Internal problem ID [6830]
Internal file name [OUTPUT/6077_Thursday_July_28_2022_04_29_23_AM_4198183/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: 11.
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], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {y^{\prime \prime }-2 y {y^{\prime }}^{3}=0} \]
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 )-2 y p \left (y \right )^{3} = 0 \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)\\ &= 2 p^{2} y \end {align*}
Where \(f(y)=2 y\) and \(g(p)=p^{2}\). Integrating both sides gives \begin{align*} \frac {1}{p^{2}} \,dp &= 2 y \,d y \\ \int { \frac {1}{p^{2}} \,dp} &= \int {2 y \,d y} \\ -\frac {1}{p}&=y^{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{p \left (y \right )}-y^{2}-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {1}{y^{\prime }}-y^{2}-c_{1} = 0 \end {align*}
Integrating both sides gives \begin {align*} \int \left (-y^{2}-c_{1} \right )d y &= x +c_{2}\\ -\frac {1}{3} y^{3}-c_{1} y&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}-\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\\ y_2&=-\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{4}+\frac {c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}+\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\right )}{2}\\ y_3&=-\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{4}+\frac {c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}+\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\right )}{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}-\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}} \\ \tag{2} y &= -\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{4}+\frac {c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}+\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\right )}{2} \\ \tag{3} y &= -\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{4}+\frac {c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}+\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\right )}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}-\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}} \] Verified OK.
\[ y = -\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{4}+\frac {c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}+\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\right )}{2} \] Verified OK.
\[ y = -\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{4}+\frac {c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (\frac {\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}+\frac {2 c_{1}}{\left (-12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}\right )}{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-2 y {y^{\prime }}^{3}=0 \\ \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} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-2 y u \left (y \right )^{3}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=2 y u \left (y \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )^{2}}=2 y \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )^{2}}d y =\int 2 y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (y \right )}=y^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {1}{y^{2}+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {1}{y^{2}+c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {1}{y^{2}+c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {1}{y^{2}+c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime } \left (y^{2}+c_{1} \right )=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime } \left (y^{2}+c_{1} \right )d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {y^{3}}{3}+c_{1} y=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left (12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}-18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}}{2}-\frac {2 c_{1}}{\left (12 c_{2} -12 x +4 \sqrt {4 c_{1}^{3}+9 c_{2}^{2}-18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}} \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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-2*_a*_b(_a)^3 = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 324
dsolve(diff(y(x),x$2)=2*y(x)*diff(y(x),x)^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= \frac {\left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {2}{3}}+4 c_{1}}{2 \left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= \frac {-i \sqrt {3}\, \left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {2}{3}}+4 i \sqrt {3}\, c_{1} -\left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {2}{3}}-4 c_{1}}{4 \left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= -\frac {-i \sqrt {3}\, \left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {2}{3}}+4 i \sqrt {3}\, c_{1} +\left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {2}{3}}+4 c_{1}}{4 \left (-12 c_{2} -12 x +4 \sqrt {-4 c_{1}^{3}+9 c_{2}^{2}+18 c_{2} x +9 x^{2}}\right )^{\frac {1}{3}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 7.768 (sec). Leaf size: 351
DSolve[y''[x]==2*y[x]*(y'[x])^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt [3]{2} c_1}{\sqrt [3]{\sqrt {9 x^2+18 c_2 x+4 c_1{}^3+9 c_2{}^2}+3 x+3 c_2}}-\frac {\sqrt [3]{\sqrt {9 x^2+18 c_2 x+4 c_1{}^3+9 c_2{}^2}+3 x+3 c_2}}{\sqrt [3]{2}} \\ y(x)\to \frac {2^{2/3} \left (1-i \sqrt {3}\right ) \left (\sqrt {9 x^2+18 c_2 x+4 c_1{}^3+9 c_2{}^2}+3 x+3 c_2\right ){}^{2/3}+\sqrt [3]{2} \left (-2-2 i \sqrt {3}\right ) c_1}{4 \sqrt [3]{\sqrt {9 x^2+18 c_2 x+4 c_1{}^3+9 c_2{}^2}+3 x+3 c_2}} \\ y(x)\to \frac {2^{2/3} \left (1+i \sqrt {3}\right ) \left (\sqrt {9 x^2+18 c_2 x+4 c_1{}^3+9 c_2{}^2}+3 x+3 c_2\right ){}^{2/3}+2 i \sqrt [3]{2} \left (\sqrt {3}+i\right ) c_1}{4 \sqrt [3]{\sqrt {9 x^2+18 c_2 x+4 c_1{}^3+9 c_2{}^2}+3 x+3 c_2}} \\ y(x)\to 0 \\ \end{align*}