Internal problem ID [15446]
Internal file name [OUTPUT/15447_Wednesday_May_08_2024_03_59_01_PM_47068444/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 16. The method of isoclines for differential
equations of the second order. Exercises page 158
Problem number: 703.
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]]
\[ \boxed {x^{\prime \prime }+x {x^{\prime }}^{2}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(x\) an independent variable. Using \begin {align*} x' &= p(x) \end {align*}
Then \begin {align*} x'' &= \frac {dp}{dt}\\ &= \frac {dx}{dt} \frac {dp}{dx}\\ &= p \frac {dp}{dx} \end {align*}
Hence the ode becomes \begin {align*} p \left (x \right ) \left (\frac {d}{d x}p \left (x \right )\right )+x p \left (x \right )^{2} = 0 \end {align*}
Which is now solved as first order ode for \(p(x)\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -p x \end {align*}
Where \(f(x)=-x\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= -x \,d x\\ \int { \frac {1}{p} \,dp} &= \int {-x \,d x}\\ \ln \left (p \right )&=-\frac {x^{2}}{2}+c_{1}\\ p&={\mathrm e}^{-\frac {x^{2}}{2}+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {x^{2}}{2}} \end {align*}
For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} x^{\prime } = c_{1} {\mathrm e}^{-\frac {x^{2}}{2}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {{\mathrm e}^{\frac {x^{2}}{2}}}{c_{1}}d x &= \int {dt}\\ \int _{}^{x}\frac {{\mathrm e}^{\frac {\textit {\_a}^{2}}{2}}}{c_{1}}d \textit {\_a}&= t +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x}\frac {{\mathrm e}^{\frac {\textit {\_a}^{2}}{2}}}{c_{1}}d \textit {\_a} &= t +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{x}\frac {{\mathrm e}^{\frac {\textit {\_a}^{2}}{2}}}{c_{1}}d \textit {\_a} = t +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime \prime }+x {x^{\prime }}^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (t \right )=x^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} x^{\prime \prime } \\ {} & {} & u^{\prime }\left (t \right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & x^{\prime } \left (\frac {d}{d x}u \left (x \right )\right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )=x^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} x^{\prime }=u \left (x \right ),x^{\prime \prime }=u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (x \right ) \left (\frac {d}{d x}u \left (x \right )\right )+x u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-x u \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )}=-x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )}d x =\int -x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (x \right )\right )=-\frac {x^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{-\frac {x^{2}}{2}+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )={\mathrm e}^{-\frac {x^{2}}{2}+c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (x \right )=x^{\prime },x =x \\ {} & {} & x^{\prime }={\mathrm e}^{-\frac {x^{2}}{2}+c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }={\mathrm e}^{-\frac {x^{2}}{2}+c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {x^{\prime }}{{\mathrm e}^{-\frac {x^{2}}{2}+c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {x^{\prime }}{{\mathrm e}^{-\frac {x^{2}}{2}+c_{1}}}d t =\int 1d t +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {-\frac {\mathrm {I}}{2} \sqrt {\pi }\, \sqrt {2}\, \mathrm {erf}\left (\frac {\mathrm {I}}{2} \sqrt {2}\, x\right )}{{\mathrm e}^{c_{1}}}=t +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} x \\ {} & {} & \left \{\mathrm {-I} \mathit {RootOf}\left (\mathrm {I} \,{\mathrm e}^{c_{1}} \sqrt {2}\, c_{2} +\mathrm {I} \,{\mathrm e}^{c_{1}} \sqrt {2}\, t -\mathrm {erf}\left (\textit {\_Z} \right ) \sqrt {\pi }\right ) \sqrt {2}\right \} \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: 34
dsolve(diff(x(t),t$2)+x(t)*diff(x(t),t)^2=0,x(t), singsol=all)
\[ x \left (t \right ) = -i \operatorname {RootOf}\left (i \sqrt {2}\, c_{1} t +i \sqrt {2}\, c_{2} -\operatorname {erf}\left (\textit {\_Z} \right ) \sqrt {\pi }\right ) \sqrt {2} \]
✓ Solution by Mathematica
Time used: 1.757 (sec). Leaf size: 34
DSolve[x''[t]+x[t]*x'[t]^2==0,x[t],t,IncludeSingularSolutions -> True]
\[ x(t)\to -i \sqrt {2} \text {erf}^{-1}\left (i \sqrt {\frac {2}{\pi }} c_1 (t+c_2)\right ) \]