Internal problem ID [15203]
Internal file name [OUTPUT/15204_Tuesday_April_23_2024_04_54_05_PM_26597427/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: 346.
ODE order: 2.
ODE degree: 0.
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]]
\[ \boxed {y^{\prime \prime }-y^{\prime } \ln \left (y^{\prime }\right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}
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 )-p \left (x \right ) \ln \left (p \left (x \right )\right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Since ode has form \(p^{\prime }\left (x \right )= f(p \left (x \right ))\) and initial conditions \(p = 1\) is verified to satisfy the ode, then the solution is \begin {align*} p \left (x \right )&=p_0 \\ &=1 \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = 1 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { 1\,\mathop {\mathrm {d}x}}\\ &= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=x \end {align*}
The constant \(c_{2} = 0\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \\ \end{align*}
Verification of solutions
\[ y = x \] 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 )-p \left (y \right ) \ln \left (p \left (y \right )\right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Since ode has form \(\frac {d}{d y}p \left (y \right )= f(p \left (y \right ))\) and initial conditions \(p = 1\) is verified to satisfy the ode, then the solution is \begin {align*} p \left (y \right )&=p_0 \\ &=1 \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 } = 1 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { 1\,\mathop {\mathrm {d}x}}\\ &= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=x \end {align*}
The constant \(c_{2} = 0\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \\ \end{align*}
Verification of solutions
\[ y = x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-y^{\prime } \ln \left (y^{\prime }\right )=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\right ] \\ \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 )-u \left (x \right ) \ln \left (u \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right ) \ln \left (u \left (x \right )\right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right ) \ln \left (u \left (x \right )\right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right ) \ln \left (u \left (x \right )\right )}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (\ln \left (u \left (x \right )\right )\right )=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{{\mathrm e}^{x +c_{1}}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{{\mathrm e}^{x +c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }={\mathrm e}^{{\mathrm e}^{x +c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int {\mathrm e}^{{\mathrm e}^{x +c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\mathrm {Ei}_{1}\left (-{\mathrm e}^{x +c_{1}}\right )+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\mathrm {Ei}_{1}\left (-{\mathrm e}^{x +c_{1}}\right )+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=-\mathrm {Ei}_{1}\left (-{\mathrm e}^{c_{1}}\right )+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{{\mathrm e}^{x +c_{1}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1={\mathrm e}^{{\mathrm e}^{c_{1}}} \\ {} & \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} \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)*ln(_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: 23
dsolve([diff(y(x),x$2)=diff(y(x),x)*ln(diff(y(x),x)),y(0) = 0, D(y)(0) = 1],y(x), singsol=all)
\[ y = -\operatorname {expIntegral}_{1}\left (-2 i \pi \_Z5 \,{\mathrm e}^{x}\right )+\operatorname {expIntegral}_{1}\left (-2 i \pi \_Z5 \right ) \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y''[x]==y'[x]*Log[y'[x]],{y[0]==0,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
{}