Internal problem ID [12752]
Internal file name [OUTPUT/11405_Friday_November_03_2023_06_32_36_AM_25067696/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.1, page 186
Problem number: 10.
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]]
\[ \boxed {2 y y^{\prime \prime }-{y^{\prime }}^{2}=0} \]
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*} 2 y 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 \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {p}{2 y} \end {align*}
Where \(f(y)=\frac {1}{2 y}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {1}{2 y} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {1}{2 y} \,d y}\\ \ln \left (p \right )&=\frac {\ln \left (y \right )}{2}+c_{1}\\ p&={\mathrm e}^{\frac {\ln \left (y \right )}{2}+c_{1}}\\ &=c_{1} \sqrt {y} \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} \sqrt {y} \end {align*}
Integrating both sides gives \begin{align*} \int \frac {1}{c_{1} \sqrt {y}}d y &= \int d x \\ \frac {2 \sqrt {y}}{c_{1}}&=x +c_{2} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {1}{4} c_{2}^{2} c_{1}^{2}+\frac {1}{2} c_{2} c_{1}^{2} x +\frac {1}{4} c_{1}^{2} x^{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {1}{4} c_{2}^{2} c_{1}^{2}+\frac {1}{2} c_{2} c_{1}^{2} x +\frac {1}{4} c_{1}^{2} x^{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y \left (\frac {d}{d x}y^{\prime }\right )-{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} \\ {} & {} & 2 y u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )-u \left (y \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {u \left (y \right )}{2 y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {1}{2 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 {1}{2 y}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=\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}} y}}{{\mathrm e}^{-2 c_{1}}}, u \left (y \right )=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} y}}{{\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}} y}}{{\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}} y}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {{\mathrm e}^{-2 c_{1}} y}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 c_{1}} y}}=\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}} y}}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}}{{\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 +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}} y}}{{\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}} y}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {{\mathrm e}^{-2 c_{1}} y}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {{\mathrm e}^{-2 c_{1}} y}}=-\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}} y}}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}}{{\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 +x^{2}}{4 \,{\mathrm e}^{-2 c_{1}}} \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: 0.047 (sec). Leaf size: 17
dsolve(2*y(x)*diff(y(x),x$2)-diff(y(x),x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {\left (c_{1} x +c_{2} \right )^{2}}{4} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 29
DSolve[2*y[x]*y''[x]-(y'[x])^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {(c_1 x+2 c_2){}^2}{4 c_2} \\ y(x)\to \text {Indeterminate} \\ \end{align*}