Internal problem ID [7403]
Internal file name [OUTPUT/6370_Sunday_June_05_2022_04_41_58_PM_92550606/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 14.
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} \end {align*}
Solving for \(p\) gives these solutions \begin {align*} p_1&={\mathrm e}^{-y -c_{1}}\\ &=\frac {{\mathrm e}^{-y}}{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 } = \frac {{\mathrm e}^{-y}}{c_{1}} \end {align*}
Integrating both sides gives \begin {align*} \int {\mathrm e}^{y} c_{1} d y &= c_{2} +x\\ {\mathrm e}^{y} c_{1}&=c_{2} +x \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\ln \left (\frac {c_{2} +x}{c_{1}}\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \ln \left (\frac {c_{2} +x}{c_{1}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \ln \left (\frac {c_{2} +x}{c_{1}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+{y^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}d x =\int \left (-1\right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x +c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x +c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{-x +c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{-x +c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\ln \left (-x +c_{1} \right )+c_{2} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 10
dsolve(diff(y(x),x$2)+diff(y(x),x)^2=0,y(x), singsol=all)
\[ y \left (x \right ) = \ln \left (c_{1} x +c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.205 (sec). Leaf size: 15
DSolve[y''[x]+(y'[x])^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \log (x-c_1)+c_2 \]