Internal problem ID [6859]
Internal file name [OUTPUT/6106_Friday_July_29_2022_03_09_42_AM_61425486/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 42.
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]]
\[ \boxed {3 y y^{\prime } y^{\prime \prime }-{y^{\prime }}^{3}=-1} \]
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*} 3 y p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{3} = -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^{3}-1}{3 y \,p^{2}} \end {align*}
Where \(f(y)=\frac {1}{3 y}\) and \(g(p)=\frac {p^{3}-1}{p^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{3}-1}{p^{2}}} \,dp &= \frac {1}{3 y} \,d y \\ \int { \frac {1}{\frac {p^{3}-1}{p^{2}}} \,dp} &= \int {\frac {1}{3 y} \,d y} \\ \frac {\ln \left (p^{3}-1\right )}{3}&=\frac {\ln \left (y \right )}{3}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \left (p^{3}-1\right )^{\frac {1}{3}} &= {\mathrm e}^{\frac {\ln \left (y \right )}{3}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \left (p^{3}-1\right )^{\frac {1}{3}} &= c_{2} y^{\frac {1}{3}} \end {align*}
Which simplifies to \[ \left (p \left (y \right )^{3}-1\right )^{\frac {1}{3}} = c_{2} y^{\frac {1}{3}} {\mathrm e}^{c_{1}} \] The solution is \[ \left (p \left (y \right )^{3}-1\right )^{\frac {1}{3}} = c_{2} y^{\frac {1}{3}} {\mathrm e}^{c_{1}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \left ({y^{\prime }}^{3}-1\right )^{\frac {1}{3}} = c_{2} y^{\frac {1}{3}} {\mathrm e}^{c_{1}} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {1}{3}} \tag {1} \\ y^{\prime }&=-\frac {\left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {1}{3}}}{2} \tag {2} \\ y^{\prime }&=-\frac {\left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {1}{3}}}{2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\left ({\mathrm e}^{3 c_{1}} y \,c_{2}^{3}+1\right )^{\frac {1}{3}}}d y &= \int d x \\ \frac {3 \left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {2}{3}} {\mathrm e}^{-3 c_{1}}}{2 c_{2}^{3}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int \frac {1}{-\frac {\left ({\mathrm e}^{3 c_{1}} y \,c_{2}^{3}+1\right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left ({\mathrm e}^{3 c_{1}} y \,c_{2}^{3}+1\right )^{\frac {1}{3}}}{2}}d y &= \int d x \\ \frac {3 \left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {2}{3}} {\mathrm e}^{-3 c_{1}}}{\left (i \sqrt {3}-1\right ) c_{2}^{3}}&=x +c_{4} \\ \end{align*} Solving equation (3)
Integrating both sides gives \begin{align*} \int \frac {1}{-\frac {\left ({\mathrm e}^{3 c_{1}} y \,c_{2}^{3}+1\right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left ({\mathrm e}^{3 c_{1}} y \,c_{2}^{3}+1\right )^{\frac {1}{3}}}{2}}d y &= \int d x \\ -\frac {3 \left ({\mathrm e}^{3 c_{1}} y c_{2}^{3}+1\right )^{\frac {2}{3}} {\mathrm e}^{-3 c_{1}}}{\left (1+i \sqrt {3}\right ) c_{2}^{3}}&=x +c_{5} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (-4 \left ({\mathrm e}^{3 c_{1}} c_{2}^{3} \left (x +c_{3} \right )\right )^{\frac {3}{2}}+3 \sqrt {6}\right ) \sqrt {6}\, {\mathrm e}^{-3 c_{1}}}{18 c_{2}^{3}} \\ \tag{2} y &= \frac {\left (\left (3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} c_{4} +3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} x -3 c_{4} {\mathrm e}^{3 c_{1}} c_{2}^{3}-3 \,{\mathrm e}^{3 c_{1}} c_{2}^{3} x \right )^{\frac {3}{2}}-27\right ) {\mathrm e}^{-3 c_{1}}}{27 c_{2}^{3}} \\ \tag{3} y &= \frac {\left (\left (-3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} c_{5} -3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} x -3 c_{5} {\mathrm e}^{3 c_{1}} c_{2}^{3}-3 \,{\mathrm e}^{3 c_{1}} c_{2}^{3} x \right )^{\frac {3}{2}}-27\right ) {\mathrm e}^{-3 c_{1}}}{27 c_{2}^{3}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (-4 \left ({\mathrm e}^{3 c_{1}} c_{2}^{3} \left (x +c_{3} \right )\right )^{\frac {3}{2}}+3 \sqrt {6}\right ) \sqrt {6}\, {\mathrm e}^{-3 c_{1}}}{18 c_{2}^{3}} \] Verified OK.
\[ y = \frac {\left (\left (3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} c_{4} +3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} x -3 c_{4} {\mathrm e}^{3 c_{1}} c_{2}^{3}-3 \,{\mathrm e}^{3 c_{1}} c_{2}^{3} x \right )^{\frac {3}{2}}-27\right ) {\mathrm e}^{-3 c_{1}}}{27 c_{2}^{3}} \] Verified OK.
