8.8 problem 13.2 (b)

Internal problem ID [13479]
Internal file name [OUTPUT/12651_Friday_February_16_2024_12_04_40_AM_91568826/index.tex]

Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises page 259
Problem number: 13.2 (b).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "second_order_ode_missing_y", "exact nonlinear second order ode"

Maple gives the following as the ode type

[[_2nd_order, _missing_x], [_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_poly_yn]]

\[ \boxed {y^{\prime } y^{\prime \prime }=1} \]

8.8.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int y^{\prime } y^{\prime \prime }d x &= \int 1d x\\ \frac {{y^{\prime }}^{2}}{2} = x + c_{1} \end {align*}

Which is now solved for \(y\). Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {2 c_{1} +2 x} \tag {1} \\ y^{\prime }&=-\sqrt {2 c_{1} +2 x} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Integrating both sides gives \begin {align*} y &= \int { \sqrt {2 c_{1} +2 x}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \end {align*}

Solving equation (2)

Integrating both sides gives \begin {align*} y &= \int { -\sqrt {2 c_{1} +2 x}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \\ \tag{2} y &= -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \] Verified OK.

\[ y = -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \] Verified OK.

8.8.2 Solving as second order ode missing y ode

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 \left (x \right ) p^{\prime }\left (x \right )-1 = 0 \end {align*}

Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int p d p &= x +c_{1}\\ \frac {p^{2}}{2}&=x +c_{1} \end {align*}

Solving for \(p\) gives these solutions \begin {align*} p_1&=\sqrt {2 c_{1} +2 x}\\ p_2&=-\sqrt {2 c_{1} +2 x} \end {align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \sqrt {2 c_{1} +2 x} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { \sqrt {2 c_{1} +2 x}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \end {align*}

Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\sqrt {2 c_{1} +2 x} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { -\sqrt {2 c_{1} +2 x}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \\ \tag{2} y &= -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \] Verified OK.

\[ y = -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \] Verified OK.

8.8.3 Solving as second order ode missing x ode

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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right ) = 1 \end {align*}

Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int p^{2}d p &= y +c_{1}\\ \frac {p^{3}}{3}&=y +c_{1} \end {align*}

Solving for \(p\) gives these solutions \begin {align*} p_1&=\left (3 c_{1} +3 y \right )^{\frac {1}{3}}\\ p_2&=-\frac {\left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}\\ p_3&=-\frac {\left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{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 } = \left (3 c_{1} +3 y\right )^{\frac {1}{3}} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {1}{\left (3 c_{1} +3 y \right )^{\frac {1}{3}}}d y &= \int d x \\ \frac {\left (3 c_{1} +3 y\right )^{\frac {2}{3}}}{2}&=x +c_{2} \\ \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 } = -\frac {\left (3 c_{1} +3 y\right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (3 c_{1} +3 y\right )^{\frac {1}{3}}}{2} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {1}{-\frac {\left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}}d y &= \int d x \\ -\frac {\left (3 c_{1} +3 y\right )^{\frac {2}{3}}}{1+i \sqrt {3}}&=x +c_{3} \\ \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 } = -\frac {\left (3 c_{1} +3 y\right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (3 c_{1} +3 y\right )^{\frac {1}{3}}}{2} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {1}{-\frac {\left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (3 c_{1} +3 y \right )^{\frac {1}{3}}}{2}}d y &= \int d x \\ \frac {\left (3 c_{1} +3 y\right )^{\frac {2}{3}}}{i \sqrt {3}-1}&=x +c_{4} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -c_{1} +\frac {\left (2 c_{2} +2 x \right )^{\frac {3}{2}}}{3} \\ \tag{2} y &= -c_{1} +\frac {\left (-i \sqrt {3}\, c_{3} -i \sqrt {3}\, x -c_{3} -x \right )^{\frac {3}{2}}}{3} \\ \tag{3} y &= -c_{1} +\frac {\left (i \sqrt {3}\, c_{4} +i \sqrt {3}\, x -c_{4} -x \right )^{\frac {3}{2}}}{3} \\ \end{align*}

Verification of solutions

\[ y = -c_{1} +\frac {\left (2 c_{2} +2 x \right )^{\frac {3}{2}}}{3} \] Verified OK.

\[ y = -c_{1} +\frac {\left (-i \sqrt {3}\, c_{3} -i \sqrt {3}\, x -c_{3} -x \right )^{\frac {3}{2}}}{3} \] Verified OK.

\[ y = -c_{1} +\frac {\left (i \sqrt {3}\, c_{4} +i \sqrt {3}\, x -c_{4} -x \right )^{\frac {3}{2}}}{3} \] Verified OK.

8.8.4 Solving as exact nonlinear second order ode ode

An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}

Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}

Looking at the the ode given we see that \begin {align*} a_2 &= y^{\prime }\\ a_1 &= 0\\ a_0 &= -1 \end {align*}

Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {y^{\prime }\,d y'} + \int {0\,d y} + \int {-1\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} \frac {{y^{\prime }}^{2}}{2}-x = c_{1} \end {align*}

Which is now solved Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {2 c_{1} +2 x} \tag {1} \\ y^{\prime }&=-\sqrt {2 c_{1} +2 x} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Integrating both sides gives \begin {align*} y &= \int { \sqrt {2 c_{1} +2 x}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \end {align*}

Solving equation (2)

Integrating both sides gives \begin {align*} y &= \int { -\sqrt {2 c_{1} +2 x}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \\ \tag{2} y &= -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \] Verified OK.

\[ y = -\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{3} \] Verified OK.

8.8.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d x}y^{\prime }\right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (x \right ) u^{\prime }\left (x \right )=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int u \left (x \right ) u^{\prime }\left (x \right )d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (x \right )^{2}}{2}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\sqrt {2 c_{1} +2 x}, u \left (x \right )=-\sqrt {2 c_{1} +2 x}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\sqrt {2 c_{1} +2 x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\sqrt {2 c_{1} +2 x} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \sqrt {2 c_{1} +2 x}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\sqrt {2 c_{1} +2 x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\sqrt {2 c_{1} +2 x} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\sqrt {2 c_{1} +2 x}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\left (2 c_{1} +2 x \right )^{\frac {3}{2}}}{3}+c_{2} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
`, `-> Computing symmetries using: way = 3 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = 1/_b(_a), _b(_a), HINT = [[1, 0], [_a, (1/2)*_b]]`   *** Sublevel 2 *** 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[1, 0], [_a, 1/2*_b]
 

✓ Solution by Maple

Time used: 0.094 (sec). Leaf size: 40

dsolve(diff(y(x),x)*diff(y(x),x$2)=1,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= \frac {\left (2 x +2 c_{1} \right )^{\frac {3}{2}}}{3}+c_{2} \\ y \left (x \right ) &= \frac {\left (-2 c_{1} -2 x \right ) \sqrt {2 x +2 c_{1}}}{3}+c_{2} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.014 (sec). Leaf size: 49

DSolve[y'[x]*y''[x]==1,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to c_2-\frac {2}{3} \sqrt {2} (x+c_1){}^{3/2} \\ y(x)\to \frac {2}{3} \sqrt {2} (x+c_1){}^{3/2}+c_2 \\ \end{align*}