Internal problem ID [5795]
Internal file name [OUTPUT/5043_Sunday_June_05_2022_03_18_48_PM_32788025/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: 43.
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, _Bernoulli]
\[ \boxed {2 x^{2} y^{\prime }-y^{3}-y x=0} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= \frac {y \left (y^{2}+x \right )}{2 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 \[ f(x,y) = \frac {y \left (y^{2}+x \right )}{2 x^{2}}\tag {2} \] \(m\) 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 \[ \frac {2 x u^{\prime }\left (x \right )+u \left (x \right )}{2 \sqrt {x}} = \frac {u \left (x \right ) \left (u \left (x \right )^{2}+1\right )}{2 \sqrt {x}} \] Or \[ u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{3}}{2 x} \] 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^{3}}{2 x} \end {align*}
Where \(f(x)=\frac {1}{2 x}\) and \(g(u)=u^{3}\). Integrating both sides gives \begin{align*} \frac {1}{u^{3}} \,du &= \frac {1}{2 x} \,d x \\ \int { \frac {1}{u^{3}} \,du} &= \int {\frac {1}{2 x} \,d x} \\ -\frac {1}{2 u^{2}}&=\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{2 u \left (x \right )^{2}}-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Now \(u \left (x \right )\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{\sqrt {x}}\) which results in the solution \[ -\frac {x}{2 y^{2}}-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -\frac {x}{2 y^{2}}-\frac {\ln \left (x \right )}{2}-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {x}{2 y^{2}}-\frac {\ln \left (x \right )}{2}-c_{1} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} y^{\prime }-y^{3}-y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3}+y x}{2 x^{2}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 45
dsolve(2*x^2*diff(y(x),x)=y(x)^3+x*y(x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {\left (-\ln \left (x \right )+c_{1} \right ) x}}{\ln \left (x \right )-c_{1}} \\ y \left (x \right ) &= \frac {\sqrt {\left (-\ln \left (x \right )+c_{1} \right ) x}}{-\ln \left (x \right )+c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.158 (sec). Leaf size: 49
DSolve[2*x^2*y'[x]==y[x]^3+x*y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {x}}{\sqrt {-\log (x)+c_1}} \\ y(x)\to \frac {\sqrt {x}}{\sqrt {-\log (x)+c_1}} \\ y(x)\to 0 \\ \end{align*}