2.1 problem 1

2.1.1 Solving as separable ode
2.1.2 Maple step by step solution

Internal problem ID [3140]
Internal file name [OUTPUT/2632_Sunday_June_05_2022_08_37_44_AM_69209561/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: 1.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program :

Maple gives the following as the ode type

[_separable]

\[ \boxed {x^{2} \left (1+y^{2}\right ) y^{\prime }+y^{2} \left (x^{2}+1\right )=0} \]

2.1.1 Solving as separable ode

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

Where \(f(x)=-\frac {x^{2}+1}{x^{2}}\) and \(g(y)=\frac {y^{2}}{y^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {y^{2}}{y^{2}+1}} \,dy &= -\frac {x^{2}+1}{x^{2}} \,d x \\ \int { \frac {1}{\frac {y^{2}}{y^{2}+1}} \,dy} &= \int {-\frac {x^{2}+1}{x^{2}} \,d x} \\ y -\frac {1}{y}&=-x +\frac {1}{x}+c_{1} \\ \end{align*} Which results in \begin{align*} y &= \frac {c_{1} x -x^{2}+1+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}}{2 x} \\ y &= -\frac {-c_{1} x +x^{2}+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}-1}{2 x} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} x -x^{2}+1+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}}{2 x} \\ \tag{2} y &= -\frac {-c_{1} x +x^{2}+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}-1}{2 x} \\ \end{align*}

Figure 10: Slope field plot

Verification of solutions

\[ y = \frac {c_{1} x -x^{2}+1+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}}{2 x} \] Verified OK.

\[ y = -\frac {-c_{1} x +x^{2}+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}-1}{2 x} \] Verified OK.

2.1.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (1+y^{2}\right ) y^{\prime }+y^{2} \left (x^{2}+1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2} \left (x^{2}+1\right )}{x^{2} \left (1+y^{2}\right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } \left (1+y^{2}\right )}{y^{2}}=-\frac {x^{2}+1}{x^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } \left (1+y^{2}\right )}{y^{2}}d x =\int -\frac {x^{2}+1}{x^{2}}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y-\frac {1}{y}=-x +\frac {1}{x}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{y=\frac {c_{1} x -x^{2}+1+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}}{2 x}, y=-\frac {-c_{1} x +x^{2}+\sqrt {c_{1}^{2} x^{2}-2 c_{1} x^{3}+x^{4}+2 c_{1} x +2 x^{2}+1}-1}{2 x}\right \} \end {array} \]

Maple trace

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

Solution by Maple

Time used: 0.0 (sec). Leaf size: 93

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

\begin{align*} y \left (x \right ) &= \frac {-x^{2}-c_{1} x +\sqrt {1+x^{4}+2 c_{1} x^{3}+\left (c_{1}^{2}+2\right ) x^{2}-2 c_{1} x}+1}{2 x} \\ y \left (x \right ) &= \frac {-x^{2}-c_{1} x -\sqrt {1+x^{4}+2 c_{1} x^{3}+\left (c_{1}^{2}+2\right ) x^{2}-2 c_{1} x}+1}{2 x} \\ \end{align*}

Solution by Mathematica

Time used: 1.162 (sec). Leaf size: 95

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

\begin{align*} y(x)\to -\frac {x^2+\sqrt {4 x^2+\left (-x^2+c_1 x+1\right ){}^2}-c_1 x-1}{2 x} \\ y(x)\to \frac {-x^2+\sqrt {4 x^2+\left (-x^2+c_1 x+1\right ){}^2}+c_1 x+1}{2 x} \\ y(x)\to 0 \\ \end{align*}