problem 354

Internal problem ID [15211]
Internal ﬁle name [OUTPUT/15212_Tuesday_April_23_2024_04_54_15_PM_17262081/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: 354.
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"

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

14.28.1 Solving as second order ode can be made integrable ode

Multiplying the ode by \(y^{\prime }\) gives \[ 2 y^{\prime } y^{\prime \prime }-3 y^{2} y^{\prime } = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (2 y^{\prime } y^{\prime \prime }-3 y^{2} y^{\prime }\right )d x &= 0 \\ {y^{\prime }}^{2}-y^{3} = 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 }&=\sqrt {y^{3}+c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {y^{3}+c_{1}} \tag {2} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {y^{3}+c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}

Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {y^{3}+c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a}&= c_{3} +x \end {align*}

Looking at the First solution \begin {align*} \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} = 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 \(x = -2\) in the above gives \begin {align*} \int _{}^{1}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} = -2+c_{2}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = \sqrt {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} \right )+x +c_{2} \right )}^{3}+c_{1}} \end {align*}

substituting \(y^{\prime } = -1\) and \(x = -2\) in the above gives \begin {align*} -1 = \sqrt {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} \right )-2+c_{2} \right )}^{3}+c_{1}}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.

Looking at the Second solution \begin {align*} \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} = c_{3} +x \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 \(x = -2\) in the above gives \begin {align*} -\left (\int _{}^{1}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} \right ) = c_{3} -2\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = -\sqrt {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} \right )+c_{3} +x \right )}^{3}+c_{1}} \end {align*}

substituting \(y^{\prime } = -1\) and \(x = -2\) in the above gives \begin {align*} -1 = -\sqrt {\operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {\textit {\_a}^{3}+c_{1}}}d \textit {\_a} +c_{3} -2\right )^{3}+c_{1}}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). There is no solution for the constants of integrations. This solution is removed.

Veriﬁcation of solutions N/A

14.28.2 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*} 2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-3 y^{2} = 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 {3 y^{2}}{2 p} \end {align*}

Where \(f(y)=\frac {3 y^{2}}{2}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {3 y^{2}}{2} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {3 y^{2}}{2} \,d y} \\ \frac {p^{2}}{2}&=\frac {y^{3}}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {y^{3}}{2}-c_{1} = 0 \] 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*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2}-\frac {y^{3}}{2} = 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^{3}}{2} = 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 }&=y^{\frac {3}{2}} \tag {1} \\ y^{\prime }&=-y^{\frac {3}{2}} \tag {2} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{y^{\frac {3}{2}}}d y &= \int {dx}\\ -\frac {2}{\sqrt {y}}&= x +c_{2} \end {align*}

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

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

Integrating both sides gives \begin {align*} \int -\frac {1}{y^{\frac {3}{2}}}d y &= \int {dx}\\ \frac {2}{\sqrt {y}}&= c_{3} +x \end {align*}

Initial conditions are used to solve for \(c_{3}\). Substituting \(x=-2\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{3} -2 \end {align*}

Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {2}{\sqrt {y}} = 4+x \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {4}{x^{2}} \\ \tag{2} y &= \frac {4}{\left (4+x \right )^{2}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {4}{x^{2}} \] Warning, solution could not be veriﬁed

\[ y = \frac {4}{\left (4+x \right )^{2}} \] Veriﬁed OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
<- 2nd_order WeierstrassP successful`

14.28 problem 354

14.28.1 Solving as second order ode can be made integrable ode

14.28.2 Solving as second order ode missing x ode