Internal problem ID [12172]
Internal file name [OUTPUT/10825_Thursday_September_21_2023_05_47_32_AM_45372491/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 10.
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]]
\[ \boxed {x^{3} x^{\prime \prime }=-1} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(x\) an independent variable. Using \begin {align*} x' &= p(x) \end {align*}
Then \begin {align*} x'' &= \frac {dp}{dt}\\ &= \frac {dx}{dt} \frac {dp}{dx}\\ &= p \frac {dp}{dx} \end {align*}
Hence the ode becomes \begin {align*} x^{3} p \left (x \right ) \left (\frac {d}{d x}p \left (x \right )\right ) = -1 \end {align*}
Which is now solved as first order ode for \(p(x)\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {1}{x^{3} p} \end {align*}
Where \(f(x)=-\frac {1}{x^{3}}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= -\frac {1}{x^{3}} \,d x \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {-\frac {1}{x^{3}} \,d x} \\ \frac {p^{2}}{2}&=\frac {1}{2 x^{2}}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (x \right )^{2}}{2}-\frac {1}{2 x^{2}}-c_{1} = 0 \] For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {{x^{\prime }}^{2}}{2}-\frac {1}{2 x^{2}}-c_{1} = 0 \end {align*}
Solving the given ode for \(x^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} x^{\prime }&=\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \tag {1} \\ x^{\prime }&=-\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {x}{\sqrt {2 c_{1} x^{2}+1}}d x &= \int d t \\ \frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}}&=t +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {x}{\sqrt {2 c_{1} x^{2}+1}}d x &= \int d t \\ -\frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}}&=t +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}} &= t +c_{2} \\ \tag{2} -\frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}} &= t +c_{3} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}} = t +c_{2} \] Verified OK.
\[ -\frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}} = t +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d}{d t}x^{\prime }\right )=-1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}x^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (t \right )=x^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d t}x^{\prime } \\ {} & {} & u^{\prime }\left (t \right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & x^{\prime } \left (\frac {d}{d x}u \left (x \right )\right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} x^{\prime }=u \left (x \right ),\frac {d}{d t}x^{\prime }=u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & x^{3} u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-\frac {1}{x^{3} u \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )=-\frac {1}{x^{3}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )d x =\int -\frac {1}{x^{3}}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (x \right )^{2}}{2}=\frac {1}{2 x^{2}}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {\sqrt {2 c_{1} x^{2}+1}}{x}, u \left (x \right )=-\frac {\sqrt {2 c_{1} x^{2}+1}}{x}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (x \right )=x^{\prime },x =x \\ {} & {} & x^{\prime }=\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x x^{\prime }}{\sqrt {2 c_{1} x^{2}+1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x x^{\prime }}{\sqrt {2 c_{1} x^{2}+1}}d t =\int 1d t +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}}=t +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & \left \{x=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} t +4 c_{1}^{2} t^{2}-1\right )}}{2 c_{1}}, x=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} t +4 c_{1}^{2} t^{2}-1\right )}}{2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (x \right )=x^{\prime },x =x \\ {} & {} & x^{\prime }=-\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=-\frac {\sqrt {2 c_{1} x^{2}+1}}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x x^{\prime }}{\sqrt {2 c_{1} x^{2}+1}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x x^{\prime }}{\sqrt {2 c_{1} x^{2}+1}}d t =\int \left (-1\right )d t +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 c_{1} x^{2}+1}}{2 c_{1}}=-t +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & \left \{x=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} t +4 c_{1}^{2} t^{2}-1\right )}}{2 c_{1}}, x=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} t +4 c_{1}^{2} t^{2}-1\right )}}{2 c_{1}}\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/_a^3 = 0, _b(_a), HINT = [[_a, -_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, -_b]
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 52
dsolve(x(t)^3*diff(x(t),t$2)+1=0,x(t), singsol=all)
\begin{align*} x \left (t \right ) &= \frac {\sqrt {\left (1+c_{1} \left (c_{2} +t \right )\right ) \left (-1+c_{1} \left (c_{2} +t \right )\right ) c_{1}}}{c_{1}} \\ x \left (t \right ) &= -\frac {\sqrt {\left (1+c_{1} \left (c_{2} +t \right )\right ) \left (-1+c_{1} \left (c_{2} +t \right )\right ) c_{1}}}{c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 4.287 (sec). Leaf size: 93
DSolve[x[t]^3*x''[t]+1==0,x[t],t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -\frac {\sqrt {c_1{}^2 t^2+2 c_2 c_1{}^2 t-1+c_2{}^2 c_1{}^2}}{\sqrt {c_1}} \\ x(t)\to \frac {\sqrt {c_1{}^2 t^2+2 c_2 c_1{}^2 t-1+c_2{}^2 c_1{}^2}}{\sqrt {c_1}} \\ x(t)\to \text {Indeterminate} \\ \end{align*}