Internal problem ID [2307]
Internal file name [OUTPUT/2307_Tuesday_February_27_2024_08_25_39_AM_11739347/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 35, page 157
Problem number: 35.
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]]
\[ \boxed {y y^{\prime \prime }-y^{3}-{y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2] \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 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-y^{3}-p \left (y \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Using the change of variables \(p \left (y \right ) = u \left (y \right ) y\) on the above ode results in new ode in \(u \left (y \right )\) \begin {align*} y^{2} u \left (y \right ) \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )-u \left (y \right )^{2} y^{2} = y^{3} \end {align*}
Integrating both sides gives \begin {align*} \int u d u &= y +c_{2}\\ \frac {u^{2}}{2}&=y +c_{2} \end {align*}
Solving for \(u\) gives these solutions \begin {align*} u_1&=\sqrt {2 c_{2} +2 y}\\ u_2&=-\sqrt {2 c_{2} +2 y} \end {align*}
Therefore the solution \(p \left (y \right )\) is \begin {align*} p \left (y \right )&=y u\\ &=y \sqrt {2 c_{2} +2 y} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(y=1\) and \(p=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = \sqrt {2 c_{2} +2} \end {align*}
The solutions are \begin {align*} c_{2} = 1 \end {align*}
Trying the constant \begin {align*} c_{2} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} p \left (y \right )&=\sqrt {2 y +2}\, y \end {align*}
The constant \(c_{2} = 1\) 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*} y^{\prime } = \sqrt {2 y+2}\, y \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {2 y +2}\, y}d y &= \int {dx}\\ -\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2 y +2}\, \sqrt {2}}{2}\right )&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )+\frac {i \sqrt {2}\, \pi }{2} = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = -\sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )+\frac {i \sqrt {2}\, \pi }{2} \end {align*}
Trying the constant \begin {align*} c_{3} = -\sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )+\frac {i \sqrt {2}\, \pi }{2} \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} -\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2 y +2}\, \sqrt {2}}{2}\right ) = x -\sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )+\frac {i \sqrt {2}\, \pi }{2} \end {align*}
The constant \(c_{3} = -\sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )+\frac {i \sqrt {2}\, \pi }{2}\) 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 &= \tanh \left (\frac {\left (-i \sqrt {2}\, \pi +2 \sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )-2 x \right ) \sqrt {2}}{4}\right )^{2}-1 \\ \end{align*}
Verification of solutions
\[ y = \tanh \left (\frac {\left (-i \sqrt {2}\, \pi +2 \sqrt {2}\, \operatorname {arccoth}\left (\sqrt {2}\right )-2 x \right ) \sqrt {2}}{4}\right )^{2}-1 \] Verified OK.
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)-(_a^3+_b(_a)^2)/_a = 0, _b(_a), HINT = [[_a, (3/2)*_b]]` *** Sublevel symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 3/2*_b]
✓ Solution by Maple
Time used: 18.109 (sec). Leaf size: 25
dsolve([y(x)*diff(y(x),x$2)=y(x)^3+diff(y(x),x)^2,y(0) = 1, D(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = -\operatorname {sech}\left (\frac {\sqrt {2}\, \left (x -\sqrt {2}\, \operatorname {arctanh}\left (\sqrt {2}\right )\right )}{2}\right )^{2} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y[x]*y''[x]==y[x]^3+y'[x]^2,{y[0]==1,y'[0]==2}},y[x],x,IncludeSingularSolutions -> True]
{}