Internal problem ID [13518]
Internal file name [OUTPUT/12690_Friday_February_16_2024_12_11_00_AM_55498391/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending first order concepts. Additional exercises
page 259
Problem number: 13.8 (ii).
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]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime \prime }+2 x {y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 0] \end {align*}
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*} p^{\prime }\left (x \right )+2 x p \left (x \right )^{2} = 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)\\ &= -2 x \,p^{2} \end {align*}
Where \(f(x)=-2 x\) and \(g(p)=p^{2}\). Integrating both sides gives \begin{align*} \frac {1}{p^{2}} \,dp &= -2 x \,d x \\ \int { \frac {1}{p^{2}} \,dp} &= \int {-2 x \,d x} \\ -\frac {1}{p}&=-x^{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{p \left (x \right )}+x^{2}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \textit {undefined} = 0 \end {align*}
This shows that no solution exist. 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 {1}{y^{\prime }}+x^{2}-c_{1} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{-x^{2}+c_{1}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\operatorname {arctanh}\left (\frac {x}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=3\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 3 = c_{2} \end {align*}
This solution is valid for any \(c_{1}\). Hence there are infinite number of solutions.
Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = -\frac {\operatorname {arctanh}\left (\frac {x}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 3\) and \(x = 0\) in the above gives \begin {align*} 3 = c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\frac {1}{c_{1} \left (-\frac {x^{2}}{c_{1}}+1\right )} \end {align*}
substituting \(y^{\prime } = 0\) and \(x = 0\) in the above gives \begin {align*} 0 = -\frac {1}{c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Verification of solutions N/A
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y^{\prime }+2 x {y^{\prime }}^{2}=0, y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+2 x u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-2 x u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=-2 x \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}d x =\int -2 x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-x^{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x^{2}+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x^{2}+c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{-x^{2}+c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{-x^{2}+c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\mathrm {arctanh}\left (\frac {x}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\mathrm {arctanh}\left (\frac {x}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {1}{c_{1} \left (-\frac {x^{2}}{c_{1}}+1\right )} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0 \\ {} & {} & 0=-\frac {1}{c_{1}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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 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) = -2*_a*_b(_a)^2, _b(_a), HINT = [[_a, -2*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, -2*_b]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 5
dsolve([diff(y(x),x$2)=-2*x*diff(y(x),x)^2,y(0) = 3, D(y)(0) = 0],y(x), singsol=all)
\[ y \left (x \right ) = 3 \]
✓ Solution by Mathematica
Time used: 0.949 (sec). Leaf size: 6
DSolve[{y''[x]==-2*x*y'[x]^2,{y[0]==3,y'[0]==0}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to 3 \]