Internal problem ID [15210]
Internal file name [OUTPUT/15211_Tuesday_April_23_2024_04_54_12_PM_97607109/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: 353.
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]]
\[ \boxed {3 y^{\prime } y^{\prime \prime }-2 y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 1] \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*} 3 p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )-2 y = 0 \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 {2 y}{3 p^{2}} \end {align*}
Where \(f(y)=\frac {2 y}{3}\) and \(g(p)=\frac {1}{p^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p^{2}}} \,dp &= \frac {2 y}{3} \,d y \\ \int { \frac {1}{\frac {1}{p^{2}}} \,dp} &= \int {\frac {2 y}{3} \,d y} \\ \frac {p^{3}}{3}&=\frac {y^{2}}{3}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{3}}{3}-\frac {y^{2}}{3}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=1\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{3}}{3}-\frac {y^{2}}{3} = 0 \end {align*}
The constant \(c_{1} = 0\) 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^{\prime }}^{3}}{3}-\frac {y^{2}}{3} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=y^{\frac {2}{3}} \tag {1} \\ y^{\prime }&=-\frac {y^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, y^{\frac {2}{3}}}{2} \tag {2} \\ y^{\prime }&=-\frac {y^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, y^{\frac {2}{3}}}{2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {1}{y^{\frac {2}{3}}}d y &= \int {dx}\\ 3 y^{\frac {1}{3}}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 3 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 3 \end {align*}
Trying the constant \begin {align*} c_{2} = 3 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} 3 y^{\frac {1}{3}} = x +3 \end {align*}
The constant \(c_{2} = 3\) gives valid solution.
Solving equation (2)
Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{\frac {2}{3}}}{2}+\frac {i \sqrt {3}\, y^{\frac {2}{3}}}{2}}d y &= \int {dx}\\ \frac {6 y^{\frac {1}{3}}}{i \sqrt {3}-1}&= c_{3} +x \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*} \frac {6}{i \sqrt {3}-1} = c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = \frac {6}{i \sqrt {3}-1} \end {align*}
Trying the constant \begin {align*} c_{3} = \frac {6}{i \sqrt {3}-1} \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {6 y^{\frac {1}{3}}}{i \sqrt {3}-1} = \frac {i \sqrt {3}\, x -x +6}{i \sqrt {3}-1} \end {align*}
The constant \(c_{3} = \frac {6}{i \sqrt {3}-1}\) gives valid solution.
Solving equation (3)
Integrating both sides gives \begin {align*} \int \frac {1}{-\frac {y^{\frac {2}{3}}}{2}-\frac {i \sqrt {3}\, y^{\frac {2}{3}}}{2}}d y &= \int {dx}\\ -\frac {6 y^{\frac {1}{3}}}{1+i \sqrt {3}}&= x +c_{4} \end {align*}
Initial conditions are used to solve for \(c_{4}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -\frac {6}{1+i \sqrt {3}} = c_{4} \end {align*}
The solutions are \begin {align*} c_{4} = -\frac {6}{1+i \sqrt {3}} \end {align*}
Trying the constant \begin {align*} c_{4} = -\frac {6}{1+i \sqrt {3}} \end {align*}
Substituting \(c_{4}\) found above in the general solution gives \begin {align*} -\frac {6 y^{\frac {1}{3}}}{1+i \sqrt {3}} = \frac {i \sqrt {3}\, x +x -6}{1+i \sqrt {3}} \end {align*}
The constant \(c_{4} = -\frac {6}{1+i \sqrt {3}}\) gives valid solution.
The solution \[ -\frac {6 y^{\frac {1}{3}}}{1+i \sqrt {3}} = \frac {i \sqrt {3}\, x +x -6}{1+i \sqrt {3}} \] can be simplified to \[ -6 y^{\frac {1}{3}} = i \sqrt {3}\, x +x -6 \] 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 {1}{27} x^{3}+\frac {1}{3} x^{2}+x +1 \\ \tag{2} y &= -\frac {i \left (2 i x^{3}-9 i x^{2}+9 \sqrt {3}\, x^{2}-27 i x -27 \sqrt {3}\, x +54 i\right )}{54} \\ \tag{3} y &= \frac {i x^{2} \sqrt {3}}{6}-\frac {i \sqrt {3}\, x}{2}+\frac {x^{3}}{27}-\frac {x^{2}}{6}-\frac {x}{2}+1 \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{27} x^{3}+\frac {1}{3} x^{2}+x +1 \] Verified OK.
\[ y = -\frac {i \left (2 i x^{3}-9 i x^{2}+9 \sqrt {3}\, x^{2}-27 i x -27 \sqrt {3}\, x +54 i\right )}{54} \] Warning, solution could not be verified
\[ y = \frac {i x^{2} \sqrt {3}}{6}-\frac {i \sqrt {3}\, x}{2}+\frac {x^{3}}{27}-\frac {x^{2}}{6}-\frac {x}{2}+1 \] Warning, solution could not be verified
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** 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)-(2/3)*_a/_b(_a) = 0, _b(_a), HINT = [[_a, (2/3)*_b]]` *** Sublev symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 2/3*_b]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 11
dsolve([3*diff(y(x),x)*diff(y(x),x$2)=2*y(x),y(0) = 1, D(y)(0) = 1],y(x), singsol=all)
\[ y = \frac {\left (x +3\right )^{3}}{27} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{3*y'[x]*y''[x]==2*y[x],{y[0]==1,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
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