4.23 problem Problem 3.38

Internal problem ID [5897]
Internal file name [OUTPUT/5145_Sunday_June_05_2022_03_26_14_PM_80465364/index.tex]

Book: THEORY OF DIFFERENTIAL EQUATIONS IN ENGINEERING AND MECHANICS. K.T. CHAU, CRC Press. Boca Raton, FL. 2018
Section: Chapter 3. Ordinary Differential Equations. Section 3.6 Summary and Problems. Page 218
Problem number: Problem 3.38.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_ode_missing_x"

Maple gives the following as the ode type

[[_2nd_order, _missing_x], [_2nd_order, _with_potential_symmetries], [_2nd_order, _reducible, _mu_xy]]

\[ \boxed {y y^{\prime \prime }-{y^{\prime }}^{2}-y^{\prime } y^{2}=0} \]

4.23.1 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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-p \left (y \right )-y^{2}\right ) p \left (y \right ) = 0 \end {align*}

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

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d y}p \left (y \right ) + p(y)p \left (y \right ) &= q(y) \end {align*}

Where here \begin {align*} p(y) &=-\frac {1}{y}\\ q(y) &=y \end {align*}

Hence the ode is \begin {align*} \frac {d}{d y}p \left (y \right )-\frac {p \left (y \right )}{y} = y \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{y}d y} \\ &= \frac {1}{y} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu p\right ) &= \left (\mu \right ) \left (y\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (\frac {p}{y}\right ) &= \left (\frac {1}{y}\right ) \left (y\right )\\ \mathrm {d} \left (\frac {p}{y}\right ) &= \mathrm {d} y \end {align*}

Integrating gives \begin {align*} \frac {p}{y} &= \int {\mathrm {d} y}\\ \frac {p}{y} &= y + c_{1} \end {align*}

Dividing both sides by the integrating factor \(\mu =\frac {1}{y}\) results in \begin {align*} p \left (y \right ) &= c_{1} y +y^{2} \end {align*}

which simplifies to \begin {align*} p \left (y \right ) &= y \left (y +c_{1} \right ) \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 } = y \left (y+c_{1} \right ) \end {align*}

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

Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {c_{1} {\mathrm e}^{c_{1} c_{2} +c_{1} x}}{-1+{\mathrm e}^{c_{1} c_{2} +c_{1} x}}\\ &=-\frac {c_{1} c_{2} {\mathrm e}^{c_{1} x}}{-1+c_{2} {\mathrm e}^{c_{1} x}} \end {align*}

Summary

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

Verification of solutions

\[ y = -\frac {c_{1} c_{2} {\mathrm e}^{c_{1} x}}{-1+c_{2} {\mathrm e}^{c_{1} x}} \] Verified OK.

4.23.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y y^{\prime \prime }+\left (-y^{\prime }-y^{2}\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),y^{\prime \prime }=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 )+\left (-u \left (y \right )-y^{2}\right ) u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {-u \left (y \right )-y^{2}}{y} \\ \bullet & {} & \textrm {Collect w.r.t.}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )}{y}+y \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} u \left (y \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )-\frac {u \left (y \right )}{y}=y \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (y \right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )-\frac {u \left (y \right )}{y}\right )=\mu \left (y \right ) y \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right ) \\ {} & {} & \mu \left (y \right ) \left (\frac {d}{d y}u \left (y \right )-\frac {u \left (y \right )}{y}\right )=\left (\frac {d}{d y}u \left (y \right )\right ) \mu \left (y \right )+u \left (y \right ) \left (\frac {d}{d y}\mu \left (y \right )\right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \frac {d}{d y}\mu \left (y \right ) \\ {} & {} & \frac {d}{d y}\mu \left (y \right )=-\frac {\mu \left (y \right )}{y} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (y \right )=\frac {1}{y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \left (\frac {d}{d y}\left (u \left (y \right ) \mu \left (y \right )\right )\right )d y =\int \mu \left (y \right ) y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & u \left (y \right ) \mu \left (y \right )=\int \mu \left (y \right ) y d y +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\int \mu \left (y \right ) y d y +c_{1}}{\mu \left (y \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (y \right )=\frac {1}{y} \\ {} & {} & u \left (y \right )=y \left (\int 1d y +c_{1} \right ) \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & u \left (y \right )=y \left (y +c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=y \left (y +c_{1} \right ) \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=y \left (y+c_{1} \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y \left (y+c_{1} \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y \left (y+c_{1} \right )}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y \left (y+c_{1} \right )}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (y+c_{1} \right )}{c_{1}}+\frac {\ln \left (y\right )}{c_{1}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {c_{1} {\mathrm e}^{c_{2} c_{1} +c_{1} x}}{-1+{\mathrm e}^{c_{2} c_{1} +c_{1} x}} \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))*_b(_a)-_b(_a)*(_a^2+_b(_a))/_a = 0, _b(_a), HINT = [[_a, 2*_b]]`   *** Sublevel 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[_a, 2*_b]
 

✓ Solution by Maple

Time used: 0.063 (sec). Leaf size: 27

dsolve(y(x)*diff(y(x),x$2)-(diff(y(x),x))^2-y(x)^2*diff(y(x),x)=0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= -\frac {c_{1} {\mathrm e}^{\left (x +c_{2} \right ) c_{1}}}{-1+{\mathrm e}^{\left (x +c_{2} \right ) c_{1}}} \\ \end{align*}

✓ Solution by Mathematica

Time used: 1.53 (sec). Leaf size: 43

DSolve[y[x]*y''[x]-(y'[x])^2-y[x]^2*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\frac {c_1 e^{c_1 (x+c_2)}}{-1+e^{c_1 (x+c_2)}} \\ y(x)\to -\frac {1}{x+c_2} \\ \end{align*}