Internal problem ID [15216]
Internal file name [OUTPUT/15217_Tuesday_April_23_2024_04_54_24_PM_12687429/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 14. Differential equations admitting of
depression of their order. Exercises page 107
Problem number: 359.
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]]
\[ \boxed {y^{3} y^{\prime \prime }=-1} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = 0] \end {align*}
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*} y^{3} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = -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 {1}{y^{3} p} \end {align*}
Where \(f(y)=-\frac {1}{y^{3}}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= -\frac {1}{y^{3}} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {-\frac {1}{y^{3}} \,d y} \\ \frac {p^{2}}{2}&=\frac {1}{2 y^{2}}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {1}{2 y^{2}}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=1\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} -\frac {1}{2} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = -{\frac {1}{2}} \end {align*}
Trying the constant \begin {align*} c_{1} = -{\frac {1}{2}} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2} y^{2}+y^{2}-1}{2 y^{2}} = 0 \end {align*}
The constant \(c_{1} = -{\frac {1}{2}}\) gives valid solution.
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 {y^{2} {y^{\prime }}^{2}+y^{2}-1}{2 y^{2}} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {1-y^{2}}}{y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {1-y^{2}}}{y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=1 \end {align*}
Solving equation (2)
Since ode has form \(y^{\prime }= f(y)\) and initial conditions \(y = 1\) is verified to satisfy the ode, then the solution is \begin {align*} y&=y_0 \\ &=1 \end {align*}
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= 1 \\ \end{align*}
Verification of solutions
\[ y = 1 \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{3} y^{\prime \prime }=-1, y \left (1\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0\right ] \\ \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} \\ {} & {} & y^{3} u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {1}{y^{3} u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=-\frac {1}{y^{3}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int -\frac {1}{y^{3}}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=\frac {1}{2 y^{2}}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {2 c_{1} y^{2}+1}}{y}, u \left (y \right )=-\frac {\sqrt {2 c_{1} y^{2}+1}}{y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {2 c_{1} y^{2}+1}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {2 y^{2} c_{1} +1}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {2 y^{2} c_{1} +1}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {2 y^{2} c_{1} +1}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {2 y^{2} c_{1} +1}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 y^{2} c_{1} +1}}{2 c_{1}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}}, y=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {2 c_{1} y^{2}+1}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {2 y^{2} c_{1} +1}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2 y^{2} c_{1} +1}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {2 y^{2} c_{1} +1}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {2 y^{2} c_{1} +1}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2 y^{2} c_{1} +1}}{2 c_{1}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}}, y=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}}\right \} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2}\, \left (-8 c_{1}^{2} c_{2} +8 c_{1}^{2} x \right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=-\frac {\sqrt {2}\, \left (-8 c_{1}^{2} c_{2} +8 c_{1}^{2}\right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{2}, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {-x \left (x -2\right )} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {2}\, \left (-8 c_{1}^{2} c_{2} +8 c_{1}^{2} x \right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=\frac {\sqrt {2}\, \left (-8 c_{1}^{2} c_{2} +8 c_{1}^{2}\right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}-8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=-\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2}\, \left (8 c_{1}^{2} c_{2} +8 c_{1}^{2} x \right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=-\frac {\sqrt {2}\, \left (8 c_{1}^{2} c_{2} +8 c_{1}^{2}\right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {1}{2}, c_{2} =-1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {-x \left (x -2\right )} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 1=\frac {\sqrt {2}\, \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}}{2 c_{1}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {2}\, \left (8 c_{1}^{2} c_{2} +8 c_{1}^{2} x \right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} x +4 c_{1}^{2} x^{2}-1\right )}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=\frac {\sqrt {2}\, \left (8 c_{1}^{2} c_{2} +8 c_{1}^{2}\right )}{4 \sqrt {c_{1} \left (4 c_{1}^{2} c_{2}^{2}+8 c_{1}^{2} c_{2} +4 c_{1}^{2}-1\right )}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\sqrt {-x \left (x -2\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
dsolve([y(x)^3*diff(y(x),x$2)=-1,y(1) = 1, D(y)(1) = 0],y(x), singsol=all)
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 0.051 (sec). Leaf size: 15
DSolve[{y[x]^3*y''[x]==-1,{y[1]==1,y'[1]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {-((x-2) x)} \]