1}{2}+\frac {i x}{2}\right )^{i \sqrt {-1+i}}-\frac {\operatorname {HeunCPrime}\left (0, i \sqrt {-1+i}, -1, 0, \frac {1}{2}-\frac {i}{2}, \frac {x -i}{x +i}\right ) \left (\frac {1}{2}+\frac {i x}{2}\right )^{i \sqrt {-1+i}}}{4}\right ) \left (i x +1\right )}{\left (4 \left (\frac {1}{2}-\frac {i x}{2}\right )^{\frac {\sqrt {2+2 \sqrt {2}}}{2}} c_{1} \operatorname {hypergeom}\left (\left [\frac {\sqrt {2+2 \sqrt {2}}}{2}, \frac {\sqrt {2+2 \sqrt {2}}}{2}+1\right ], \left [-i \sqrt {-1+i}+1\right ], \frac {1}{2}+\frac {i x}{2}\right ) \left (-\frac {1}{2}+\frac {i x}{2}\right )^{i \sqrt {-1+i}}-\left (\frac {1}{2}-\frac {i x}{2}\right )^{\frac {i \sqrt {-2+2 \sqrt {2}}}{2}} \operatorname {hypergeom}\left (\left [\frac {i \sqrt {-2+2 \sqrt {2}}}{2}, \frac {i \sqrt {-1+i}}{2}+\frac {\sqrt {1+i}}{2}+1\right ], \left [i \sqrt {-1+i}+1\right ], \frac {1}{2}+\frac {i x}{2}\right ) \left (\frac {1}{2}+\frac {i x}{2}\right )^{i \sqrt {-1+i}}\right ) \left (x +i\right )}d x \right )+c_{2} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[(1+x^2)*y''[x]+1+(y'[x])^2==x,y[x],x,IncludeSingularSolutions -> True]
Not solved