Internal problem ID [10023]
Internal file name [OUTPUT/8970_Monday_June_06_2022_06_05_53_AM_67193416/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1702 (book 6.111).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime } y-{y^{\prime }}^{2}=1} \]
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 ) y -p \left (y \right )^{2} = 1 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {p^{2}+1}{p y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=\frac {p^{2}+1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}+1}{p}} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}+1}{p}} \,dp} &= \int {\frac {1}{y} \,d y} \\ \frac {\ln \left (p^{2}+1\right )}{2}&=\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}+1} &= {\mathrm e}^{\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}+1} &= c_{2} y \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}+1} = c_{2} {\mathrm e}^{c_{1}} y \] The solution is \[ \sqrt {p \left (y \right )^{2}+1} = c_{2} {\mathrm e}^{c_{1}} y \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {{y^{\prime }}^{2}+1} = c_{2} {\mathrm e}^{c_{1}} y \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}d y &= \int d x \\ \frac {\operatorname {arctanh}\left (\frac {\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2} y}\right )}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}d y &= \int d x \\ -\frac {\operatorname {arctanh}\left (\frac {\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2} y}\right )}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2}}&=x +c_{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\operatorname {arctanh}\left (\frac {\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2} y}\right )}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2}} &= x +c_{3} \\ \tag{2} -\frac {\operatorname {arctanh}\left (\frac {\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2} y}\right )}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2}} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \frac {\operatorname {arctanh}\left (\frac {\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2} y}\right )}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2}} = x +c_{3} \] Verified OK.
\[ -\frac {\operatorname {arctanh}\left (\frac {\sqrt {-1+c_{2}^{2} {\mathrm e}^{2 c_{1}} y^{2}}}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2} y}\right )}{\sqrt {{\mathrm e}^{2 c_{1}}}\, c_{2}} = x +c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) y-{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 {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right ) y -u \left (y \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )^{2}+1}{u \left (y \right ) y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{u \left (y \right )^{2}+1}=\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{u \left (y \right )^{2}+1}d y =\int \frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (u \left (y \right )^{2}+1\right )}{2}=\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}, u \left (y \right )=-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (\frac {\left ({\mathrm e}^{c_{1}}\right )^{2} y}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}\right )}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}\, \left (\left ({\mathrm e}^{c_{2} \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}+x \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}}\right )^{2}+1\right )}{2 \,{\mathrm e}^{c_{2} \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}+x \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}} \left ({\mathrm e}^{c_{1}}\right )^{2}} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (\frac {\left ({\mathrm e}^{c_{1}}\right )^{2} y}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}}+\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2} y^{2}-1}\right )}{\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}\, \left (\left ({\mathrm e}^{c_{2} \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}-x \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}}\right )^{2}+1\right )}{2 \,{\mathrm e}^{c_{2} \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}-x \sqrt {\left ({\mathrm e}^{c_{1}}\right )^{2}}} \left ({\mathrm e}^{c_{1}}\right )^{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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(_b(_a)^2+1)/_a = 0, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 55
dsolve(diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2-1=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {c_{1} \left ({\mathrm e}^{\frac {c_{2} +x}{c_{1}}}+{\mathrm e}^{\frac {-c_{2} -x}{c_{1}}}\right )}{2} \\ y \left (x \right ) &= \frac {c_{1} \left ({\mathrm e}^{\frac {c_{2} +x}{c_{1}}}+{\mathrm e}^{\frac {-c_{2} -x}{c_{1}}}\right )}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.197 (sec). Leaf size: 80
DSolve[-1 - y'[x]^2 + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {e^{-c_1} \tanh \left (e^{c_1} (x+c_2)\right )}{\sqrt {-\text {sech}^2\left (e^{c_1} (x+c_2)\right )}} \\ y(x)\to \frac {e^{-c_1} \tanh \left (e^{c_1} (x+c_2)\right )}{\sqrt {-\text {sech}^2\left (e^{c_1} (x+c_2)\right )}} \\ \end{align*}