Internal problem ID [4658]
Internal file name [OUTPUT/4151_Sunday_June_05_2022_12_29_37_PM_65462958/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.8, page 504.
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"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime }-\frac {3 k y^{2}}{2}=0} \]
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }-\frac {3 k y^{2} y^{\prime }}{2} = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }-\frac {3 k y^{2} y^{\prime }}{2}\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\frac {k y^{3}}{2} = 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 {k y^{3}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {k y^{3}+2 c_{1}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {k \,y^{3}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {k \,y^{3}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} = x +c_{3} \] Verified OK.
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 )-\frac {3 k \,y^{2}}{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 k \,y^{2}}{2 p} \end {align*}
Where \(f(y)=\frac {3 k \,y^{2}}{2}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {3 k \,y^{2}}{2} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {3 k \,y^{2}}{2} \,d y} \\ \frac {p^{2}}{2}&=\frac {k \,y^{3}}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {k \,y^{3}}{2}-c_{1} = 0 \] 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 {k y^{3}}{2}-c_{1} = 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 }&=\sqrt {k y^{3}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {k y^{3}+2 c_{1}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {1}{\sqrt {k \,y^{3}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {k \,y^{3}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{3} k +2 c_{1}}}d \textit {\_a} = x +c_{3} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP <- 2nd_order WeierstrassP successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 15
dsolve(diff(y(x),x$2)=3/2*k*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {4 \operatorname {WeierstrassP}\left (x +c_{1} , 0, c_{2}\right )}{k} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y''[x]==3/2*(k*y[x]^2),y[x],x,IncludeSingularSolutions -> True]
Not solved