3.23 problem 23

Internal problem ID [7861]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 23
Date solved : Monday, October 21, 2024 at 04:30:19 PM
CAS classification : [[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]

Solve

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

3.23.1 Solved as second order missing y ode

Time used: 0.366 (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 )+x p \left (x \right )^{2} = 0 \end{align*}

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

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

Where

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

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx}\\ \int { \frac {1}{p^{2}}\,dp} &= \int { -\frac {x}{x^{2}+1} \,dx}\\ -\frac {1}{p \left (x \right )}&=\ln \left (\frac {1}{\sqrt {x^{2}+1}}\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 \(p^{2}=0\) for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=0 \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*} -\frac {1}{p \left (x \right )} = \ln \left (\frac {1}{\sqrt {x^{2}+1}}\right )+c_1\\ p \left (x \right ) = 0 \end{align*}

Solving for \(p \left (x \right )\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} p \left (x \right )&=0\\ p \left (x \right )&=-\frac {1}{\ln \left (\frac {1}{\sqrt {x^{2}+1}}\right )+c_1} \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 } = 0 \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 {0\, dx} + c_2 \\ y &= 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 } = -\frac {1}{-\frac {\ln \left (x^{2}+1\right )}{2}+c_1} \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 {\frac {2}{\ln \left (x^{2}+1\right )-2 c_1}\, dx}\\ y &= \int \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}d x + c_3 \end{align*}

\begin{align*} y&= \int \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}d x +c_3 \end{align*}

Will add steps showing solving for IC soon.

3.23.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d}{d x}y^{\prime }\right )+1+x {y^{\prime }}^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\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 )+1+x u \left (x \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {x u \left (x \right )^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )^{2}}=-\frac {x}{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}}d x =\int -\frac {x}{x^{2}+1}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-\frac {\ln \left (x^{2}+1\right )}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2}{\ln \left (x^{2}+1\right )-2 \mathit {C1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2}{\ln \left (x^{2}+1\right )-2 \mathit {C1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {2}{\ln \left (x^{2}+1\right )-2 \mathit {C1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {2}{\ln \left (x^{2}+1\right )-2 \mathit {C1}}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\int \frac {2}{\ln \left (x^{2}+1\right )-2 \mathit {C1}}d x +\mathit {C2} \end {array} \]

3.23.3 Maple trace

Methods for second order ODEs:
 

3.23.4 Maple dsolve solution

Solving time : 0.014 (sec)
Leaf size : 22

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

\[ y = 2 \left (\int \frac {1}{\ln \left (x^{2}+1\right )+2 c_1}d x \right )+c_2 \]

3.23.5 Mathematica DSolve solution

Solving time : 60.266 (sec)
Leaf size : 33

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

\[ y(x)\to \int _1^x-\frac {2}{2 c_1-\log \left (K[1]^2+1\right )}dK[1]+c_2 \]