Internal problem ID [2299]
Internal file name [OUTPUT/2299_Tuesday_February_27_2024_08_24_16_AM_22539582/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: 27.
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]]
\[ \boxed {y^{\prime \prime }+y^{\prime }-{y^{\prime }}^{3}=0} \]
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 (1-p \left (x \right )^{2}\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 {1}{\sqrt {1-{\mathrm e}^{2 x +2 c_{1}}}}\\ &=\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\\ p_2&=-\frac {1}{\sqrt {1-{\mathrm e}^{2 x +2 c_{1}}}}\\ &=-\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}} \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 {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\,\mathop {\mathrm {d}x}}\\ &= -\operatorname {arctanh}\left (\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\right )+c_{2} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\,\mathop {\mathrm {d}x}}\\ &= \operatorname {arctanh}\left (\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\right )+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\operatorname {arctanh}\left (\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\right )+c_{2} \\ \tag{2} y &= \operatorname {arctanh}\left (\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\right )+c_{3} \\ \end{align*}
Verification of solutions
\[ y = -\operatorname {arctanh}\left (\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\right )+c_{2} \] Verified OK.
\[ y = \operatorname {arctanh}\left (\frac {1}{\sqrt {1-{\mathrm e}^{2 x} c_{1}^{2}}}\right )+c_{3} \] 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 (1-p \left (y \right )^{2}\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}\\ -\operatorname {arctanh}\left (p \right )&=y +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=-\tanh \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 } = -\tanh \left (y+c_{1} \right ) \end {align*}
Integrating both sides gives \begin {align*} \int -\frac {1}{\tanh \left (y +c_{1} \right )}d y &= x +c_{2}\\ -\ln \left (\tanh \left (y +c_{1} \right )\right )+\ln \left (1-\tanh \left (y +c_{1} \right )\right )+y&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x +c_{2} -\ln \left (-1-\sqrt {1+{\mathrm e}^{2 x +2 c_{2} +2 c_{1}}}\right ) \\ \tag{2} y &= x +c_{2} -\ln \left (-1+\sqrt {1+{\mathrm e}^{2 x +2 c_{2} +2 c_{1}}}\right ) \\ \end{align*}
Verification of solutions
\[ y = x +c_{2} -\ln \left (-1-\sqrt {1+{\mathrm e}^{2 x +2 c_{2} +2 c_{1}}}\right ) \] Verified OK.
\[ y = x +c_{2} -\ln \left (-1+\sqrt {1+{\mathrm e}^{2 x +2 c_{2} +2 c_{1}}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\left (1-{y^{\prime }}^{2}\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 (1-u \left (x \right )^{2}\right ) u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\left (1-u \left (x \right )^{2}\right ) u \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (1-u \left (x \right )^{2}\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 (1-u \left (x \right )^{2}\right ) u \left (x \right )}d x =\int \left (-1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (x \right )+1\right )}{2}-\frac {\ln \left (u \left (x \right )-1\right )}{2}+\ln \left (u \left (x \right )\right )=-x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-1+{\mathrm e}^{-2 x +2 c_{1}}}, u \left (x \right )=-\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-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 {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-1+{\mathrm e}^{-2 x +2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-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 {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-1+{\mathrm e}^{-2 x +2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right )^{2}+{\mathrm e}^{-2 x +2 c_{1}}-1}}{2}-\frac {\ln \left ({\mathrm e}^{-2 x +2 c_{1}}-\frac {1}{2}+\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right )^{2}+{\mathrm e}^{-2 x +2 c_{1}}-1}\right )}{4}+\frac {\sqrt {\left ({\mathrm e}^{-2 x +2 c_{1}}\right )^{2}-{\mathrm e}^{-2 x +2 c_{1}}}}{2}-\frac {\ln \left ({\mathrm e}^{-2 x +2 c_{1}}-\frac {1}{2}+\sqrt {\left ({\mathrm e}^{-2 x +2 c_{1}}\right )^{2}-{\mathrm e}^{-2 x +2 c_{1}}}\right )}{4}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-1+{\mathrm e}^{-2 x +2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-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 {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right ) {\mathrm e}^{-2 x +2 c_{1}}}}{-1+{\mathrm e}^{-2 x +2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right )^{2}+{\mathrm e}^{-2 x +2 c_{1}}-1}}{2}+\frac {\ln \left ({\mathrm e}^{-2 x +2 c_{1}}-\frac {1}{2}+\sqrt {\left (-1+{\mathrm e}^{-2 x +2 c_{1}}\right )^{2}+{\mathrm e}^{-2 x +2 c_{1}}-1}\right )}{4}-\frac {\sqrt {\left ({\mathrm e}^{-2 x +2 c_{1}}\right )^{2}-{\mathrm e}^{-2 x +2 c_{1}}}}{2}+\frac {\ln \left ({\mathrm e}^{-2 x +2 c_{1}}-\frac {1}{2}+\sqrt {\left ({\mathrm e}^{-2 x +2 c_{1}}\right )^{2}-{\mathrm e}^{-2 x +2 c_{1}}}\right )}{4}+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: 0.031 (sec). Leaf size: 37
dsolve(diff(y(x),x$2)+diff(y(x),x)=diff(y(x),x)^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\operatorname {arctanh}\left (\sqrt {-c_{1} {\mathrm e}^{2 x}+1}\right )+c_{2} \\ y \left (x \right ) &= \operatorname {arctanh}\left (\sqrt {-c_{1} {\mathrm e}^{2 x}+1}\right )+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.1 (sec). Leaf size: 47
DSolve[y''[x]+y'[x]==y'[x]^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\text {arctanh}\left (\sqrt {1+e^{2 (x+c_1)}}\right ) \\ y(x)\to \text {arctanh}\left (\sqrt {1+e^{2 (x+c_1)}}\right )+c_2 \\ \end{align*}