2.1.65 problem 65

Internal problem ID [8453]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 65
Date solved : Thursday, December 12, 2024 at 09:10:07 AM
CAS classification : [[_2nd_order, _quadrature]]

Solve

\begin{align*} y^{2} y^{\prime \prime }&=0 \end{align*}

Solved as second order missing x ode

Time used: 0.100 (sec)

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 \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

Since the ode has the form \(p^{\prime }=f(y)\), then we only need to integrate \(f(y)\).

\begin{align*} \int {dp} &= \int {0\, dy} + c_1 \\ p &= c_1 \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 } = c_1 \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {c_1\, dx}\\ y &= c_1 x + c_2 \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 x +c_2 \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right )^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\mathit {C1} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful`
 

Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 13

dsolve(y(x)^2*diff(diff(y(x),x),x) = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= 0 \\ y &= c_{1} x +c_{2} \\ \end{align*}

Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 17

DSolve[{y[x]^2*D[y[x],{x,2}]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to 0 \\ y(x)\to c_2 x+c_1 \\ \end{align*}