2.1.65 problem 65
Internal
problem
ID
[8203]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
65
Date
solved
:
Sunday, November 10, 2024 at 03:09:43 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*}