Internal problem ID [6241]
Internal file name [OUTPUT/5489_Sunday_June_05_2022_03_41_18_PM_49576723/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. Section 1.9. Reduction of Order. Page
38
Problem number: 1(e).
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], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {2 y y^{\prime \prime }-{y^{\prime }}^{2}=1} \]
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*} 2 y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{2} = 1 \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 {p^{2}+1}{2 y p} \end {align*}
Where \(f(y)=\frac {1}{2 y}\) and \(g(p)=\frac {p^{2}+1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}+1}{p}} \,dp &= \frac {1}{2 y} \,d y \\ \int { \frac {1}{\frac {p^{2}+1}{p}} \,dp} &= \int {\frac {1}{2 y} \,d y} \\ \frac {\ln \left (p^{2}+1\right )}{2}&=\frac {\ln \left (y \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}+1} &= {\mathrm e}^{\frac {\ln \left (y \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}+1} &= c_{2} \sqrt {y} \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}+1} = c_{2} \sqrt {y}\, {\mathrm e}^{c_{1}} \] The solution is \[ \sqrt {p \left (y \right )^{2}+1} = c_{2} \sqrt {y}\, {\mathrm e}^{c_{1}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {1+{y^{\prime }}^{2}} = c_{2} \sqrt {y}\, {\mathrm e}^{c_{1}} \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 }&=\sqrt {-1+c_{2}^{2} y \,{\mathrm e}^{2 c_{1}}} \tag {1} \\ y^{\prime }&=-\sqrt {-1+c_{2}^{2} y \,{\mathrm e}^{2 c_{1}}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {-1+c_{2}^{2} y \,{\mathrm e}^{2 c_{1}}}}d y &= \int d x \\ \frac {2 \sqrt {-1+c_{2}^{2} y \,{\mathrm e}^{2 c_{1}}}\, {\mathrm e}^{-2 c_{1}}}{c_{2}^{2}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {-1+c_{2}^{2} y \,{\mathrm e}^{2 c_{1}}}}d y &= \int d x \\ -\frac {2 \sqrt {-1+c_{2}^{2} y \,{\mathrm e}^{2 c_{1}}}\, {\mathrm e}^{-2 c_{1}}}{c_{2}^{2}}&=x +c_{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{3} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +{\mathrm e}^{4 c_{1}} c_{2}^{4} x^{2}+4\right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \\ \tag{2} y &= \frac {\left (c_{4}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{4} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +{\mathrm e}^{4 c_{1}} c_{2}^{4} x^{2}+4\right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{3} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +{\mathrm e}^{4 c_{1}} c_{2}^{4} x^{2}+4\right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \] Verified OK.
\[ y = \frac {\left (c_{4}^{2} {\mathrm e}^{4 c_{1}} c_{2}^{4}+2 c_{4} {\mathrm e}^{4 c_{1}} c_{2}^{4} x +{\mathrm e}^{4 c_{1}} c_{2}^{4} x^{2}+4\right ) {\mathrm e}^{-2 c_{1}}}{4 c_{2}^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y y^{\prime \prime }-{y^{\prime }}^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \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 ),y^{\prime \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} \\ {} & {} & 2 y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-u \left (y \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )^{2}+1}{2 y u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{u \left (y \right )^{2}+1}=\frac {1}{2 y} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{u \left (y \right )^{2}+1}d y =\int \frac {1}{2 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (u \left (y \right )^{2}+1\right )}{2}=\frac {\ln \left (y \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y \right )}}{{\mathrm e}^{-2 c_{1}}}, u \left (y \right )=-\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y \right )}}{{\mathrm e}^{-2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y \right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}=\frac {1}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}d x =\int \frac {1}{{\mathrm e}^{-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 \sqrt {{\mathrm e}^{-2 c_{1}} y-\left ({\mathrm e}^{-2 c_{1}}\right )^{2}}}{{\mathrm e}^{-2 c_{1}}}=\frac {x}{{\mathrm e}^{-2 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {c_{2}^{2} \left ({\mathrm e}^{-2 c_{1}}\right )^{2}+2 c_{2} {\mathrm e}^{-2 c_{1}} x +4 \left ({\mathrm e}^{-2 c_{1}}\right )^{2}+x^{2}}{4 \,{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y \right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}=-\frac {1}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-y\right )}}d x =\int -\frac {1}{{\mathrm e}^{-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {2 \sqrt {{\mathrm e}^{-2 c_{1}} y-\left ({\mathrm e}^{-2 c_{1}}\right )^{2}}}{{\mathrm e}^{-2 c_{1}}}=-\frac {x}{{\mathrm e}^{-2 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {c_{2}^{2} \left ({\mathrm e}^{-2 c_{1}}\right )^{2}-2 c_{2} {\mathrm e}^{-2 c_{1}} x +4 \left ({\mathrm e}^{-2 c_{1}}\right )^{2}+x^{2}}{4 \,{\mathrm e}^{-2 c_{1}}} \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 2nd order ODE linearizable_by_differentiation -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x), y(x)` *** Sublevel 2 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful <- 2nd order ODE linearizable_by_differentiation successful`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 22
dsolve(2*y(x)*diff(y(x),x$2)=1+(diff(y(x),x))^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (c_{1}^{2}+1\right ) x^{2}}{4 c_{2}}+c_{1} x +c_{2} \]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 34
DSolve[2*y[x]*y''[x]==1+(y'[x])^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\left (1+c_1{}^2\right ) x^2}{4 c_2}+c_1 x+c_2 \\ y(x)\to \text {Indeterminate} \\ \end{align*}