Internal problem ID [2281]
Internal file name [OUTPUT/2281_Tuesday_February_27_2024_08_23_59_AM_24740622/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 35, page 157
Problem number: 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]]
\[ \boxed {y^{\prime \prime }-{y^{\prime }}^{3}-y^{\prime }=0} \]
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 )+\left (-p \left (x \right )^{2}-1\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{p \left (p^{2}+1\right )}d p &= x +c_{1}\\ \ln \left (p \right )-\frac {\ln \left (p^{2}+1\right )}{2}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\frac {\sqrt {-\left ({\mathrm e}^{2 x +2 c_{1}}-1\right ) {\mathrm e}^{2 x +2 c_{1}}}}{{\mathrm e}^{2 x +2 c_{1}}-1}\\ &=\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}}{{\mathrm e}^{2 x} c_{1}^{2}-1}\\ p_2&=-\frac {\sqrt {-\left ({\mathrm e}^{2 x +2 c_{1}}-1\right ) {\mathrm e}^{2 x +2 c_{1}}}}{{\mathrm e}^{2 x +2 c_{1}}-1}\\ &=-\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}}{{\mathrm e}^{2 x} c_{1}^{2}-1} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}}{{\mathrm e}^{2 x} c_{1}^{2}-1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}}{{\mathrm e}^{2 x} c_{1}^{2}-1}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}\, c_{1}^{2} \arctan \left (\frac {\sqrt {c_{1}^{4}}\, {\mathrm e}^{x}}{\sqrt {-\left (c_{1} {\mathrm e}^{x}-1\right ) \left (1+c_{1} {\mathrm e}^{x}\right ) c_{1}^{2}}}\right ) {\mathrm e}^{-x}}{\sqrt {-c_{1}^{2} \left ({\mathrm e}^{2 x} c_{1}^{2}-1\right )}\, \sqrt {c_{1}^{4}}}+c_{2} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}}{{\mathrm e}^{2 x} c_{1}^{2}-1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}}{{\mathrm e}^{2 x} c_{1}^{2}-1}\,\mathop {\mathrm {d}x}}\\ &= \frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}\, c_{1}^{2} \arctan \left (\frac {\sqrt {c_{1}^{4}}\, {\mathrm e}^{x}}{\sqrt {-\left (c_{1} {\mathrm e}^{x}-1\right ) \left (1+c_{1} {\mathrm e}^{x}\right ) c_{1}^{2}}}\right ) {\mathrm e}^{-x}}{\sqrt {-c_{1}^{2} \left ({\mathrm e}^{2 x} c_{1}^{2}-1\right )}\, \sqrt {c_{1}^{4}}}+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}\, c_{1}^{2} \arctan \left (\frac {\sqrt {c_{1}^{4}}\, {\mathrm e}^{x}}{\sqrt {-\left (c_{1} {\mathrm e}^{x}-1\right ) \left (1+c_{1} {\mathrm e}^{x}\right ) c_{1}^{2}}}\right ) {\mathrm e}^{-x}}{\sqrt {-c_{1}^{2} \left ({\mathrm e}^{2 x} c_{1}^{2}-1\right )}\, \sqrt {c_{1}^{4}}}+c_{2} \\ \tag{2} y &= \frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}\, c_{1}^{2} \arctan \left (\frac {\sqrt {c_{1}^{4}}\, {\mathrm e}^{x}}{\sqrt {-\left (c_{1} {\mathrm e}^{x}-1\right ) \left (1+c_{1} {\mathrm e}^{x}\right ) c_{1}^{2}}}\right ) {\mathrm e}^{-x}}{\sqrt {-c_{1}^{2} \left ({\mathrm e}^{2 x} c_{1}^{2}-1\right )}\, \sqrt {c_{1}^{4}}}+c_{3} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}\, c_{1}^{2} \arctan \left (\frac {\sqrt {c_{1}^{4}}\, {\mathrm e}^{x}}{\sqrt {-\left (c_{1} {\mathrm e}^{x}-1\right ) \left (1+c_{1} {\mathrm e}^{x}\right ) c_{1}^{2}}}\right ) {\mathrm e}^{-x}}{\sqrt {-c_{1}^{2} \left ({\mathrm e}^{2 x} c_{1}^{2}-1\right )}\, \sqrt {c_{1}^{4}}}+c_{2} \] Verified OK.
