Internal problem ID [2306]
Internal file name [OUTPUT/2306_Tuesday_February_27_2024_08_25_39_AM_65165895/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: 34.
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]]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }+{y^{\prime }}^{2}=-1} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, 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*} \left (x^{2}+1\right ) p^{\prime }\left (x \right )+1+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)\\ &= \frac {-p^{2}-1}{x^{2}+1} \end {align*}
Where \(f(x)=\frac {1}{x^{2}+1}\) and \(g(p)=-p^{2}-1\). Integrating both sides gives \begin{align*} \frac {1}{-p^{2}-1} \,dp &= \frac {1}{x^{2}+1} \,d x \\ \int { \frac {1}{-p^{2}-1} \,dp} &= \int {\frac {1}{x^{2}+1} \,d x} \\ -\arctan \left (p \right )&=\arctan \left (x \right )+c_{1} \\ \end{align*} The solution is \[ -\arctan \left (p \left (x \right )\right )-\arctan \left (x \right )-c_{1} = 0 \] 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*} -\frac {\pi }{4}-c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = -\frac {\pi }{4} \end {align*}
Trying the constant \begin {align*} c_{1} = -\frac {\pi }{4} \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} -\arctan \left (p \right )-\arctan \left (x \right )+\frac {\pi }{4} = 0 \end {align*}
The constant \(c_{1} = -\frac {\pi }{4}\) 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*} -\arctan \left (y^{\prime }\right )-\arctan \left (x \right )+\frac {\pi }{4} = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \cot \left (\arctan \left (x \right )+\frac {\pi }{4}\right )\,\mathop {\mathrm {d}x}}\\ &= -x +2 \ln \left (x +1\right )+c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = 1 \end {align*}
Trying the constant \begin {align*} c_{2} = 1 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=-x +2 \ln \left (x +1\right )+1 \end {align*}
But this does not satisfy the initial conditions. Hence no solution can be found. The constant \(c_{2} = 1\) does not give valid solution.
Which is valid for any constant of integration. Therefore keeping the constant in place. Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +2 \ln \left (x +1\right )+1 \\ \end{align*}
Verification of solutions
\[ y = -x +2 \ln \left (x +1\right )+1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (x^{2}+1\right ) y^{\prime \prime }+{y^{\prime }}^{2}=-1, y \left (0\right )=1, 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} \\ {} & {} & \left (x^{2}+1\right ) u^{\prime }\left (x \right )+u \left (x \right )^{2}=-1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=\frac {-u \left (x \right )^{2}-1}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{-u \left (x \right )^{2}-1}=\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 )^{2}-1}d x =\int \frac {1}{x^{2}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (u \left (x \right )\right )=\arctan \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\tan \left (\arctan \left (x \right )+c_{1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\tan \left (\arctan \left (x \right )+c_{1} \right )d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\frac {\mathrm {I} \,{\mathrm e}^{4 \,\mathrm {I} c_{1}} x}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} c_{1}} \ln \left (\left (-{\mathrm e}^{2 \,\mathrm {I} c_{1}}+1\right ) x +\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} c_{1}}+\mathrm {I}\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {\mathrm {I} x}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=\frac {\mathrm {I} {\mathrm e}^{4 \,\mathrm {I} c_{1}} x}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {4 {\mathrm e}^{2 \,\mathrm {I} c_{1}} \ln \left (\left (-{\mathrm e}^{2 \,\mathrm {I} c_{1}}+1\right ) x +\mathrm {I} {\mathrm e}^{2 \,\mathrm {I} c_{1}}+\mathrm {I}\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {\mathrm {I} x}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} c_{1}} \ln \left (\mathrm {I}+\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} c_{1}}\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\mathrm {I} \,{\mathrm e}^{4 \,\mathrm {I} c_{1}}}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} c_{1}} \left (-{\mathrm e}^{2 \,\mathrm {I} c_{1}}+1\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2} \left (\left (-{\mathrm e}^{2 \,\mathrm {I} c_{1}}+1\right ) x +\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} c_{1}}+\mathrm {I}\right )}-\frac {\mathrm {I}}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=\frac {\mathrm {I} \,{\mathrm e}^{4 \,\mathrm {I} c_{1}}}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} c_{1}} \left (-{\mathrm e}^{2 \,\mathrm {I} c_{1}}+1\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2} \left (\mathrm {I}+\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} c_{1}}\right )}-\frac {\mathrm {I}}{\left ({\mathrm e}^{2 \,\mathrm {I} c_{1}}-1\right )^{2}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =-\frac {\pi }{4}, c_{2} =1-2 \ln \left (1+\mathrm {I}\right )\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-x +2 \ln \left (\left (1+\mathrm {I}\right ) \left (x +1\right )\right )+1-\ln \left (2\right )-\frac {\mathrm {I} \pi }{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-x +2 \ln \left (\left (1+\mathrm {I}\right ) \left (x +1\right )\right )+1-\ln \left (2\right )-\frac {\mathrm {I} \pi }{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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 21
dsolve([(1+x^2)*diff(y(x),x$2)+1+diff(y(x),x)^2=0,y(0) = 1, D(y)(0) = 1],y(x), singsol=all)
\[ y \left (x \right ) = -x +2 \ln \left (-x -1\right )-2 i \pi +1 \]
✓ Solution by Mathematica
Time used: 6.806 (sec). Leaf size: 23
DSolve[{(1+x^2)*y''[x]+1+y'[x]^2==0,{y[0]==1,y'[0]==1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -x+2 \log (-x-1)-2 i \pi +1 \]