Internal problem ID [13513]
Internal file name [OUTPUT/12685_Friday_February_16_2024_12_10_55_AM_88937239/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises
page 259
Problem number: 13.6 (h).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1], [_2nd_order, _reducible, _mu_poly_yn]]
\[ \boxed {2 x y^{\prime } y^{\prime \prime }-{y^{\prime }}^{2}=-1} \] With initial conditions \begin {align*} \left [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = \sqrt {3}\right ] \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*} 2 x p \left (x \right ) p^{\prime }\left (x \right )-p \left (x \right )^{2}+1 = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {p^{2}-1}{2 x p} \end {align*}
Where \(f(x)=\frac {1}{2 x}\) and \(g(p)=\frac {p^{2}-1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}-1}{p}} \,dp &= \frac {1}{2 x} \,d x \\ \int { \frac {1}{\frac {p^{2}-1}{p}} \,dp} &= \int {\frac {1}{2 x} \,d x} \\ \frac {\ln \left (p^{2}-1\right )}{2}&=\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}-1} &= {\mathrm e}^{\frac {\ln \left (x \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}-1} &= c_{2} \sqrt {x} \end {align*}
Which can be simplified to become \[ \sqrt {p \left (x \right )^{2}-1} = c_{2} \sqrt {x}\, {\mathrm e}^{c_{1}} \] The solution is \[ \sqrt {p \left (x \right )^{2}-1} = c_{2} \sqrt {x}\, {\mathrm e}^{c_{1}} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(p=\sqrt {3}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \sqrt {2} = c_{2} {\mathrm e}^{c_{1}} \end {align*}
The solutions are \begin {align*} c_{1} = \frac {\ln \left (\frac {2}{c_{2}^{2}}\right )}{2} \end {align*}
Trying the constant \begin {align*} c_{1} = \frac {\ln \left (\frac {2}{c_{2}^{2}}\right )}{2} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \sqrt {p^{2}-1} = c_{2} \sqrt {x}\, \sqrt {2}\, \sqrt {\frac {1}{c_{2}^{2}}} \end {align*}
The constant \(c_{1} = \frac {\ln \left (\frac {2}{c_{2}^{2}}\right )}{2}\) 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*} \sqrt {{y^{\prime }}^{2}-1} = \sqrt {x}\, \sqrt {2}\, \operatorname {csgn}\left (\frac {1}{c_{2}}\right ) \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 }&=\sqrt {2 \operatorname {csgn}\left (\frac {1}{c_{2}}\right )^{2} x +1} \tag {1} \\ y^{\prime }&=-\sqrt {2 \operatorname {csgn}\left (\frac {1}{c_{2}}\right )^{2} x +1} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { \sqrt {2 \operatorname {csgn}\left (\frac {1}{c_{2}}\right )^{2} x +1}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}+c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=1\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = \sqrt {3}+c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = -\sqrt {3} \end {align*}
Trying the constant \begin {align*} c_{3} = -\sqrt {3} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}-\sqrt {3} \end {align*}
The constant \(c_{3} = -\sqrt {3}\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { -\sqrt {2 \operatorname {csgn}\left (\frac {1}{c_{2}}\right )^{2} x +1}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}+c_{4} \end {align*}
Initial conditions are used to solve for \(c_{4}\). Substituting \(x=1\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = -\sqrt {3}+c_{4} \end {align*}
The solutions are \begin {align*} c_{4} = \sqrt {3} \end {align*}
Trying the constant \begin {align*} c_{4} = \sqrt {3} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}+\sqrt {3} \end {align*}
The constant \(c_{4} = \sqrt {3}\) 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 &= \frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}-\sqrt {3} \\ \tag{2} y &= -\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}+\sqrt {3} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}-\sqrt {3} \] Verified OK.
\[ y = -\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}+\sqrt {3} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 x y^{\prime } \left (\frac {d}{d x}y^{\prime }\right )-{y^{\prime }}^{2}=-1, y \left (1\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=\sqrt {3}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & 2 x u \left (x \right ) u^{\prime }\left (x \right )-u \left (x \right )^{2}=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {u \left (x \right )^{2}-1}{2 x u \left (x \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right ) u \left (x \right )}{u \left (x \right )^{2}-1}=\frac {1}{2 x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right ) u \left (x \right )}{u \left (x \right )^{2}-1}d x =\int \frac {1}{2 x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (u \left (x \right )-1\right )}{2}+\frac {\ln \left (u \left (x \right )+1\right )}{2}=\frac {\ln \left (x \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}}, u \left (x \right )=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {2 \left (x +{\mathrm e}^{-2 c_{1}}\right ) \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {\sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{{\mathrm e}^{-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {2 \left (x +{\mathrm e}^{-2 c_{1}}\right ) \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {2 \left (x +{\mathrm e}^{-2 c_{1}}\right ) \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{3 {\mathrm e}^{-2 c_{1}}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=-\frac {2 \left (1+{\mathrm e}^{-2 c_{1}}\right ) \sqrt {{\mathrm e}^{-2 c_{1}} \left (1+{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}-\frac {x +{\mathrm e}^{-2 c_{1}}}{3 \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=\sqrt {3} \\ {} & {} & \sqrt {3}=-\frac {2 \sqrt {{\mathrm e}^{-2 c_{1}} \left (1+{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}-\frac {1+{\mathrm e}^{-2 c_{1}}}{3 \sqrt {{\mathrm e}^{-2 c_{1}} \left (1+{\mathrm e}^{-2 c_{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 {2 \left (x +{\mathrm e}^{-2 c_{1}}\right ) \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{3 {\mathrm e}^{-2 c_{1}}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=\frac {2 \left (1+{\mathrm e}^{-2 c_{1}}\right ) \sqrt {{\mathrm e}^{-2 c_{1}} \left (1+{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}+\frac {x +{\mathrm e}^{-2 c_{1}}}{3 \sqrt {{\mathrm e}^{-2 c_{1}} \left (x +{\mathrm e}^{-2 c_{1}}\right )}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=\sqrt {3} \\ {} & {} & \sqrt {3}=\frac {2 \sqrt {{\mathrm e}^{-2 c_{1}} \left (1+{\mathrm e}^{-2 c_{1}}\right )}}{3 \,{\mathrm e}^{-2 c_{1}}}+\frac {1+{\mathrm e}^{-2 c_{1}}}{3 \sqrt {{\mathrm e}^{-2 c_{1}} \left (1+{\mathrm e}^{-2 c_{1}}\right )}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\frac {\ln \left (2\right )}{2}, c_{2} =-\sqrt {3}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}-\sqrt {3} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}-\sqrt {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 `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = (1/2)*(_b(_a)^2-1)/(_b(_a)*_a), _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.188 (sec). Leaf size: 19
dsolve([2*x*diff(y(x),x)*diff(y(x),x$2)=diff(y(x),x)^2-1,y(1) = 0, D(y)(1) = sqrt(3)],y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (2 x +1\right )^{\frac {3}{2}}}{3}-\sqrt {3} \]
✓ Solution by Mathematica
Time used: 0.121 (sec). Leaf size: 26
DSolve[{2*x*y'[x]*y''[x]==y'[x]^2-1,{y[1]==0,y'[1]==Sqrt[3]}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{3} \left ((2 x+1)^{3/2}-3 \sqrt {3}\right ) \]