Internal problem ID [7217]
Internal file name [OUTPUT/6203_Sunday_June_05_2022_04_32_03_PM_48103369/index.tex]
Book: Own collection of miscellaneous problems
Section: section 3.0
Problem number: 27.
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]]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }+{y^{\prime }}^{3}=0} \]
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*} \left (x^{2}+1\right ) p^{\prime }\left (x \right )+p \left (x \right )^{3} = 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}}{x^{2}+1} \end {align*}
Where \(f(x)=-\frac {1}{x^{2}+1}\) and \(g(p)=p^{3}\). Integrating both sides gives \begin{align*} \frac {1}{p^{3}} \,dp &= -\frac {1}{x^{2}+1} \,d x \\ \int { \frac {1}{p^{3}} \,dp} &= \int {-\frac {1}{x^{2}+1} \,d x} \\ -\frac {1}{2 p^{2}}&=-\arctan \left (x \right )+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{2 p \left (x \right )^{2}}+\arctan \left (x \right )-c_{1} = 0 \] 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}}+\arctan \left (x \right )-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 {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}} \tag {1} \\ y^{\prime }&=\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}} \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 {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\sqrt {2 \arctan \left (x \right )-2 c_{1}}\, \left (x +i\right )}{2 \left (\arctan \left (x \right )-c_{1} \right )}-2 i \left (\int _{}^{\arctan \left (x \right )}\frac {1}{\left (2 \textit {\_a} -2 c_{1} \right )^{\frac {3}{2}} \left ({\mathrm e}^{2 i \textit {\_a}}+1\right )}d \textit {\_a} \right )+c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { \frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}\,\mathop {\mathrm {d}x}}\\ &= \frac {x +i}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}+2 i \left (\int _{}^{\arctan \left (x \right )}\frac {1}{\left (2 \textit {\_a} -2 c_{1} \right )^{\frac {3}{2}} \left ({\mathrm e}^{2 i \textit {\_a}}+1\right )}d \textit {\_a} \right )+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sqrt {2 \arctan \left (x \right )-2 c_{1}}\, \left (x +i\right )}{2 \left (\arctan \left (x \right )-c_{1} \right )}-2 i \left (\int _{}^{\arctan \left (x \right )}\frac {1}{\left (2 \textit {\_a} -2 c_{1} \right )^{\frac {3}{2}} \left ({\mathrm e}^{2 i \textit {\_a}}+1\right )}d \textit {\_a} \right )+c_{2} \\ \tag{2} y &= \frac {x +i}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}+2 i \left (\int _{}^{\arctan \left (x \right )}\frac {1}{\left (2 \textit {\_a} -2 c_{1} \right )^{\frac {3}{2}} \left ({\mathrm e}^{2 i \textit {\_a}}+1\right )}d \textit {\_a} \right )+c_{3} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\sqrt {2 \arctan \left (x \right )-2 c_{1}}\, \left (x +i\right )}{2 \left (\arctan \left (x \right )-c_{1} \right )}-2 i \left (\int _{}^{\arctan \left (x \right )}\frac {1}{\left (2 \textit {\_a} -2 c_{1} \right )^{\frac {3}{2}} \left ({\mathrm e}^{2 i \textit {\_a}}+1\right )}d \textit {\_a} \right )+c_{2} \] Verified OK.
\[ y = \frac {x +i}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}+2 i \left (\int _{}^{\arctan \left (x \right )}\frac {1}{\left (2 \textit {\_a} -2 c_{1} \right )^{\frac {3}{2}} \left ({\mathrm e}^{2 i \textit {\_a}}+1\right )}d \textit {\_a} \right )+c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) y^{\prime \prime }+{y^{\prime }}^{3}=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} \\ {} & {} & \left (x^{2}+1\right ) u^{\prime }\left (x \right )+u \left (x \right )^{3}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {u \left (x \right )^{3}}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{3}}=-\frac {1}{x^{2}+1} \\ \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 {1}{x^{2}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 u \left (x \right )^{2}}=-\arctan \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}, u \left (x \right )=-\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\int \frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\int -\frac {1}{\sqrt {2 \arctan \left (x \right )-2 c_{1}}}d x +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) = -_b(_a)^3/(_a^2+1), _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: 0.016 (sec). Leaf size: 33
dsolve((1+x^2)*diff(y(x),x$2)+diff(y(x),x)^3=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \int \frac {1}{\sqrt {c_{1} +2 \arctan \left (x \right )}}d x +c_{2} \\ y \left (x \right ) &= -\left (\int \frac {1}{\sqrt {c_{1} +2 \arctan \left (x \right )}}d x \right )+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 62.161 (sec). Leaf size: 59
DSolve[(1+x^2)*y''[x]+y'[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \int _1^x-\frac {1}{\sqrt {2 \arctan (K[1])-2 c_1}}dK[1]+c_2 \\ y(x)\to \int _1^x\frac {1}{\sqrt {2 \arctan (K[2])-2 c_1}}dK[2]+c_2 \\ \end{align*}