Internal problem ID [5798]
Internal file name [OUTPUT/5046_Sunday_June_05_2022_03_18_55_PM_27794649/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: 46.
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, _special]]
\[ \boxed {y^{\prime }-y^{2}=-\frac {2}{x^{2}}} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= \frac {y^{2} x^{2}-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 \[ f(x,y) = \frac {y^{2} x^{2}-2}{x^{2}}\tag {2} \] \(m\) 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 \[ \frac {u^{\prime }\left (x \right ) x -u \left (x \right )}{x^{2}} = \frac {u \left (x \right )^{2}-2}{x^{2}} \] Or \[ u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{2}+u \left (x \right )-2}{x} \] 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}+u -2}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u^{2}+u -2\). Integrating both sides gives \begin{align*} \frac {1}{u^{2}+u -2} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u^{2}+u -2} \,du} &= \int {\frac {1}{x} \,d x} \\ -\frac {\ln \left (u +2\right )}{3}+\frac {\ln \left (u -1\right )}{3}&=\ln \left (x \right )+c_{1} \\ \end{align*} The above can be written as \begin {align*} \left (-{\frac {1}{3}}\right ) \left (\ln \left (u +2\right )-\ln \left (u -1\right )\right ) &= 2 c_{1} +\ln \left (x \right )\\ \ln \left (u +2\right )-\ln \left (u -1\right ) &= \left (-3\right ) \left (2 c_{1} +\ln \left (x \right )\right ) \\ &= -6 c_{1} -3 \ln \left (x \right ) \end {align*}
Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (u +2\right )-\ln \left (u -1\right )} &= {\mathrm e}^{-3 \ln \left (x \right )-3 c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {u +2}{u -1} &= -\frac {3 c_{1}}{x^{3}}\\ &= \frac {c_{2}}{x^{3}} \end {align*}
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 \[ y = \frac {2 x^{3}+c_{2}}{x \left (-x^{3}+c_{2} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 x^{3}+c_{2}}{x \left (-x^{3}+c_{2} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 x^{3}+c_{2}}{x \left (-x^{3}+c_{2} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=-\frac {2}{x^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-\frac {2}{x^{2}} \end {array} \]
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: 0.328 (sec). Leaf size: 24
dsolve(diff(y(x),x)=y(x)^2-2/x^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 x^{3}+c_{1}}{x \left (-x^{3}+c_{1} \right )} \]
✓ Solution by Mathematica
Time used: 0.14 (sec). Leaf size: 32
DSolve[y'[x]==y[x]^2-2/x^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {-2 x^3+c_1}{x \left (x^3+c_1\right )} \\ y(x)\to \frac {1}{x} \\ \end{align*}