Internal problem ID [2304]
Internal file name [OUTPUT/2304_Tuesday_February_27_2024_08_25_37_AM_70736289/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 35, page 157
Problem number: 32.
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]]
\[ \boxed {y^{\prime \prime }-{y^{\prime }}^{2} \cos \left (x \right )=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 2, y^{\prime }\left (0\right ) = 1] \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 )-p \left (x \right )^{2} \cos \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)\\ &= p^{2} \cos \left (x \right ) \end {align*}
Where \(f(x)=\cos \left (x \right )\) and \(g(p)=p^{2}\). Integrating both sides gives \begin{align*} \frac {1}{p^{2}} \,dp &= \cos \left (x \right ) \,d x \\ \int { \frac {1}{p^{2}} \,dp} &= \int {\cos \left (x \right ) \,d x} \\ -\frac {1}{p}&=\sin \left (x \right )+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{p \left (x \right )}-\sin \left (x \right )-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1-c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = -1 \end {align*}
Trying the constant \begin {align*} c_{1} = -1 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\frac {\sin \left (x \right ) p +1-p}{p} = 0 \end {align*}
The constant \(c_{1} = -1\) 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 {-\sin \left (x \right ) y^{\prime }-1+y^{\prime }}{y^{\prime }} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{\sin \left (x \right )-1}\,\mathop {\mathrm {d}x}}\\ &= -\frac {2}{\tan \left (\frac {x}{2}\right )-1}+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 2 = c_{2} +2 \end {align*}
The solutions are \begin {align*} c_{2} = 0 \end {align*}
Trying the constant \begin {align*} c_{2} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-\frac {2}{\tan \left (\frac {x}{2}\right )-1} \end {align*}
The constant \(c_{2} = 0\) 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 &= -\frac {2}{\tan \left (\frac {x}{2}\right )-1} \\ \end{align*}
Verification of solutions
\[ y = -\frac {2}{\tan \left (\frac {x}{2}\right )-1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-{y^{\prime }}^{2} \cos \left (x \right )=0, y \left (0\right )=2, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1\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 )-u \left (x \right )^{2} \cos \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=u \left (x \right )^{2} \cos \left (x \right ) \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=\cos \left (x \right ) \\ \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 \cos \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=\sin \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\sin \left (x \right )+c_{1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\sin \left (x \right )+c_{1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{\sin \left (x \right )+c_{1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{\sin \left (x \right )+c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {2 \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) c_{1} +2}{2 \sqrt {c_{1}^{2}-1}}\right )}{\sqrt {c_{1}^{2}-1}}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-\frac {2 \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) c_{1} +2}{2 \sqrt {c_{1}^{2}-1}}\right )}{\sqrt {c_{1}^{2}-1}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=2 \\ {} & {} & 2=-\frac {2 \arctan \left (\frac {1}{\sqrt {c_{1}^{2}-1}}\right )}{\sqrt {c_{1}^{2}-1}}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {2 \left (\frac {1}{2}+\frac {\tan \left (\frac {x}{2}\right )^{2}}{2}\right ) c_{1}}{\left (c_{1}^{2}-1\right ) \left (\frac {\left (2 \tan \left (\frac {x}{2}\right ) c_{1} +2\right )^{2}}{4 \left (c_{1}^{2}-1\right )}+1\right )} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-\frac {c_{1}}{\left (c_{1}^{2}-1\right ) \left (\frac {1}{c_{1}^{2}-1}+1\right )} \\ {} & \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 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, diff(_b(_a), _a) = _b(_a)^2*cos(_a), _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✗ Solution by Maple
dsolve([diff(y(x),x$2)=diff(y(x),x)^2*cos(x),y(0) = 2, D(y)(0) = 1],y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{y''[x]==y'[x]^2*Cos[x],{y[0]==2,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
{}