Internal problem ID [6266]
Internal file name [OUTPUT/5514_Sunday_June_05_2022_03_42_24_PM_11812817/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a differential equation. Problems for Review and Discovery. Page
53
Problem number: 4(b).
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]]
\[ \boxed {x y^{\prime \prime }-y^{\prime }+2 {y^{\prime }}^{3}=0} \]
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 (-1+2 p \left (x \right )^{2}\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 {2 p^{3}-p}{x} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(p)=2 p^{3}-p\). Integrating both sides gives \begin{align*} \frac {1}{2 p^{3}-p} \,dp &= -\frac {1}{x} \,d x \\ \int { \frac {1}{2 p^{3}-p} \,dp} &= \int {-\frac {1}{x} \,d x} \\ -\ln \left (p \right )+\frac {\ln \left (p^{2}-\frac {1}{2}\right )}{2}&=-\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (p \right )+\frac {\ln \left (p^{2}-\frac {1}{2}\right )}{2}} &= {\mathrm e}^{-\ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {\sqrt {2}\, \sqrt {2 p^{2}-1}}{2 p} &= \frac {c_{2}}{x} \end {align*}
The solution is \[ \frac {\sqrt {2}\, \sqrt {-1+2 p \left (x \right )^{2}}}{2 p \left (x \right )} = \frac {c_{2}}{x} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {\sqrt {2}\, \sqrt {-1+2 {y^{\prime }}^{2}}}{2 y^{\prime }} = \frac {c_{2}}{x} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {\sqrt {2}\, x}{2 \sqrt {-c_{2}^{2}+x^{2}}} \tag {1} \\ y^{\prime }&=\frac {\sqrt {2}\, x}{2 \sqrt {-c_{2}^{2}+x^{2}}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { -\frac {\sqrt {2}\, x}{2 \sqrt {-c_{2}^{2}+x^{2}}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\sqrt {2}\, \sqrt {-c_{2}^{2}+x^{2}}}{2}+c_{3} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { \frac {\sqrt {2}\, x}{2 \sqrt {-c_{2}^{2}+x^{2}}}\,\mathop {\mathrm {d}x}}\\ &= \frac {\sqrt {2}\, \sqrt {-c_{2}^{2}+x^{2}}}{2}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sqrt {2}\, \sqrt {-c_{2}^{2}+x^{2}}}{2}+c_{3} \\ \tag{2} y &= \frac {\sqrt {2}\, \sqrt {-c_{2}^{2}+x^{2}}}{2}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\sqrt {2}\, \sqrt {-c_{2}^{2}+x^{2}}}{2}+c_{3} \] Verified OK.
\[ y = \frac {\sqrt {2}\, \sqrt {-c_{2}^{2}+x^{2}}}{2}+c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (-1+2 {y^{\prime }}^{2}\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \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 (-1+2 u \left (x \right )^{2}\right ) u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {\left (-1+2 u \left (x \right )^{2}\right ) u \left (x \right )}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (-1+2 u \left (x \right )^{2}\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 (-1+2 u \left (x \right )^{2}\right ) u \left (x \right )}d x =\int -\frac {1}{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (-1+2 u \left (x \right )^{2}\right )}{2}-\ln \left (u \left (x \right )\right )=-\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}}, u \left (x \right )=-\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}}{2}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {x}{\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2 x^{2}-\left ({\mathrm e}^{c_{1}}\right )^{2}}}{2}+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 a change of variables {x -> y(x), y(x) -> x} 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 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -_b(_a)*(2*_b(_a)^2-1)/_a, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 37
dsolve(x*diff(y(x),x$2)=diff(y(x),x)-2*(diff(y(x),x))^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {2 x^{2}-c_{1}}}{2}+c_{2} \\ y \left (x \right ) &= -\frac {\sqrt {2 x^{2}-c_{1}}}{2}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.628 (sec). Leaf size: 96
DSolve[x*y''[x]==y'[x]-2*(y'[x])^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\frac {1}{2} \sqrt {2 x^2+e^{2 c_1}} \\ y(x)\to \frac {1}{2} \sqrt {2 x^2+e^{2 c_1}}+c_2 \\ y(x)\to -\frac {\sqrt {x^2}}{\sqrt {2}}+c_2 \\ y(x)\to \frac {\sqrt {x^2}}{\sqrt {2}}+c_2 \\ \end{align*}