problem 38

Internal problem ID [2310]
Internal ﬁle name [OUTPUT/2310_Tuesday_February_27_2024_08_25_49_AM_38444091/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: 38.
ODE order: 2.
ODE degree: 1.

\[ \boxed {2 y y^{\prime \prime }-y^{3}-2 {y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = -1, y^{\prime }\left (0\right ) = 0] \end {align*}

18.38.1 Solving as second order ode missing x ode

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*} 2 y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-y^{3}-2 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*} 2 y^{2} u \left (y \right ) \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )-2 u \left (y \right )^{2} y^{2} = y^{3} \end {align*}

Integrating both sides gives \begin {align*} \int 2 u d u &= y +c_{2}\\ u^{2}&=y +c_{2} \end {align*}

Solving for \(u\) gives these solutions \begin {align*} u_1&=\sqrt {y +c_{2}}\\ u_2&=-\sqrt {y +c_{2}} \end {align*}

Therefore the solution \(p \left (y \right )\) is \begin {align*} p \left (y \right )&=y u\\ &=y \sqrt {y +c_{2}} \end {align*}

Initial conditions are used to solve for \(c_{2}\). Substituting \(y=-1\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = -\sqrt {c_{2} -1} \end {align*}

Substituting this in the general solution gives \begin {align*} p \left (y \right )&=\sqrt {y +1}\, y \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 } = \sqrt {1+y}\, y \end {align*}

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*}

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

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)-(1/2)*(_a^3+2*_b(_a)^2)/_a = 0, _b(_a), HINT = [[_a, (3/2)*_b]]`   *** S 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[_a, 3/2*_b]

\[ y \left (x \right ) = \operatorname {RootOf}\left (2 \,\operatorname {arctanh}\left (\sqrt {\textit {\_Z} +1}\right )+x \right ) \]

18.38 problem 38

18.38.1 Solving as second order ode missing x ode