Internal problem ID [13499]
Internal file name [OUTPUT/12671_Friday_February_16_2024_12_10_39_AM_40535847/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 (d).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {x y^{\prime \prime }-{y^{\prime }}^{2}+y^{\prime }=0} \]
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*} x p^{\prime }\left (x \right )+\left (-p \left (x \right )+1\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {p \left (p -1\right )}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(p)=p \left (p -1\right )\). Integrating both sides gives \begin{align*} \frac {1}{p \left (p -1\right )} \,dp &= \frac {1}{x} \,d x \\ \int { \frac {1}{p \left (p -1\right )} \,dp} &= \int {\frac {1}{x} \,d x} \\ \ln \left (p -1\right )-\ln \left (p \right )&=\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (p -1\right )-\ln \left (p \right )} &= {\mathrm e}^{\ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {p -1}{p} &= c_{2} x \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {1}{c_{2} x -1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{c_{2} x -1}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\ln \left (c_{2} x -1\right )}{c_{2}}+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\ln \left (c_{2} x -1\right )}{c_{2}}+c_{3} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\ln \left (c_{2} x -1\right )}{c_{2}}+c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \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 {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & x u^{\prime }\left (x \right )+\left (-u \left (x \right )+1\right ) u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {\left (-u \left (x \right )+1\right ) u \left (x \right )}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )+1\right ) u \left (x \right )}=-\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (-u \left (x \right )+1\right ) u \left (x \right )}d x =\int -\frac {1}{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\ln \left (u \left (x \right )-1\right )+\ln \left (u \left (x \right )\right )=-\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{c_{1}}-x} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{c_{1}}-x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{c_{1}}-x} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{c_{1}}-x}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{c_{1}} \ln \left ({\mathrm e}^{c_{1}}-x \right )+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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 17
dsolve(x*diff(y(x),x$2)=diff(y(x),x)^2-diff(y(x),x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\ln \left (c_{1} x -1\right )}{c_{1}}+c_{2} \]
✓ Solution by Mathematica
Time used: 0.338 (sec). Leaf size: 38
DSolve[x*y''[x]==y'[x]^2-y'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to e^{-c_1} \log \left (1+e^{c_1} x\right )+c_2 \\ y(x)\to c_2 \\ y(x)\to x+c_2 \\ \end{align*}