Internal problem ID [12169]
Internal file name [OUTPUT/10822_Thursday_September_21_2023_05_47_30_AM_81996745/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 7.
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], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y^{\prime \prime }+\frac {2 {y^{\prime }}^{2}}{1-y}=0} \]
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*} \left (-1+y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-2 p \left (y \right )^{2} = 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 {2 p}{-1+y} \end {align*}
Where \(f(y)=\frac {2}{-1+y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {2}{-1+y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {2}{-1+y} \,d y}\\ \ln \left (p \right )&=2 \ln \left (-1+y \right )+c_{1}\\ p&={\mathrm e}^{2 \ln \left (-1+y \right )+c_{1}}\\ &=c_{1} \left (-1+y \right )^{2} \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} \left (-1+y\right )^{2} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{1} \left (-1+y \right )^{2}}d y &= x +c_{2}\\ -\frac {1}{\left (-1+y \right ) c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {c_{1} c_{2} +c_{1} x -1}{\left (x +c_{2} \right ) c_{1}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} c_{2} +c_{1} x -1}{\left (x +c_{2} \right ) c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} c_{2} +c_{1} x -1}{\left (x +c_{2} \right ) c_{1}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-1+y\right ) \left (\frac {d}{d x}y^{\prime }\right )-2 {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 {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} \\ {} & {} & \left (-1+y \right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-2 u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {2 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 )}=\frac {2}{-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 )}d y =\int \frac {2}{-1+y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=2 \ln \left (-1+y \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} \left (-1+y \right )^{2} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} \left (-1+y \right )^{2} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \left (-1+y\right )^{2} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} \left (-1+y\right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\left (-1+y\right )^{2}}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\left (-1+y\right )^{2}}d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{-1+y}={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{c_{1}} x +c_{2} -1}{{\mathrm e}^{c_{1}} x +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: 0.015 (sec). Leaf size: 18
dsolve(diff(y(x),x$2)+2/(1-y(x))*diff(y(x),x)^2=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} x +c_{2} -1}{c_{1} x +c_{2}} \]
✓ Solution by Mathematica
Time used: 0.298 (sec). Leaf size: 37
DSolve[y''[x]+2/(1-y[x])*y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {c_1 x-1+c_2 c_1}{c_1 (x+c_2)} \\ y(x)\to 1 \\ y(x)\to \text {Indeterminate} \\ \end{align*}