2.8.31 Problem 13.5 (g)

Internal problem ID [16412]
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 : 13.5 (g)
Date solved : Wednesday, December 10, 2025 at 09:35:50 PM
CAS classification : [[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]

2.8.31.1 second order ode missing x

0.243 (sec)

\begin{align*} y y^{\prime \prime }&=2 {y^{\prime }}^{2} \\ \end{align*}

Entering second order ode missing \(x\) solverThis 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 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = 2 p \left (y \right )^{2} \end{align*}

Which is now solved as first order ode for \(p(y)\).

2.8.31.2 Solved by factoring the differential equation

Time used: 0.031 (sec)

\begin{align*} y p p^{\prime }&=2 p^{2} \\ \end{align*}

Writing the ode as

\begin{align*} \left (p\right )\left (p^{\prime } y -2 p\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} p &= 0 \\ \tag{2} p^{\prime } y -2 p &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Entering zero order ode solverSolving for \(p\) from

\begin{align*} p = 0 \end{align*}

Solving gives

\begin{align*} p &= 0 \\ \end{align*}

Solving equation (2)

Entering first order ode linear solverIn canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(y) &=-\frac {2}{y}\\ p(y) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dy}}\\ &= {\mathrm e}^{\int -\frac {2}{y}d y}\\ &= \frac {1}{y^{2}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \mu p &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (\frac {p}{y^{2}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} \frac {p}{y^{2}}&= \int {0 \,dy} + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{y^{2}}\) gives the final solution

\[ p = y^{2} c_1 \]

Summary of solutions found

\begin{align*} p &= 0 \\ p &= y^{2} 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 } = 0 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + c_2 \\ y &= 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 } = y^{2} c_1 \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{y^{2} c_1}d y &= dx\\ -\frac {1}{y c_1}&= x +c_3 \end{align*}

Singular solutions are found by solving

\begin{align*} y^{2} c_1&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = 0 \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= -\frac {1}{\left (x +c_3 \right ) c_1} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= c_2 \\ y &= -\frac {1}{\left (x +c_3 \right ) c_1} \\ \end{align*}

2.8.31.3 SSolved using second order ode arccos transformation

0.857 (sec)

\begin{align*} y y^{\prime \prime }&=2 {y^{\prime }}^{2} \\ \end{align*}

Applying change of variable \(x = \arccos \left (\tau \right )\) to the above ode results in the following new ode

\[ \left (-\tau ^{2}+1\right ) y \left (\tau \right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )-y \left (\tau \right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right ) \tau = \left (-2 \tau ^{2}+2\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )^{2} \]

Which is now solved for \(y \left (\tau \right )\). Entering second order ode flip role solverreversing the roles of the dependent and independent variables, the ode becomes

\begin{align*} \frac {d^{2}}{d y^{2}}\tau \left (y \right ) = -\frac {\left (-\frac {2 \tau \left (y \right )^{2}}{\left (\frac {d}{d y}\tau \left (y \right )\right )^{2}}+\frac {y \tau \left (y \right )}{\frac {d}{d y}\tau \left (y \right )}+\frac {2}{\left (\frac {d}{d y}\tau \left (y \right )\right )^{2}}\right ) \left (\frac {d}{d y}\tau \left (y \right )\right )^{3}}{\left (-\tau \left (y \right )^{2}+1\right ) y} \end{align*}

Which is now solved for \(\tau \left (y \right )\) instead for \(y \left (\tau \right )\) Entering second order nonlinear solved by mainardi lioville method solverThe ode has the Liouville form given by

\begin{align*} \frac {d^{2}}{d y^{2}}\tau \left (y \right )+ f(y) \frac {d}{d y}\tau \left (y \right ) + g(\tau ) \left (\frac {d}{d y}\tau \left (y \right )\right )^{2} &= 0 \tag {1A} \end{align*}

