Internal problem ID [7400]
Internal file name [OUTPUT/6367_Sunday_June_05_2022_04_41_49_PM_91902991/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 11.
ODE order: 2.
ODE degree: 3.
The type(s) of ODE detected by this program : "second_order_ode_high_degree", "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _quadrature]]
\[ \boxed {{y^{\prime \prime }}^{3}=0} \]
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 the given ode for \(p^{\prime }\left (x \right )\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} p^{\prime }\left (x \right )&=0 \tag {1} \\ p^{\prime }\left (x \right )&=0 \tag {2} \\ p^{\prime }\left (x \right )&=0 \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} p \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{1} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} p \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{2} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} p \left (x \right ) &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{3} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = c_{1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{1}\,\mathop {\mathrm {d}x}}\\ &= c_{1} x +c_{4} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = c_{2} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{2}\,\mathop {\mathrm {d}x}}\\ &= c_{2} x +c_{5} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = c_{3} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { c_{3}\,\mathop {\mathrm {d}x}}\\ &= c_{3} x +c_{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{4} \\ \tag{2} y &= c_{2} x +c_{5} \\ \tag{3} y &= c_{3} x +c_{6} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{4} \] Verified OK.
\[ y = c_{2} x +c_{5} \] Verified OK.
\[ y = c_{3} x +c_{6} \] 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 )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end {align*}
Gives the following equations \begin {align*} p \left (y \right )^{3} = 0\tag {1} \\ \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Since \(p \left (y \right )^{3} = 0\), is missing derivative in \(p\) then it is an algebraic equation. Solving for \(p \left (y \right )\). \begin {align*} \end {align*}
Solving ODE (2) Solving the given ode for \(\frac {d}{d y}p \left (y \right )\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} \frac {d}{d y}p \left (y \right )&=0 \tag {1} \\ \frac {d}{d y}p \left (y \right )&=0 \tag {2} \\ \frac {d}{d y}p \left (y \right )&=0 \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { 0\,\mathop {\mathrm {d}y}}\\ &= c_{1} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { 0\,\mathop {\mathrm {d}y}}\\ &= c_{2} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} p \left (y \right ) &= \int { 0\,\mathop {\mathrm {d}y}}\\ &= 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*}
Integrating both sides gives \begin {align*} y &= \int { c_{1}\,\mathop {\mathrm {d}x}}\\ &= 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*}
Integrating both sides gives \begin {align*} y &= \int { c_{2}\,\mathop {\mathrm {d}x}}\\ &= 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*}
Integrating both sides gives \begin {align*} y &= \int { c_{3}\,\mathop {\mathrm {d}x}}\\ &= c_{3} x +c_{6} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{4} \\ \tag{2} y &= c_{2} x +c_{5} \\ \tag{3} y &= c_{3} x +c_{6} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{4} \] Verified OK.
\[ y = c_{2} x +c_{5} \] Verified OK.
\[ y = c_{3} x +c_{6} \] Verified OK.
Solving for \(y^{\prime \prime }\) from the ode gives \begin{align*} \tag{1} y^{\prime \prime } &= 0 \\ \end{align*} Now each ode is solved. Integrating twice gives the solution \[ y= c_{1} x + c_{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right )^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=0 \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (x \right )\hspace {3pt}\textrm {by}\hspace {3pt} x \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=x \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{2} x +c_{1} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 9
dsolve(diff(y(x),x$2)^3=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x +c_{2} \]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 12
DSolve[(y''[x])^3==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 x+c_1 \]