Internal problem ID [5796]
Internal file name [OUTPUT/5044_Sunday_June_05_2022_03_18_51_PM_9301031/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: 44.
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, [_Abel, `2nd type`, `class B`]]
\[ \boxed {y+x \left (1+2 y x \right ) y^{\prime }=0} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= -\frac {y}{x \left (1+2 y x \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 \[ f(x,y) = -\frac {y}{x \left (1+2 y x \right )}\tag {2} \] \(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives \[ m = -1 \] Since the ode is isobaric of order \(m=-1\), then the substitution \begin {align*} y&=x u^m \\ &=\frac {u}{x} \end {align*}
Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives \[ \frac {u^{\prime }\left (x \right ) x -u \left (x \right )}{x^{2}} = -\frac {u \left (x \right )}{x^{2} \left (1+2 u \left (x \right )\right )} \] Or \[ u^{\prime }\left (x \right ) = \frac {2 u \left (x \right )^{2}}{x \left (1+2 u \left (x \right )\right )} \] 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 {2 u^{2}}{x \left (1+2 u \right )} \end {align*}
Where \(f(x)=\frac {2}{x}\) and \(g(u)=\frac {u^{2}}{1+2 u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}}{1+2 u}} \,du &= \frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}}{1+2 u}} \,du} &= \int {\frac {2}{x} \,d x} \\ 2 \ln \left (u \right )-\frac {1}{u}&=2 \ln \left (x \right )+c_{1} \\ \end{align*} The solution is \[ 2 \ln \left (u \left (x \right )\right )-\frac {1}{u \left (x \right )}-2 \ln \left (x \right )-c_{1} = 0 \] Now \(u \left (x \right )\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{\frac {1}{x}}\) which results in the solution \[ 2 \ln \left (y x \right )-\frac {1}{y x}-2 \ln \left (x \right )-c_{1} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} 2 \ln \left (y x \right )-\frac {1}{y x}-2 \ln \left (x \right )-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ 2 \ln \left (y x \right )-\frac {1}{y x}-2 \ln \left (x \right )-c_{1} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y+x \left (1+2 y x \right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y}{x \left (1+2 y x \right )} \end {array} \]
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: 18
dsolve(y(x)+x*(2*x*y(x)+1)*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {1}{2 \operatorname {LambertW}\left (\frac {c_{1}}{2 x}\right ) x} \]
✓ Solution by Mathematica
Time used: 60.506 (sec). Leaf size: 36
DSolve[y[x]+x*(2*x*y[x]+1)*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2 x W\left (\frac {e^{\frac {1}{2} \left (-2-9 \sqrt [3]{-2} c_1\right )}}{x}\right )} \]