Internal problem ID [11343]
Internal file name [OUTPUT/10330_Friday_January_27_2023_02_37_06_AM_1081674/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter IX, Miscellaneous methods for solving equations of higher order than first.
Article 62. Summary. Page 144
Problem number: Ex 9.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }+{y^{\prime }}^{2}=-1} \]
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*} 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. Integrating both sides gives \begin {align*} \int \frac {1}{-p^{2}-1}d p &= x +c_{1}\\ -\arctan \left (p \right )&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=-\tan \left (x +c_{1} \right ) \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\tan \left (x +c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\tan \left (x +c_{1} \right )\,\mathop {\mathrm {d}x}}\\ &= -\frac {\ln \left (1+\tan \left (x +c_{1} \right )^{2}\right )}{2}+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (1+\tan \left (x +c_{1} \right )^{2}\right )}{2}+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\ln \left (1+\tan \left (x +c_{1} \right )^{2}\right )}{2}+c_{2} \] Verified OK.
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 \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{2} = -1 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int -\frac {p}{p^{2}+1}d p &= y +c_{1}\\ -\frac {\ln \left (p^{2}+1\right )}{2}&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\sqrt {-1+{\mathrm e}^{-2 y -2 c_{1}}}\\ &=\sqrt {-1+\frac {{\mathrm e}^{-2 y}}{c_{1}^{2}}}\\ p_2&=-\sqrt {-1+{\mathrm e}^{-2 y -2 c_{1}}}\\ &=-\sqrt {-1+\frac {{\mathrm e}^{-2 y}}{c_{1}^{2}}} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = \sqrt {-1+\frac {{\mathrm e}^{-2 y}}{c_{1}^{2}}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {-1+\frac {{\mathrm e}^{-2 y}}{c_{1}^{2}}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {-1+\frac {{\mathrm e}^{-2 \textit {\_a}}}{c_{1}^{2}}}}d \textit {\_a}&= x +c_{2} \end {align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = -\sqrt {-1+\frac {{\mathrm e}^{-2 y}}{c_{1}^{2}}} \end {align*}
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {-1+\frac {{\mathrm e}^{-2 y}}{c_{1}^{2}}}}d y &= \int d x \\ -\frac {{\mathrm e}^{-y} \sqrt {-\frac {{\mathrm e}^{2 y} c_{1}^{2}-1}{c_{1}^{2}}}\, \arctan \left (\frac {{\mathrm e}^{y}}{\sqrt {-\frac {{\mathrm e}^{2 y} c_{1}^{2}-1}{c_{1}^{2}}}}\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 y} c_{1}^{2}-1\right ) {\mathrm e}^{-2 y}}{c_{1}^{2}}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {-1+\frac {{\mathrm e}^{-2 \textit {\_a}}}{c_{1}^{2}}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} -\frac {{\mathrm e}^{-y} \sqrt {-\frac {{\mathrm e}^{2 y} c_{1}^{2}-1}{c_{1}^{2}}}\, \arctan \left (\frac {{\mathrm e}^{y}}{\sqrt {-\frac {{\mathrm e}^{2 y} c_{1}^{2}-1}{c_{1}^{2}}}}\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 y} c_{1}^{2}-1\right ) {\mathrm e}^{-2 y}}{c_{1}^{2}}}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {-1+\frac {{\mathrm e}^{-2 \textit {\_a}}}{c_{1}^{2}}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ -\frac {{\mathrm e}^{-y} \sqrt {-\frac {{\mathrm e}^{2 y} c_{1}^{2}-1}{c_{1}^{2}}}\, \arctan \left (\frac {{\mathrm e}^{y}}{\sqrt {-\frac {{\mathrm e}^{2 y} c_{1}^{2}-1}{c_{1}^{2}}}}\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 y} c_{1}^{2}-1\right ) {\mathrm e}^{-2 y}}{c_{1}^{2}}}} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+{y^{\prime }}^{2}=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+u \left (x \right )^{2}=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-u \left (x \right )^{2}-1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{-u \left (x \right )^{2}-1}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{-u \left (x \right )^{2}-1}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (u \left (x \right )\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\tan \left (x +c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\tan \left (x +c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\ln \left (1+\tan \left (x +c_{1} \right )^{2}\right )}{2}+c_{2} \end {array} \]
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 a quadrature checking if the LODE has constant coefficients <- constant coefficients successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 15
dsolve(diff(y(x),x$2)+diff(y(x),x)^2+1=0,y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (-c_{1} \sin \left (x \right )+c_{2} \cos \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 3.113 (sec). Leaf size: 16
DSolve[y''[x]+y'[x]^2+1==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \log (\cos (x-c_1))+c_2 \]