Internal problem ID [13813]
Internal file name [OUTPUT/12985_Sunday_February_18_2024_08_02_30_AM_44100249/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 25. Review exercises for part III. page 447
Problem number: 16.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }-{y^{\prime }}^{2}=0} \]
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*} p^{\prime }\left (x \right )-p \left (x \right )^{2} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Integrating both sides gives \begin {align*} \int \frac {1}{p^{2}}d p &= x +c_{1}\\ -\frac {1}{p}&=x +c_{1} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&=-\frac {1}{x +c_{1}} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -\frac {1}{x +c_{1}} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{x +c_{1}}\,\mathop {\mathrm {d}x}}\\ &= -\ln \left (x +c_{1} \right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\ln \left (x +c_{1} \right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\ln \left (x +c_{1} \right )+c_{2} \] Verified OK.
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 ) \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Integrating both sides gives \begin {align*} \int \frac {1}{p}d p &= y +c_{1}\\ \ln \left (p \right )&=y +c_{1}\\ p&={\mathrm e}^{y +c_{1}}\\ p&=c_{1} {\mathrm e}^{y} \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 } = c_{1} {\mathrm e}^{y} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {{\mathrm e}^{-y}}{c_{1}}d y &= x +c_{2}\\ -\frac {{\mathrm e}^{-y}}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\ln \left (-\frac {1}{\left (x +c_{2} \right ) c_{1}}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (-\frac {1}{\left (x +c_{2} \right ) c_{1}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (-\frac {1}{\left (x +c_{2} \right ) c_{1}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-{y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )-u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}d x =\int 1d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{x +c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{x +c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{x +c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{x +c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\ln \left (x +c_{1} \right )+c_{2} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 15
dsolve(diff(y(x),x$2)=diff(y(x),x)^2,y(x), singsol=all)
\[ y \left (x \right ) = -\ln \left (-c_{1} x -c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.217 (sec). Leaf size: 15
DSolve[y''[x]==y'[x]^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2-\log (x+c_1) \]