Internal problem ID [13501]
Internal file name [OUTPUT/12673_Friday_February_16_2024_12_10_41_AM_29512102/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: 13.5 (f).
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, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y y^{\prime \prime }-{y^{\prime }}^{2}-y^{\prime }=0} \]
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 )-1\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {p +1}{y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=p +1\). Integrating both sides gives \begin{align*} \frac {1}{p +1} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{p +1} \,dp} &= \int {\frac {1}{y} \,d y} \\ \ln \left (p +1\right )&=\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} p +1 &= {\mathrm e}^{\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} p +1 &= c_{2} y \end {align*}
Which simplifies to \[ p \left (y \right ) = c_{2} y \,{\mathrm e}^{c_{1}}-1 \] 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_{2} y \,{\mathrm e}^{c_{1}}-1 \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{2} y \,{\mathrm e}^{c_{1}}-1}d y &= x +c_{3}\\ \frac {\ln \left (c_{2} y \,{\mathrm e}^{c_{1}}-1\right ) {\mathrm e}^{-c_{1}}}{c_{2}}&=x +c_{3} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\left ({\mathrm e}^{c_{3} c_{2} {\mathrm e}^{c_{1}}+x c_{2} {\mathrm e}^{c_{1}}}+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}}\\ &=\frac {\left (c_{3} {\mathrm e}^{x c_{2} {\mathrm e}^{c_{1}}}+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} {\mathrm e}^{x c_{2} {\mathrm e}^{c_{1}}}+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} {\mathrm e}^{x c_{2} {\mathrm e}^{c_{1}}}+1\right ) {\mathrm e}^{-c_{1}}}{c_{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (\frac {d}{d x}y^{\prime }\right )+\left (-y^{\prime }-1\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\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 ),\frac {d}{d x}y^{\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 )-1\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 )-1}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{-u \left (y \right )-1}=-\frac {1}{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 )-1}d y =\int -\frac {1}{y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (-u \left (y \right )-1\right )=-\ln \left (y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {{\mathrm e}^{c_{1}}+y}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {{\mathrm e}^{c_{1}}+y}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{c_{1}}+y}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{c_{1}}+y}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{{\mathrm e}^{c_{1}}+y}=-\frac {1}{{\mathrm e}^{c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{{\mathrm e}^{c_{1}}+y}d x =\int -\frac {1}{{\mathrm e}^{c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left ({\mathrm e}^{c_{1}}+y\right )=-\frac {x}{{\mathrm e}^{c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\frac {c_{2} {\mathrm e}^{c_{1}}-x}{{\mathrm e}^{c_{1}}}}-{\mathrm e}^{c_{1}} \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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 20
dsolve(y(x)*diff(y(x),x$2)-diff(y(x),x)^2=diff(y(x),x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {{\mathrm e}^{c_{1} \left (c_{2} +x \right )}+1}{c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.761 (sec). Leaf size: 26
DSolve[y[x]*y''[x]-y'[x]^2==y'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1+e^{c_1 (x+c_2)}}{c_1} \\ y(x)\to \text {Indeterminate} \\ \end{align*}