problem 47

Internal problem ID [5799]
Internal ﬁle name [OUTPUT/5047_Sunday_June_05_2022_03_18_58_PM_62390225/index.tex]

Book: Ordinary diﬀerential equations and calculus of variations. Makarets and Reshetnyak. Wold Scientiﬁc. Singapore. 1995
Section: Chapter 1. First order diﬀerential equations. Section 1.2 Homogeneous equations problems. page 12
Problem number: 47.
ODE order: 1.
ODE degree: 1.

2.51.1 Solving as isobaric ode

Solving for \(y'\) gives \begin{align*} \tag{1} y' &= \frac {y \left (y \sqrt {x -x^{2} y^{2}}-1\right )}{2 x} \\ \end{align*} Each of the above ode’s is now solved

An ode \(y^{\prime }=f(x,y)\) is isobaric if \[ f(t x, t^m y) = t^{m-1} f(x,y)\tag {1} \] Where here \[ f(x,y) = \frac {y \left (y \sqrt {x -x^{2} y^{2}}-1\right )}{2 x}\tag {2} \] \(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives \[ m = -{\frac {1}{2}} \] Since the ode is isobaric of order \(m=-{\frac {1}{2}}\), then the substitution \begin {align*} y&=x u^m \\ &=\frac {u}{\sqrt {x}} \end {align*}

Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives \[ \frac {2 u^{\prime }\left (x \right ) x -u \left (x \right )}{2 x^{\frac {3}{2}}} = \frac {u \left (x \right ) \left (u \left (x \right ) \sqrt {x -x u \left (x \right )^{2}}-\sqrt {x}\right )}{2 x^{2}} \] Or \[ u^{\prime }\left (x \right ) = \frac {\sqrt {x -x u \left (x \right )^{2}}\, u \left (x \right )^{2}}{2 x^{\frac {3}{2}}} \] Simplifying the above ode, assuming \(x>0\) gives \[ u^{\prime }\left (x \right ) = \frac {\sqrt {1-u \left (x \right )^{2}}\, u \left (x \right )^{2}}{2 x} \] Which is now solved as separable in \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {\sqrt {-u^{2}+1}\, u^{2}}{2 x} \end {align*}

Where \(f(x)=\frac {1}{2 x}\) and \(g(u)=\sqrt {-u^{2}+1}\, u^{2}\). Integrating both sides gives \begin{align*} \frac {1}{\sqrt {-u^{2}+1}\, u^{2}} \,du &= \frac {1}{2 x} \,d x \\ \int { \frac {1}{\sqrt {-u^{2}+1}\, u^{2}} \,du} &= \int {\frac {1}{2 x} \,d x} \\ -\frac {\sqrt {-u^{2}+1}}{u}&=\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {\sqrt {1-u \left (x \right )^{2}}}{u \left (x \right )}-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Now \(u \left (x \right )\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{\frac {1}{\sqrt {x}}}\) which results in the solution \[ -\frac {\sqrt {1-y^{2} x}}{y \sqrt {x}}-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\sqrt {1-y^{2} x}}{y \sqrt {x}}-\frac {\ln \left (x \right )}{2}-c_{1} &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ -\frac {\sqrt {1-y^{2} x}}{y \sqrt {x}}-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Veriﬁed OK. {0 < x}

2.51.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime } x +y-y^{2} \sqrt {x -x^{2} y^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y+y^{2} \sqrt {x -x^{2} y^{2}}}{2 x} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying homogeneous types: 
trying homogeneous G 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful 
<- homogeneous successful`

\[ -\frac {-1+x y \left (x \right )^{2}}{y \left (x \right ) \sqrt {-x \left (-1+x y \left (x \right )^{2}\right )}}+\frac {\ln \left (x \right )}{2}-c_{1} = 0 \]

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

2.51 problem 47

2.51.1 Solving as isobaric ode

2.51.2 Maple step by step solution