Where in this problem

\begin{align*} f(y) &= \frac {2}{y}\\ g(\tau ) &= -\frac {\tau }{\tau ^{2}-1} \end{align*}

Dividing through by \(\frac {d}{d y}\tau \left (y \right )\) then Eq (1A) becomes

\begin{align*} \frac {\frac {d^{2}}{d y^{2}}\tau \left (y \right )}{\frac {d}{d y}\tau \left (y \right )}+ f + g \frac {d}{d y}\tau \left (y \right ) &= 0 \tag {2A} \end{align*}

But the first term in Eq (2A) can be written as

\begin{align*} \frac {\frac {d^{2}}{d y^{2}}\tau \left (y \right )}{\frac {d}{d y}\tau \left (y \right )}&= \frac {d}{dy} \ln \left ( \frac {d}{d y}\tau \left (y \right ) \right )\tag {3A} \end{align*}

And the last term in Eq (2A) can be written as

\begin{align*} g \frac {d\tau }{dy}&= \left ( \frac {d}{d\tau } \int g d \tau \right ) \frac {d\tau }{dy} \\ &= \frac {d}{dy} \int g d \tau \tag {4A} \end{align*}

Substituting (3A,4A) back into (2A) gives

\begin{align*} \frac {d}{dy} \ln \left ( \frac {d}{d y}\tau \left (y \right ) \right ) + \frac {d}{dy} \int g d \tau &= -f \tag {5A} \end{align*}

Integrating the above w.r.t. \(y\) gives

\begin{align*} \ln \left ( \frac {d}{d y}\tau \left (y \right ) \right ) + \int g d \tau &= - \int f d y + c_1 \end{align*}

Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives

\begin{align*} \frac {d}{d y}\tau \left (y \right ) &= c_2 e^{\int -g d \tau }\, e^{\int -f d y}\tag {6A} \end{align*}

Where \(c_2\) is a new arbitrary constant. But since \(g=-\frac {\tau }{\tau ^{2}-1}\) and \(f=\frac {2}{y}\), then

\begin{align*} \int -g d \tau &= \int \frac {\tau }{\tau ^{2}-1}d \tau \\ &= \frac {\ln \left (\tau \left (y \right )^{2}-1\right )}{2}\\ \int -f d y &= \int -\frac {2}{y}d y\\ &= -2 \ln \left (y \right ) \end{align*}

Substituting the above into Eq(6A) gives

\[ \frac {d}{d y}\tau \left (y \right ) = \frac {c_2 \sqrt {\tau \left (y \right )^{2}-1}}{y^{2}} \]

Which is now solved as first order separable ode. The ode

\begin{equation} \frac {d}{d y}\tau \left (y \right ) = \frac {c_2 \sqrt {\left (\tau \left (y \right )-1\right ) \left (\tau \left (y \right )+1\right )}}{y^{2}} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d y}\tau \left (y \right )&= \frac {c_2 \sqrt {\left (\tau \left (y \right )-1\right ) \left (\tau \left (y \right )+1\right )}}{y^{2}}\\ &= f(y) g(\tau ) \end{align*}

Where

\begin{align*} f(y) &= \frac {c_2}{y^{2}}\\ g(\tau ) &= \sqrt {\left (\tau -1\right ) \left (\tau +1\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(\tau )} \,d\tau } &= \int { f(y) \,dy} \\ \int { \frac {1}{\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}}\,d\tau } &= \int { \frac {c_2}{y^{2}} \,dy} \\ \end{align*}

\[ \ln \left (\tau \left (y \right )+\sqrt {\tau \left (y \right )^{2}-1}\right )=-\frac {c_2}{y}+c_3 \]

We now need to find the singular solutions, these are found by finding for what values \(g(\tau )\) is zero, since we had to divide by this above. Solving \(g(\tau )=0\) or

\[ \sqrt {\left (\tau -1\right ) \left (\tau +1\right )}=0 \]

for \(\tau \left (y \right )\) gives

\begin{align*} \tau \left (y \right )&=-1\\ \tau \left (y \right )&=1 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (\tau \left (y \right )+\sqrt {\tau \left (y \right )^{2}-1}\right ) &= -\frac {c_2}{y}+c_3 \\ \tau \left (y \right ) &= -1 \\ \tau \left (y \right ) &= 1 \\ \end{align*}

