Internal problem ID [5793]
Internal file name [OUTPUT/5041_Sunday_June_05_2022_03_18_44_PM_19033746/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: Example 6.
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`], _rational]
\[ \boxed {2 x y^{\prime } \left (x -y^{2}\right )+y^{3}=0} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= \frac {y^{3}}{2 x \left (-x +y^{2}\right )} \\ \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 \[ f(x,y) = \frac {y^{3}}{2 x \left (-x +y^{2}\right )}\tag {2} \] \(m\) 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 \\ &=u \sqrt {x} \end {align*}
Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives \[ \frac {2 x u^{\prime }\left (x \right )+u \left (x \right )}{2 \sqrt {x}} = \frac {u \left (x \right )^{3}}{\sqrt {x}\, \left (2 u \left (x \right )^{2}-2\right )} \] Or \[ u^{\prime }\left (x \right ) = \frac {u \left (x \right )}{2 x u \left (x \right )^{2}-2 x} \] 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 {u}{2 x \left (u^{2}-1\right )} \end {align*}
Where \(f(x)=\frac {1}{2 x}\) and \(g(u)=\frac {u}{u^{2}-1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u}{u^{2}-1}} \,du &= \frac {1}{2 x} \,d x \\ \int { \frac {1}{\frac {u}{u^{2}-1}} \,du} &= \int {\frac {1}{2 x} \,d x} \\ \frac {u^{2}}{2}-\ln \left (u \right )&=\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {u \left (x \right )^{2}}{2}-\ln \left (u \left (x \right )\right )-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Now \(u \left (x \right )\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{\sqrt {x}}\) which results in the solution \[ \frac {y^{2}}{2 x}-\ln \left (\frac {y}{\sqrt {x}}\right )-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2 x}-\ln \left (\frac {y}{\sqrt {x}}\right )-\frac {\ln \left (x \right )}{2}-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2 x}-\ln \left (\frac {y}{\sqrt {x}}\right )-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x y^{\prime } \left (x -y^{2}\right )+y^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{3}}{2 x \left (x -y^{2}\right )} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous G <- homogeneous successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 28
dsolve(2*x*diff(y(x),x)*(x-y(x)^2)+y(x)^3=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\frac {c_{1}}{2}}}{\sqrt {-\frac {{\mathrm e}^{c_{1}}}{x \operatorname {LambertW}\left (-\frac {{\mathrm e}^{c_{1}}}{x}\right )}}} \]
✓ Solution by Mathematica
Time used: 2.287 (sec). Leaf size: 60
DSolve[2*x*y'[x]*(x-y[x]^2)+y[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -i \sqrt {x} \sqrt {W\left (-\frac {e^{c_1}}{x}\right )} \\ y(x)\to i \sqrt {x} \sqrt {W\left (-\frac {e^{c_1}}{x}\right )} \\ y(x)\to 0 \\ \end{align*}