1.11 problem 11

Internal problem ID [8048]
Book : Second order enumerated odes
Section : section 1
Problem number : 11
Date solved : Monday, October 21, 2024 at 04:44:46 PM
CAS classification : [[_2nd_order, _quadrature]]

Solve

\begin{align*} {y^{\prime \prime }}^{3}&=0 \end{align*}

1.11.1 Solved as second order missing y ode

Time used: 0.154 (sec)

This is second order ode with missing dependent variable \(y\). Let

\begin{align*} p(x) &= y^{\prime } \end{align*}

Then

\begin{align*} p'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} {p^{\prime }\left (x \right )}^{3} = 0 \end{align*}

Which is now solve for \(p(x)\) as first order ode. Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} p^{\prime }\left (x \right )&=0 \\ \tag{2} p^{\prime }\left (x \right )&=0 \\ \tag{3} p^{\prime }\left (x \right )&=0 \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

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

\begin{align*} \int {dp} &= \int {0\, dx} + c_1 \\ p \left (x \right ) &= c_1 \end{align*}

Solving Eq. (2)

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

\begin{align*} \int {dp} &= \int {0\, dx} + c_2 \\ p \left (x \right ) &= c_2 \end{align*}

Solving Eq. (3)

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

\begin{align*} \int {dp} &= \int {0\, dx} + c_3 \\ p \left (x \right ) &= c_3 \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_4 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = c_2 \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_2\, dx}\\ y &= c_2 x + c_5 \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = c_3 \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_3\, dx}\\ y &= c_3 x + c_6 \end{align*}

Will add steps showing solving for IC soon.

1.11.2 Maple step by step solution

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

1.11.3 Maple trace

Methods for second order ODEs:
 

1.11.4 Maple dsolve solution

Solving time : 0.004 (sec)
Leaf size : 9

dsolve(diff(diff(y(x),x),x)^3 = 0, 
       y(x),singsol=all)
 

\[ y = c_1 x +c_2 \]

1.11.5 Mathematica DSolve solution

Solving time : 0.002 (sec)
Leaf size : 12

DSolve[{(D[y[x],{x,2}])^3==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to c_2 x+c_1 \]