\[ y = \frac {\left (\left (-3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} c_{5} -3 i \sqrt {3}\, {\mathrm e}^{3 c_{1}} c_{2}^{3} x -3 c_{5} {\mathrm e}^{3 c_{1}} c_{2}^{3}-3 \,{\mathrm e}^{3 c_{1}} c_{2}^{3} x \right )^{\frac {3}{2}}-27\right ) {\mathrm e}^{-3 c_{1}}}{27 c_{2}^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 y y^{\prime \prime } y^{\prime }-{y^{\prime }}^{3}=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \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 ),y^{\prime \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} \\ {} & {} & 3 y u \left (y \right )^{2} \left (\frac {d}{d y}u \left (y \right )\right )-u \left (y \right )^{3}=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )^{3}-1}{3 u \left (y \right )^{2} y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )^{2}}{u \left (y \right )^{3}-1}=\frac {1}{3 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 )^{2}}{u \left (y \right )^{3}-1}d y =\int \frac {1}{3 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (u \left (y \right )^{3}-1\right )}{3}=\frac {\ln \left (y \right )}{3}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\left (\left ({\mathrm e}^{-3 c_{1}}+y \right ) \left ({\mathrm e}^{-3 c_{1}}\right )^{2}\right )}^{\frac {1}{3}}}{{\mathrm e}^{-3 c_{1}}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {{\left (\left ({\mathrm e}^{-3 c_{1}}+y \right ) \left ({\mathrm e}^{-3 c_{1}}\right )^{2}\right )}^{\frac {1}{3}}}{{\mathrm e}^{-3 c_{1}}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {{\left (\left ({\mathrm e}^{-3 c_{1}}+y\right ) \left ({\mathrm e}^{-3 c_{1}}\right )^{2}\right )}^{\frac {1}{3}}}{{\mathrm e}^{-3 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\left (\left ({\mathrm e}^{-3 c_{1}}+y\right ) \left ({\mathrm e}^{-3 c_{1}}\right )^{2}\right )}^{\frac {1}{3}}}{{\mathrm e}^{-3 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\left (\left ({\mathrm e}^{-3 c_{1}}+y\right ) \left ({\mathrm e}^{-3 c_{1}}\right )^{2}\right )}^{\frac {1}{3}}}=\frac {1}{{\mathrm e}^{-3 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\left (\left ({\mathrm e}^{-3 c_{1}}+y\right ) \left ({\mathrm e}^{-3 c_{1}}\right )^{2}\right )}^{\frac {1}{3}}}d x =\int \frac {1}{{\mathrm e}^{-3 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {3 {\left (\left ({\mathrm e}^{-3 c_{1}}\right )^{2} y+\left ({\mathrm e}^{-3 c_{1}}\right )^{3}\right )}^{\frac {2}{3}}}{2 \left ({\mathrm e}^{-3 c_{1}}\right )^{2}}=\frac {x}{{\mathrm e}^{-3 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\mathit {RootOf}\left (2 c_{2} \left ({\mathrm e}^{-3 c_{1}}\right )^{2}+2 x \,{\mathrm e}^{-3 c_{1}}-3 {\left (\left ({\mathrm e}^{-3 c_{1}}\right )^{2} \textit {\_Z} +\left ({\mathrm e}^{-3 c_{1}}\right )^{3}\right )}^{\frac {2}{3}}\right )\right \} \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)-(1/3)*(_b(_a)^3-1)/(_b(_a)*_a) = 0, _b(_a), HINT = [[_a, 0]]` *** Subl symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 119
dsolve(3*y(x)*diff(y(x),x)*diff(y(x),x$2)=diff(y(x),x)^3-1,y(x), singsol=all)
\begin{align*} \frac {3 \left (c_{1} y \left (x \right )+1\right )^{\frac {2}{3}}+\left (-2 x -2 c_{2} \right ) c_{1}}{2 c_{1}} &= 0 \\ \frac {-i \left (x +c_{2} \right ) c_{1} \sqrt {3}+\left (-x -c_{2} \right ) c_{1} -3 \left (c_{1} y \left (x \right )+1\right )^{\frac {2}{3}}}{c_{1} \left (1+i \sqrt {3}\right )} &= 0 \\ \frac {-3 i \left (c_{1} y \left (x \right )+1\right )^{\frac {2}{3}}+\left (-x -c_{2} \right ) c_{1} \sqrt {3}-i \left (x +c_{2} \right ) c_{1}}{c_{1} \left (\sqrt {3}+i\right )} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 45.036 (sec). Leaf size: 126
DSolve[3*y[x]*y'[x]*y''[x]==(y'[x])^3-1,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{9} e^{-3 c_1} \left (-9+2 \sqrt {6} \left (e^{3 c_1} (x+c_2)\right ){}^{3/2}\right ) \\ y(x)\to \frac {1}{9} e^{-3 c_1} \left (-9+2 \sqrt {6} \left (-\sqrt [3]{-1} e^{3 c_1} (x+c_2)\right ){}^{3/2}\right ) \\ y(x)\to \frac {1}{9} e^{-3 c_1} \left (-9+2 \sqrt {6} \left ((-1)^{2/3} e^{3 c_1} (x+c_2)\right ){}^{3/2}\right ) \\ \end{align*}