Internal problem ID [5805]
Internal file name [OUTPUT/5053_Sunday_June_05_2022_03_19_20_PM_20550564/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: 53.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_homogeneous, `class G`]]
\[ \boxed {y \left (1+\sqrt {x^{2} y^{4}-1}\right )+2 x y^{\prime }=0} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= -\frac {y \left (1+\sqrt {x^{2} y^{4}-1}\right )}{2 x} \\ \end{align*} Each of the above ode’s is now solved
Solving ode 1
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 \[ f(x,y) = -\frac {y \left (1+\sqrt {x^{2} y^{4}-1}\right )}{2 x}\tag {2} \] \(m\) 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 \[ \frac {2 u^{\prime }\left (x \right ) x -u \left (x \right )}{2 x^{\frac {3}{2}}} = -\frac {u \left (x \right ) \left (1+\sqrt {u \left (x \right )^{4}-1}\right )}{2 x^{\frac {3}{2}}} \] Or \[ u^{\prime }\left (x \right ) = -\frac {u \left (x \right ) \sqrt {u \left (x \right )^{4}-1}}{2 x} \] 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 {u \sqrt {u^{4}-1}}{2 x} \end {align*}
Where \(f(x)=-\frac {1}{2 x}\) and \(g(u)=u \sqrt {u^{4}-1}\). Integrating both sides gives \begin{align*} \frac {1}{u \sqrt {u^{4}-1}} \,du &= -\frac {1}{2 x} \,d x \\ \int { \frac {1}{u \sqrt {u^{4}-1}} \,du} &= \int {-\frac {1}{2 x} \,d x} \\ -\frac {\arctan \left (\frac {1}{\sqrt {u^{4}-1}}\right )}{2}&=-\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {\arctan \left (\frac {1}{\sqrt {u \left (x \right )^{4}-1}}\right )}{2}+\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] 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 \[ -\frac {\arctan \left (\frac {1}{\sqrt {x^{2} y^{4}-1}}\right )}{2}+\frac {\ln \left (x \right )}{2}-c_{1} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\arctan \left (\frac {1}{\sqrt {x^{2} y^{4}-1}}\right )}{2}+\frac {\ln \left (x \right )}{2}-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {\arctan \left (\frac {1}{\sqrt {x^{2} y^{4}-1}}\right )}{2}+\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (1+\sqrt {x^{2} y^{4}-1}\right )+2 x y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y \left (1+\sqrt {x^{2} y^{4}-1}\right )}{2 x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G <- homogeneous successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 32
dsolve(y(x)*(1+sqrt(x^2*y(x)^4-1))+2*x*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {RootOf}\left (-\ln \left (x \right )+c_{1} -2 \left (\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a} \sqrt {\textit {\_a}^{4}-1}}d \textit {\_a} \right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.188 (sec). Leaf size: 40
DSolve[y[x]*(1+Sqrt[x^2*y[x]^4-1])+2*x*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\arctan \left (\sqrt {x^2 y(x)^4-1}\right )+\frac {1}{2} \log \left (x^2 y(x)^4\right )-2 \log (y(x))=c_1,y(x)\right ] \]