2.1.2 problem 2
Internal
problem
ID
[8749]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
2
Date
solved
:
Thursday, December 12, 2024 at 09:42:50 AM
CAS
classification
:
[[_2nd_order, _quadrature]]
Solve
\begin{align*} {y^{\prime \prime }}^{2}&=0 \end{align*}
Solved as second order missing y ode
Time used: 0.147 (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 )}^{2} = 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 \\
\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*}
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_3 \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_4 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 x +c_3 \\
y &= c_2 x +c_4 \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C2} x +\mathit {C1} \end {array} \]
Maple trace
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful`
Maple dsolve solution
Solving time : 0.024 (sec)
Leaf size : 9
dsolve(diff(diff(y(x),x),x)^2 = 0,
y(x),singsol=all)
\[
y = c_{1} x +c_{2}
\]
Mathematica DSolve solution
Solving time : 0.002 (sec)
Leaf size : 12
DSolve[{(D[y[x],{x,2}])^2==0,{}},
y[x],x,IncludeSingularSolutions->True]
\[
y(x)\to c_2 x+c_1
\]