Internal problem ID [6848]
Internal file name [OUTPUT/6095_Thursday_July_28_2022_04_30_21_AM_39831181/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: 31.
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_y_y1]]
\[ \boxed {y y^{\prime \prime }-{y^{\prime }}^{2} \left (1-y^{\prime } \sin \left (y\right )-y y^{\prime } \cos \left (y\right )\right )=0} \]
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*} y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (\cos \left (y \right ) y p \left (y \right )^{2}+\sin \left (y \right ) p \left (y \right )^{2}-p \left (y \right )\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). Using the change of variables \(p \left (y \right ) = u \left (y \right ) y\) on the above ode results in new ode in \(u \left (y \right )\) \begin {align*} y^{2} u \left (y \right ) \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )+\left (\cos \left (y \right ) y^{3} u \left (y \right )^{2}+\sin \left (y \right ) u \left (y \right )^{2} y^{2}-u \left (y \right ) y \right ) u \left (y \right ) y = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= u^{2} \left (-\cos \left (y \right ) y -\sin \left (y \right )\right ) \end {align*}
Where \(f(y)=-\cos \left (y \right ) y -\sin \left (y \right )\) and \(g(u)=u^{2}\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}} \,du &= -\cos \left (y \right ) y -\sin \left (y \right ) \,d y \\ \int { \frac {1}{u^{2}} \,du} &= \int {-\cos \left (y \right ) y -\sin \left (y \right ) \,d y} \\ -\frac {1}{u}&=-y \sin \left (y \right )+c_{2} \\ \end{align*} The solution is \[ -\frac {1}{u \left (y \right )}+y \sin \left (y \right )-c_{2} = 0 \] Replacing \(u(y)\) in the above solution by \(\frac {p \left (y \right )}{y}\) results in the solution for \(p \left (y \right )\) in implicit form \begin {align*} -\frac {y}{p \left (y \right )}+y \sin \left (y \right )-c_{2} = 0\\ -\frac {y}{p \left (y \right )}+y \sin \left (y \right )-c_{2} = 0 \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*} -\frac {y}{y^{\prime }}+y \sin \left (y\right )-c_{2} = 0 \end {align*}
Integrating both sides gives \begin {align*} \int \frac {y \sin \left (y \right )-c_{2}}{y}d y &= \int {dx}\\ \int _{}^{y}\frac {\textit {\_a} \sin \left (\textit {\_a} \right )-c_{2}}{\textit {\_a}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {\textit {\_a} \sin \left (\textit {\_a} \right )-c_{2}}{\textit {\_a}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {\textit {\_a} \sin \left (\textit {\_a} \right )-c_{2}}{\textit {\_a}}d \textit {\_a} = x +c_{3} \] Verified OK.
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 a change of variables {x -> y(x), y(x) -> x} 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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)^2*(_b(_a)*cos(_a)*_a+_b(_a)*sin(_a)-1)/_a = 0, _b(_a)` *** Subl 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
Time used: 0.14 (sec). Leaf size: 24
dsolve(y(x)*diff(y(x),x$2)=diff(y(x),x)^2*(1-diff(y(x),x)*sin(y(x))-y(x)*diff(y(x),x)*cos(y(x)) ),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \\ -\cos \left (y \left (x \right )\right )+c_{1} \ln \left (y \left (x \right )\right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.489 (sec). Leaf size: 69
DSolve[y[x]*y''[x]==(y'[x])^2*(1-y'[x]*Sin[y[x]]-y[x]*y'[x]*Cos[y[x]] ),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}[-\cos (\text {$\#$1})+c_1 \log (\text {$\#$1})\&][x+c_2] \\ y(x)\to \text {InverseFunction}[-\cos (\text {$\#$1})-c_1 \log (\text {$\#$1})\&][x+c_2] \\ y(x)\to \text {InverseFunction}[-\cos (\text {$\#$1})+c_1 \log (\text {$\#$1})\&][x+c_2] \\ \end{align*}