problem 31

Internal problem ID [2303]
Internal ﬁle name [OUTPUT/2303_Tuesday_February_27_2024_08_25_35_AM_38217236/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: 31.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable"

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

18.31.1 Solving as second order ode can be made integrable ode

Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }-y^{3} y^{\prime } = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }-y^{3} y^{\prime }\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\frac {y^{4}}{4} = c_2 \end {align*}

Which is now solved for \(y\). 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 }&=\frac {\sqrt {2 y^{4}+8 c_{1}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2 y^{4}+8 c_{1}}}{2} \tag {2} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {2}{\sqrt {2 y^{4}+8 c_{1}}}d y &= \int d x \\ \frac {\sqrt {2}\, \sqrt {4-\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{2 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {2 y^{4}+8 c_{1}}}&=x +c_{2} \\ \end{align*} Solving equation (2)

Integrating both sides gives \begin{align*} \int -\frac {2}{\sqrt {2 y^{4}+8 c_{1}}}d y &= \int d x \\ -\frac {\sqrt {2}\, \sqrt {4-\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{2 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {2 y^{4}+8 c_{1}}}&=x +c_{3} \\ \end{align*} Initial conditions are used to solve for the constants of integration.

Looking at the First solution \begin {align*} \frac {\sqrt {2}\, \sqrt {4-\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{2 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {2 y^{4}+8 c_{1}}} = x +c_{2} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = -1\) and \(x = 0\) in the above gives \begin {align*} \frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\frac {2 \sqrt {c_{1}}-i}{\sqrt {c_{1}}}}\, \sqrt {\frac {2 \sqrt {c_{1}}+i}{\sqrt {c_{1}}}}\, \sqrt {2}\, c_{1}^{\frac {1}{4}} \operatorname {EllipticF}\left (\frac {\frac {1}{2}-\frac {i}{2}}{c_{1}^{\frac {1}{4}}}, i\right )}{\sqrt {1+4 c_{1}}} = c_{2}\tag {1A} \end {align*}

Looking at the Second solution \begin {align*} -\frac {\sqrt {2}\, \sqrt {4-\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 i y^{2}}{\sqrt {c_{1}}}}\, \operatorname {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {i}{\sqrt {c_{1}}}}}{2}, i\right )}{2 \sqrt {\frac {i}{\sqrt {c_{1}}}}\, \sqrt {2 y^{4}+8 c_{1}}} = x +c_{3} \tag {2} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = -1\) and \(x = 0\) in the above gives \begin {align*} \frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {2 \sqrt {c_{1}}-i}{\sqrt {c_{1}}}}\, \sqrt {\frac {2 \sqrt {c_{1}}+i}{\sqrt {c_{1}}}}\, \sqrt {2}\, c_{1}^{\frac {1}{4}} \operatorname {EllipticF}\left (\frac {\frac {1}{2}-\frac {i}{2}}{c_{1}^{\frac {1}{4}}}, i\right )}{\sqrt {1+4 c_{1}}} = c_{3}\tag {1A} \end {align*}

Veriﬁcation of solutions N/A

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

Where \(f(y)=y^{3}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= y^{3} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {y^{3} \,d y} \\ \frac {p^{2}}{2}&=\frac {y^{4}}{4}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {y^{4}}{4}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=-1\) and \(p=\frac {\sqrt {2}}{2}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} = 0 \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2}-\frac {y^{4}}{4} = 0 \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*} \frac {{y^{\prime }}^{2}}{2}-\frac {y^{4}}{4} = 0 \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 }&=\frac {\sqrt {2}\, y^{2}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2}\, y^{2}}{2} \tag {2} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {\sqrt {2}}{y^{2}}d y &= \int {dx}\\ -\frac {\sqrt {2}}{y}&= 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*} \sqrt {2} = c_{2} \end {align*}

Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -\frac {\sqrt {2}}{y} = x +\sqrt {2} \end {align*}

Integrating both sides gives \begin {align*} \int -\frac {\sqrt {2}}{y^{2}}d y &= \int {dx}\\ \frac {\sqrt {2}}{y}&= 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} = c_{3} \end {align*}

Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {\sqrt {2}}{y} = x -\sqrt {2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sqrt {2}}{x +\sqrt {2}} \\ \tag{2} y &= -\frac {\sqrt {2}}{\sqrt {2}-x} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {\sqrt {2}}{x +\sqrt {2}} \] Veriﬁed OK.

\[ y = -\frac {\sqrt {2}}{\sqrt {2}-x} \] 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 
<- 2nd_order JacobiSN successful`

18.31 problem 31

18.31.1 Solving as second order ode can be made integrable ode

18.31.2 Solving as second order ode missing x ode