Internal problem ID [5804]
Internal file name [OUTPUT/5052_Sunday_June_05_2022_03_19_17_PM_88188863/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: 52.
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]
\[ \boxed {\left (x^{2}-y^{4}\right ) y^{\prime }-y x=0} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= -\frac {y x}{y^{4}-x^{2}} \\ \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 \[ f(x,y) = -\frac {y x}{y^{4}-x^{2}}\tag {2} \] \(m\) 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 \[ \frac {2 x u^{\prime }\left (x \right )+u \left (x \right )}{2 \sqrt {x}} = \frac {u \left (x \right )}{\left (-u \left (x \right )^{4}+1\right ) \sqrt {x}} \] Or \[ u^{\prime }\left (x \right ) = -\frac {u \left (x \right ) \left (u \left (x \right )^{4}+1\right )}{2 u \left (x \right )^{4} x -2 x} \] 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 \left (u^{4}+1\right )}{2 \left (u^{4}-1\right ) x} \end {align*}
Where \(f(x)=-\frac {1}{2 x}\) and \(g(u)=\frac {u \left (u^{4}+1\right )}{u^{4}-1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u \left (u^{4}+1\right )}{u^{4}-1}} \,du &= -\frac {1}{2 x} \,d x \\ \int { \frac {1}{\frac {u \left (u^{4}+1\right )}{u^{4}-1}} \,du} &= \int {-\frac {1}{2 x} \,d x} \\ \frac {\ln \left (u^{4}+1\right )}{2}-\ln \left (u \right )&=-\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\frac {\ln \left (u^{4}+1\right )}{2}-\ln \left (u \right )} &= {\mathrm e}^{-\frac {\ln \left (x \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {\sqrt {u^{4}+1}}{u} &= \frac {c_{2}}{\sqrt {x}} \end {align*}
The solution is \[ \frac {\sqrt {u \left (x \right )^{4}+1}}{u \left (x \right )} = \frac {c_{2}}{\sqrt {x}} \] 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 \[ \frac {\sqrt {\frac {y^{4}}{x^{2}}+1}\, \sqrt {x}}{y} = \frac {c_{2}}{\sqrt {x}} \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {\frac {y^{4}}{x^{2}}+1}\, \sqrt {x}}{y} &= \frac {c_{2}}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {\frac {y^{4}}{x^{2}}+1}\, \sqrt {x}}{y} = \frac {c_{2}}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-y^{4}\right ) y^{\prime }-y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y x}{x^{2}-y^{4}} \end {array} \]
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: 0.125 (sec). Leaf size: 97
dsolve((x^2-y(x)^4)*diff(y(x),x)-x*y(x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {\sqrt {-2 \sqrt {c_{1}^{2}-4 x^{2}}+2 c_{1}}}{2} \\ y \left (x \right ) &= \frac {\sqrt {-2 \sqrt {c_{1}^{2}-4 x^{2}}+2 c_{1}}}{2} \\ y \left (x \right ) &= -\frac {\sqrt {2 \sqrt {c_{1}^{2}-4 x^{2}}+2 c_{1}}}{2} \\ y \left (x \right ) &= \frac {\sqrt {2 \sqrt {c_{1}^{2}-4 x^{2}}+2 c_{1}}}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 5.14 (sec). Leaf size: 122
DSolve[(x^2-y[x]^4)*y'[x]-x*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {-\sqrt {-x^2+c_1{}^2}-c_1} \\ y(x)\to \sqrt {-\sqrt {-x^2+c_1{}^2}-c_1} \\ y(x)\to -\sqrt {\sqrt {-x^2+c_1{}^2}-c_1} \\ y(x)\to \sqrt {\sqrt {-x^2+c_1{}^2}-c_1} \\ y(x)\to 0 \\ \end{align*}