\[ y = \frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{1}^{2}-1\right ) {\mathrm e}^{2 x} c_{1}^{2}}\, c_{1}^{2} \arctan \left (\frac {\sqrt {c_{1}^{4}}\, {\mathrm e}^{x}}{\sqrt {-\left (c_{1} {\mathrm e}^{x}-1\right ) \left (1+c_{1} {\mathrm e}^{x}\right ) c_{1}^{2}}}\right ) {\mathrm e}^{-x}}{\sqrt {-c_{1}^{2} \left ({\mathrm e}^{2 x} c_{1}^{2}-1\right )}\, \sqrt {c_{1}^{4}}}+c_{3} \] 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 )+\left (-p \left (y \right )^{2}-1\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int \frac {1}{p^{2}+1}d p &= y +c_{1}\\ \arctan \left (p \right )&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=\tan \left (y +c_{1} \right ) \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 } = \tan \left (y+c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{\tan \left (y +c_{1} \right )}d y &= x +c_{2}\\ -\frac {\ln \left (1+\tan \left (y +c_{1} \right )^{2}\right )}{2}+\ln \left (\tan \left (y +c_{1} \right )\right )&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-c_{1} -\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x +2 c_{2}}-1\right ) {\mathrm e}^{2 x +2 c_{2}}}}{{\mathrm e}^{2 x +2 c_{2}}-1}\right )\\ &=-c_{1} -\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{2}^{2}-1\right ) {\mathrm e}^{2 x} c_{2}^{2}}}{{\mathrm e}^{2 x} c_{2}^{2}-1}\right )\\ y_2&=-c_{1} +\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x +2 c_{2}}-1\right ) {\mathrm e}^{2 x +2 c_{2}}}}{{\mathrm e}^{2 x +2 c_{2}}-1}\right )\\ &=-c_{1} +\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{2}^{2}-1\right ) {\mathrm e}^{2 x} c_{2}^{2}}}{{\mathrm e}^{2 x} c_{2}^{2}-1}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} -\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{2}^{2}-1\right ) {\mathrm e}^{2 x} c_{2}^{2}}}{{\mathrm e}^{2 x} c_{2}^{2}-1}\right ) \\ \tag{2} y &= -c_{1} +\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{2}^{2}-1\right ) {\mathrm e}^{2 x} c_{2}^{2}}}{{\mathrm e}^{2 x} c_{2}^{2}-1}\right ) \\ \end{align*}
Verification of solutions
\[ y = -c_{1} -\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{2}^{2}-1\right ) {\mathrm e}^{2 x} c_{2}^{2}}}{{\mathrm e}^{2 x} c_{2}^{2}-1}\right ) \] Verified OK.
\[ y = -c_{1} +\arctan \left (\frac {\sqrt {-\left ({\mathrm e}^{2 x} c_{2}^{2}-1\right ) {\mathrm e}^{2 x} c_{2}^{2}}}{{\mathrm e}^{2 x} c_{2}^{2}-1}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\left (-{y^{\prime }}^{2}-1\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \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 )+\left (-u \left (x \right )^{2}-1\right ) u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\left (-u \left (x \right )^{2}-1\right ) u \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )^{2}-1\right ) u \left (x \right )}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )^{2}-1\right ) u \left (x \right )}d x =\int \left (-1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (u \left (x \right )\right )+\frac {\ln \left (u \left (x \right )^{2}+1\right )}{2}=-x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}}, u \left (x \right )=-\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\arctan \left (\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}\right )+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\arctan \left (\sqrt {-1+{\mathrm e}^{-2 x +2 c_{1}}}\right )+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 a change of variables {x -> y(x), y(x) -> x} 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)^3+_b(_a), _b(_a), HINT = [[1, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 73
dsolve(diff(y(x),x$2)=diff(y(x),x)^3+diff(y(x),x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {\arctan \left (\frac {2 c_{1} {\mathrm e}^{2 x}-1}{2 \sqrt {-\left (c_{1} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{2 x} c_{1}}}\right )}{2}+c_{2} \\ y \left (x \right ) &= \frac {\arctan \left (\frac {2 c_{1} {\mathrm e}^{2 x}-1}{2 \sqrt {-\left (c_{1} {\mathrm e}^{2 x}-1\right ) {\mathrm e}^{2 x} c_{1}}}\right )}{2}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.102 (sec). Leaf size: 71
DSolve[y''[x]==y'[x]^3+y'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-i \log \left (\sqrt {-1+e^{2 (x+c_1)}}-e^{x+c_1}\right ) \\ y(x)\to i \log \left (\sqrt {-1+e^{2 (x+c_1)}}-e^{x+c_1}\right )+c_2 \\ \end{align*}