Solving for \(\tau \left (y \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} \tau \left (y \right )&=-1\\ \tau \left (y \right )&=1\\ \tau \left (y \right )&=\frac {\left ({\mathrm e}^{-\frac {2 \left (-c_3 y +c_2 \right )}{y}}+1\right ) {\mathrm e}^{\frac {-c_3 y +c_2}{y}}}{2} \end{align*}

Now that the reversed roles ode was solved, we will change back to the original roles. This results in the above solution becoming the following.

\begin{align*} \tau &= \frac {\left ({\mathrm e}^{-\frac {2 \left (-c_3 y+c_2 \right )}{y}}+1\right ) {\mathrm e}^{\frac {-c_3 y+c_2}{y}}}{2} \\ \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=-\frac {c_2}{-c_3 +\ln \left (\tau -\sqrt {\tau ^{2}-1}\right )}\\ y&=-\frac {c_2}{-c_3 +\ln \left (\tau +\sqrt {\tau ^{2}-1}\right )} \end{align*}

Applying change of variable \(\tau = \cos \left (x \right )\) to the solutions above gives

\begin{align*} y \left (x \right ) &= -\frac {c_2}{-c_3 +\ln \left (\cos \left (x \right )-\sqrt {\cos \left (x \right )^{2}-1}\right )} \\ y \left (x \right ) &= -\frac {c_2}{-c_3 +\ln \left (\cos \left (x \right )+\sqrt {\cos \left (x \right )^{2}-1}\right )} \\ \end{align*}

2.8.31.4 ✓ Maple. Time used: 0.012 (sec). Leaf size: 17

ode:=diff(diff(y(x),x),x)*y(x) = 2*diff(y(x),x)^2; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= 0 \\ y &= -\frac {1}{c_1 x +c_2} \\ \end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
<- 2nd_order Liouville successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )=2 \left (\frac {d}{d x}y \left (x \right )\right )^{2} \\ \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 {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=\frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} \frac {d}{d x}y \left (x \right )=u \left (y \right ),\frac {d^{2}}{d x^{2}}y \left (x \right )=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=2 u \left (y \right )^{2} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {2 u \left (y \right )}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {2}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int \frac {2}{y}d y +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=2 \ln \left (y \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{\mathit {C1}} y^{2} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & u \left (y \right )=\mathit {C1} \,y^{2} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\mathit {C1} \,y^{2} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=\frac {d}{d x}y \left (x \right ),y =y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\mathit {C1} y \left (x \right )^{2} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\mathit {C1} y \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )^{2}}=\mathit {C1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{y \left (x \right )^{2}}d x =\int \mathit {C1} d x +\mathit {C2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y \left (x \right )}=\mathit {C1} x +\mathit {C2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & y \left (x \right )=-\frac {1}{\mathit {C1} x +\mathit {C2}} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & \left \{y \left (x \right )=\frac {\mathit {C1}}{x +\mathit {C2}}, y \left (x \right )=\frac {\mathit {C2}}{\mathit {C1} x +1}\right \} \end {array} \]

2.8.31.5 ✓ Mathematica. Time used: 0.083 (sec). Leaf size: 19

ode=y[x]*D[y[x],{x,2}]==2*D[y[x],x]^2; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {c_2}{x+c_1}\\ y(x)&\to 0 \end{align*}

2.8.31.6 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)*Derivative(y(x), (x, 2)) - 2*Derivative(y(x), x)**2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -sqrt(2)*sqrt(y(x)*Derivative(y(x), (x, 2)))/2 + Derivative(y(x), x) cannot be solved by the factorable group method