Internal problem ID [6835]
Internal file name [OUTPUT/6082_Thursday_July_28_2022_04_29_34_AM_6396199/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 16.
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]]
\[ \boxed {y^{\prime \prime }-x {y^{\prime }}^{2}=0} \] With initial conditions \begin {align*} \left [y \left (2\right ) = \frac {\pi }{4}, y^{\prime }\left (2\right ) = -{\frac {1}{4}}\right ] \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 )-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)\\ &= x \,p^{2} \end {align*}
Where \(f(x)=x\) and \(g(p)=p^{2}\). Integrating both sides gives \begin{align*} \frac {1}{p^{2}} \,dp &= x \,d x \\ \int { \frac {1}{p^{2}} \,dp} &= \int {x \,d x} \\ -\frac {1}{p}&=\frac {x^{2}}{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{p \left (x \right )}-\frac {x^{2}}{2}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=2\) and \(p=-{\frac {1}{4}}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} +2 = 0 \end {align*}
The solutions are \begin {align*} c_{1} = 2 \end {align*}
Trying the constant \begin {align*} c_{1} = 2 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {x^{2} p +4 p +2}{2 p} = 0 \end {align*}
The constant \(c_{1} = 2\) gives valid solution.
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 {-y^{\prime } x^{2}-4 y^{\prime }-2}{2 y^{\prime }} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {2}{x^{2}+4}\,\mathop {\mathrm {d}x}}\\ &= -\arctan \left (\frac {x}{2}\right )+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=2\) and \(y=\frac {\pi }{4}\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \frac {\pi }{4} = -\frac {\pi }{4}+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = \frac {\pi }{2} \end {align*}
Trying the constant \begin {align*} c_{2} = \frac {\pi }{2} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\arctan \left (\frac {x}{2}\right )+\frac {\pi }{2} \end {align*}
The constant \(c_{2} = \frac {\pi }{2}\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\arctan \left (\frac {x}{2}\right )+\frac {\pi }{2} \\ \end{align*}
Verification of solutions
\[ y = -\arctan \left (\frac {x}{2}\right )+\frac {\pi }{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-x {y^{\prime }}^{2}=0, y \left (2\right )=\frac {\pi }{4}, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}2\right \}}}}=-\frac {1}{4}\right ] \\ \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} \\ {} & {} & u^{\prime }\left (x \right )-x u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=x u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{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 x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=\frac {x^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {2}{x^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {2}{x^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {2}{x^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {2}{x^{2}+2 c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {\sqrt {2}\, \arctan \left (\frac {x \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {\sqrt {2}\, \arctan \left (\frac {x \sqrt {2}}{2 \sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (2\right )=\frac {\pi }{4} \\ {} & {} & \frac {\pi }{4}=-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}}{\sqrt {c_{1}}}\right )}{\sqrt {c_{1}}}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {1}{c_{1} \left (\frac {x^{2}}{2 c_{1}}+1\right )} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}2\right \}}}}=-\frac {1}{4} \\ {} & {} & -\frac {1}{4}=-\frac {1}{c_{1} \left (\frac {2}{c_{1}}+1\right )} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =2, c_{2} =\frac {\pi }{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\mathrm {arccot}\left (\frac {x}{2}\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\mathrm {arccot}\left (\frac {x}{2}\right ) \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) = _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: 0.094 (sec). Leaf size: 8
dsolve([diff(y(x),x$2)=x*diff(y(x),x)^2,y(2) = 1/4*Pi, D(y)(2) = -1/4],y(x), singsol=all)
\[ y \left (x \right ) = \operatorname {arccot}\left (\frac {x}{2}\right ) \]
✓ Solution by Mathematica
Time used: 1.241 (sec). Leaf size: 19
DSolve[{y''[x]==x*(y'[x])^2,{y[2]==1/4*Pi,y'[2]==-1/4}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2} \left (\pi -2 \arctan \left (\frac {x}{2}\right )\right ) \]