Internal problem ID [10001]
Internal file name [OUTPUT/8948_Monday_June_06_2022_06_02_04_AM_12336822/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1680 (book 6.89).
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], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }+{y^{\prime }}^{2}=-1} \]
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 )+p \left (x \right )^{2}+1 = 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)\\ &= f( x) g(p)\\ &= \frac {-p^{2}-1}{x^{2}+1} \end {align*}
Where \(f(x)=\frac {1}{x^{2}+1}\) and \(g(p)=-p^{2}-1\). Integrating both sides gives \begin{align*} \frac {1}{-p^{2}-1} \,dp &= \frac {1}{x^{2}+1} \,d x \\ \int { \frac {1}{-p^{2}-1} \,dp} &= \int {\frac {1}{x^{2}+1} \,d x} \\ -\arctan \left (p \right )&=\arctan \left (x \right )+c_{1} \\ \end{align*} The solution is \[ -\arctan \left (p \left (x \right )\right )-\arctan \left (x \right )-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) 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} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\tan \left (\arctan \left (x \right )+c_{1} \right )\,\mathop {\mathrm {d}x}}\\ &= \frac {i {\mathrm e}^{4 i c_{1}} 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}}-\frac {i x}{\left ({\mathrm e}^{2 i c_{1}}-1\right )^{2}}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {i {\mathrm e}^{4 i c_{1}} 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}}-\frac {i x}{\left ({\mathrm e}^{2 i c_{1}}-1\right )^{2}}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {i {\mathrm e}^{4 i c_{1}} x}{\left ({\mathrm e}^{2 i c_{1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 i c_{1}} \ln \left (\left (-{\mathrm e}^{2 i c_{1}}+1\right ) x +i {\mathrm e}^{2 i c_{1}}+i\right )}{\left ({\mathrm e}^{2 i c_{1}}-1\right )^{2}}-\frac {i x}{\left ({\mathrm e}^{2 i c_{1}}-1\right )^{2}}+c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )+{y^{\prime }}^{2}=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) u^{\prime }\left (x \right )+u \left (x \right )^{2}=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {-u \left (x \right )^{2}-1}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{-u \left (x \right )^{2}-1}=\frac {1}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{-u \left (x \right )^{2}-1}d x =\int \frac {1}{x^{2}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (u \left (x \right )\right )=\arctan \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\tan \left (\arctan \left (x \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\tan \left (\arctan \left (x \right )+c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\mathrm {I} \,{\mathrm e}^{4 \,\mathrm {I} c_{1}} x}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} c_{1}} \ln \left (\left (-{\mathrm e}^{2 \,\mathrm {I} c_{1}}+1\right ) x +\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} c_{1}}+\mathrm {I}\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {\mathrm {I} x}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}+c_{2} \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) = -(_b(_a)^2+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 <- separable successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 33
dsolve((x^2+1)*diff(diff(y(x),x),x)+diff(y(x),x)^2+1=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\ln \left (c_{1} x -1\right ) c_{1}^{2}+c_{2} c_{1}^{2}+c_{1} x +\ln \left (c_{1} x -1\right )}{c_{1}^{2}} \]
✓ Solution by Mathematica
Time used: 11.847 (sec). Leaf size: 33
DSolve[1 + y'[x]^2 + (1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -x \cot (c_1)+\csc ^2(c_1) \log (-x \sin (c_1)-\cos (c_1))+c_2 \]