2.5 problem 5

Internal problem ID [3144]
Internal file name [OUTPUT/2636_Sunday_June_05_2022_08_37_49_AM_16493042/index.tex]

Book: An introduction to the solution and applications of differential equations, J.W. Searl, 1966
Section: Chapter 4, Ex. 4.2
Problem number: 5.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "bernoulli", "separable", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[_separable]

\[ \boxed {y^{\prime }-\frac {x \left (1+y^{2}\right )}{y \left (x^{2}+1\right )}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

2.5.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= \frac {x \left (y^{2}+1\right )}{y \left (x^{2}+1\right )} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=1\) is \[ \{-\infty

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=1\) is \[ \{-\infty

2.5.2 Solving as separable ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {x \left (y^{2}+1\right )}{y \left (x^{2}+1\right )} \end {align*}

Where \(f(x)=\frac {x}{x^{2}+1}\) and \(g(y)=\frac {y^{2}+1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {y^{2}+1}{y}} \,dy &= \frac {x}{x^{2}+1} \,d x \\ \int { \frac {1}{\frac {y^{2}+1}{y}} \,dy} &= \int {\frac {x}{x^{2}+1} \,d x} \\ \frac {\ln \left (y^{2}+1\right )}{2}&=\frac {\ln \left (x^{2}+1\right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {y^{2}+1} &= {\mathrm e}^{\frac {\ln \left (x^{2}+1\right )}{2}+c_{1}} \end {align*}

Which simplifies to \begin {align*} \sqrt {y^{2}+1} &= c_{2} \sqrt {x^{2}+1} \end {align*}

Which can be simplified to become \[ \sqrt {1+y^{2}} = c_{2} \sqrt {x^{2}+1}\, {\mathrm e}^{c_{1}} \] The solution is \[ \sqrt {1+y^{2}} = c_{2} \sqrt {x^{2}+1}\, {\mathrm e}^{c_{1}} \] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} \sqrt {2} = c_{2} {\mathrm e}^{c_{1}} \end {align*}

The solutions are \begin {align*} c_{1} = \frac {\ln \left (\frac {2}{c_{2}^{2}}\right )}{2} \end {align*}

Trying the constant \begin {align*} c_{1} = \frac {\ln \left (\frac {2}{c_{2}^{2}}\right )}{2} \end {align*}

Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \sqrt {y^{2}+1} = c_{2} \sqrt {x^{2}+1}\, \sqrt {2}\, \sqrt {\frac {1}{c_{2}^{2}}} \end {align*}

The constant \(c_{1} = \frac {\ln \left (\frac {2}{c_{2}^{2}}\right )}{2}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} \sqrt {1+y^{2}} &= \sqrt {x^{2}+1}\, \sqrt {2}\, \operatorname {csgn}\left (\frac {1}{c_{2}}\right ) \\ \end{align*}

Verification of solutions

\[ \sqrt {1+y^{2}} = \sqrt {x^{2}+1}\, \sqrt {2}\, \operatorname {csgn}\left (\frac {1}{c_{2}}\right ) \] Verified OK.

2.5.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {x \left (1+y^{2}\right )}{y \left (x^{2}+1\right )}=0, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x \left (1+y^{2}\right )}{y \left (x^{2}+1\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{1+y^{2}}=\frac {x}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{1+y^{2}}d x =\int \frac {x}{x^{2}+1}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (1+y^{2}\right )}{2}=\frac {\ln \left (x^{2}+1\right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left (-x^{2}+{\mathrm e}^{-2 c_{1}}-1\right )}}{{\mathrm e}^{-2 c_{1}}}, y=-\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left (-x^{2}+{\mathrm e}^{-2 c_{1}}-1\right )}}{{\mathrm e}^{-2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-1\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\frac {\ln \left (2\right )}{2} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =\frac {\ln \left (2\right )}{2}\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=\sqrt {2 x^{2}+1} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=-\frac {\sqrt {-{\mathrm e}^{-2 c_{1}} \left ({\mathrm e}^{-2 c_{1}}-1\right )}}{{\mathrm e}^{-2 c_{1}}} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =\left (\right ) \\ \bullet & {} & \textrm {Solution does not satisfy initial condition}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\sqrt {2 x^{2}+1} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
<- Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.031 (sec). Leaf size: 13

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

\[ y \left (x \right ) = \sqrt {2 x^{2}+1} \]

✓ Solution by Mathematica

Time used: 0.549 (sec). Leaf size: 16

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

\[ y(x)\to \sqrt {2 x^2+1} \]