2.3.21 problem 21

Internal problem ID [8305]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 21
Date solved : Sunday, November 10, 2024 at 03:37:19 AM
CAS classification : [[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]

Solve

\begin{align*} \left (x^{2}+1\right ) y^{\prime \prime }+1+{y^{\prime }}^{2}&=0 \end{align*}

Solved as second order missing y ode

Time used: 0.704 (sec)

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. The ode \(p^{\prime }\left (x \right ) = -\frac {p \left (x \right )^{2}+1}{x^{2}+1}\) is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= -\frac {p \left (x \right )^{2}+1}{x^{2}+1}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x^{2}+1}\\ g(p) &= -p^{2}-1 \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx}\\ \int { \frac {1}{-p^{2}-1}\,dp} &= \int { \frac {1}{x^{2}+1} \,dx}\\ -\arctan \left (p \left (x \right )\right )&=\arctan \left (x \right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or \(-p^{2}-1=0\) for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=-i\\ p \left (x \right )&=i \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\arctan \left (p \left (x \right )\right ) = \arctan \left (x \right )+c_1\\ p \left (x \right ) = -i\\ p \left (x \right ) = i \end{align*}

Solving for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right ) &= -\tan \left (\arctan \left (x \right )+c_1 \right ) \\ \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*} y^{\prime } = -i \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {-i\, dx}\\ y &= -i x + c_2 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = i \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {i\, dx}\\ y &= i x + c_3 \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\tan \left (\arctan \left (x \right )+c_1 \right ) \end{align*}

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {-\tan \left (\arctan \left (x \right )+c_1 \right )\, dx}\\ y &= \frac {i {\mathrm e}^{4 i c_1} x}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-{\mathrm e}^{2 i c_1}+1\right ) x +i {\mathrm e}^{2 i c_1}+i\right )}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}}-\frac {i x}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} + c_4 \end{align*}

\begin{align*} y&= \frac {-4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-x +i\right ) {\mathrm e}^{2 i c_1}+i+x \right )-2 \,{\mathrm e}^{2 i c_1} c_4 +\left (i x +c_4 \right ) {\mathrm e}^{4 i c_1}-i x +c_4}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= \frac {-4 \,{\mathrm e}^{2 i c_1} \ln \left (\left (-x +i\right ) {\mathrm e}^{2 i c_1}+i+x \right )-2 \,{\mathrm e}^{2 i c_1} c_4 +\left (i x +c_4 \right ) {\mathrm e}^{4 i c_1}-i x +c_4}{\left ({\mathrm e}^{2 i c_1}-1\right )^{2}} \\ y &= -i x +c_2 \\ y &= i x +c_3 \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+1+\left (\frac {d}{d x}y \left (x \right )\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d}{d x}u \left (x \right )\right )+1+u \left (x \right )^{2}=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=\frac {-1-u \left (x \right )^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{-1-u \left (x \right )^{2}}=\frac {1}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{-1-u \left (x \right )^{2}}d x =\int \frac {1}{x^{2}+1}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\arctan \left (u \left (x \right )\right )=\arctan \left (x \right )+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\tan \left (\arctan \left (x \right )+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\tan \left (\arctan \left (x \right )+\mathit {C1} \right ) \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int -\tan \left (\arctan \left (x \right )+\mathit {C1} \right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\mathrm {I} \,{\mathrm e}^{4 \,\mathrm {I} \mathit {C1}} x}{\left ({\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}-1\right )^{2}}-\frac {\mathrm {I} x}{\left ({\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}-1\right )^{2}}-\frac {4 \,{\mathrm e}^{2 \,\mathrm {I} \mathit {C1}} \ln \left (\left (-{\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}+1\right ) x +\mathrm {I} \,{\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}+\mathrm {I}\right )}{\left ({\mathrm e}^{2 \,\mathrm {I} \mathit {C1}}-1\right )^{2}}+\mathit {C2} \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+_b(_a)^2)/(_a^2+1), _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   trying separable 
   <- separable successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`
 

Maple dsolve solution

Solving time : 0.020 (sec)
Leaf size : 33

dsolve((x^2+1)*diff(diff(y(x),x),x)+1+diff(y(x),x)^2 = 0, 
       y(x),singsol=all)
 

\[ y = \frac {\ln \left (c_{1} x -1\right ) c_{1}^{2}+c_{2} c_{1}^{2}+c_{1} x +\ln \left (c_{1} x -1\right )}{c_{1}^{2}} \]

Mathematica DSolve solution

Solving time : 6.907 (sec)
Leaf size : 33

DSolve[{(1+x^2)*D[y[x],{x,2}]+1+(D[y[x],x])^2==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to -x \cot (c_1)+\csc ^2(c_1) \log (-x \sin (c_1)-\cos (c_1))+c_2 \]