Internal problem ID [5794]
Internal file name [OUTPUT/5042_Sunday_June_05_2022_03_18_46_PM_36692102/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: 42.
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, _Riccati]
\[ \boxed {x^{3} \left (y^{\prime }-x \right )-y^{2}=0} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= \frac {x^{4}+y^{2}}{x^{3}} \\ \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 \[ f(x,y) = \frac {x^{4}+y^{2}}{x^{3}}\tag {2} \] \(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives \[ m = 2 \] Since the ode is isobaric of order \(m=2\), then the substitution \begin {align*} y&=x u^m \\ &=u \,x^{2} \end {align*}
Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives \[ x \left (u^{\prime }\left (x \right ) x +2 u \left (x \right )\right ) = x \left (1+u \left (x \right )^{2}\right ) \] Or \[ u^{\prime }\left (x \right ) = \frac {\left (u \left (x \right )-1\right )^{2}}{x} \] 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}-2 u +1}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u^{2}-2 u +1\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}-2 u +1} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u^{2}-2 u +1} \,du} &= \int {\frac {1}{x} \,d x} \\ -\frac {1}{u -1}&=\ln \left (x \right )+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{u \left (x \right )-1}-\ln \left (x \right )-c_{1} = 0 \] Now \(u \left (x \right )\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x^{2}}\) which results in the solution \[ -\frac {1}{\frac {y}{x^{2}}-1}-\ln \left (x \right )-c_{1} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -\frac {1}{\frac {y}{x^{2}}-1}-\ln \left (x \right )-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {1}{\frac {y}{x^{2}}-1}-\ln \left (x \right )-c_{1} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (y^{\prime }-x \right )-y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{4}+y^{2}}{x^{3}} \end {array} \]
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: 0.015 (sec). Leaf size: 23
dsolve(x^3*(diff(y(x),x)-x)=y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{2} \left (\ln \left (x \right )-c_{1} -1\right )}{\ln \left (x \right )-c_{1}} \]
✓ Solution by Mathematica
Time used: 0.157 (sec). Leaf size: 29
DSolve[x^3*(y'[x]-x)==y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^2 (\log (x)-1+c_1)}{\log (x)+c_1} \\ y(x)\to x^2 \\ \end{align*}