Internal problem ID [6844]
Internal file name [OUTPUT/6091_Thursday_July_28_2022_04_30_12_AM_80846733/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: 26.
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]]
\[ \boxed {2 y^{\prime \prime }-{y^{\prime }}^{3} \sin \left (2 x \right )=0} \]
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*} 2 p^{\prime }\left (x \right )-p \left (x \right )^{3} \sin \left (2 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)\\ &= \frac {p^{3} \sin \left (2 x \right )}{2} \end {align*}
Where \(f(x)=\frac {\sin \left (2 x \right )}{2}\) and \(g(p)=p^{3}\). Integrating both sides gives \begin{align*} \frac {1}{p^{3}} \,dp &= \frac {\sin \left (2 x \right )}{2} \,d x \\ \int { \frac {1}{p^{3}} \,dp} &= \int {\frac {\sin \left (2 x \right )}{2} \,d x} \\ -\frac {1}{2 p^{2}}&=-\frac {\cos \left (2 x \right )}{4}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{2 p \left (x \right )^{2}}+\frac {\cos \left (2 x \right )}{4}-c_{1} = 0 \] 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}{2 {y^{\prime }}^{2}}+\frac {\cos \left (2 x \right )}{4}-c_{1} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}} \tag {1} \\ y^{\prime }&=\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { -\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { \frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}\,\mathop {\mathrm {d}x}}\\ &= \frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{2} \\ \tag{2} y &= \frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{3} \\ \end{align*}
Verification of solutions
\[ y = -\frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{2} \] Verified OK.
\[ y = \frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \operatorname {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime }-{y^{\prime }}^{3} \sin \left (2 x \right )=0 \\ \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} \\ {} & {} & 2 u^{\prime }\left (x \right )-u \left (x \right )^{3} \sin \left (2 x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {u \left (x \right )^{3} \sin \left (2 x \right )}{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{3}}=\frac {\sin \left (2 x \right )}{2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{3}}d x =\int \frac {\sin \left (2 x \right )}{2}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 u \left (x \right )^{2}}=-\frac {\cos \left (2 x \right )}{4}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=-\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}, u \left (x \right )=\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-\frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \mathrm {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {2}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {2 \sqrt {-\frac {-4 c_{1} +\cos \left (2 x \right )}{4 c_{1} -1}}\, \mathrm {InverseJacobiAM}\left (x , \sqrt {2}\, \sqrt {-\frac {1}{4 c_{1} -1}}\right )}{\sqrt {-8 c_{1} +2 \cos \left (2 x \right )}}+c_{2} \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) = (1/2)*_b(_a)^3*sin(2*_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
Time used: 0.0 (sec). Leaf size: 80
dsolve(2*diff(y(x),x$2)=diff(y(x),x)^3*sin(2*x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {-\sin \left (x \right )^{2} c_{1}^{2}+1}\, \operatorname {InverseJacobiAM}\left (x , c_{1}\right )}{\sqrt {\frac {-\sin \left (x \right )^{2} c_{1}^{2}+1}{c_{1}^{2}}}}+c_{2} \\ y \left (x \right ) &= -\frac {\sqrt {-\sin \left (x \right )^{2} c_{1}^{2}+1}\, \operatorname {InverseJacobiAM}\left (x , c_{1}\right )}{\sqrt {\frac {-\sin \left (x \right )^{2} c_{1}^{2}+1}{c_{1}^{2}}}}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 6.102 (sec). Leaf size: 120
DSolve[2*y''[x]==(y'[x])^3*Sin[2*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-\frac {\sqrt {-\frac {\cos (2 x)+1-4 c_1}{-1+2 c_1}} \operatorname {EllipticF}\left (x,\frac {1}{1-2 c_1}\right )}{\sqrt {\cos (2 x)+1-4 c_1}} \\ y(x)\to \frac {\sqrt {-\frac {\cos (2 x)+1-4 c_1}{-1+2 c_1}} \operatorname {EllipticF}\left (x,\frac {1}{1-2 c_1}\right )}{\sqrt {\cos (2 x)+1-4 c_1}}+c_2 \\ y(x)\to c_2 \\ \end{align*}