Internal problem ID [5797]
Internal file name [OUTPUT/5045_Sunday_June_05_2022_03_18_53_PM_28399984/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: 45.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries], _Chini]
\[ \boxed {2 y^{\prime }-4 \sqrt {y}=-x} \]
Solving for \(y'\) gives \begin{align*} \tag{1} y' &= -\frac {x}{2}+2 \sqrt {y} \\ \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 \[ f(x,y) = -\frac {x}{2}+2 \sqrt {y}\tag {2} \] \(m\) 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 \[ x \left (u^{\prime }\left (x \right ) x +2 u \left (x \right )\right ) = -\frac {x}{2}+2 \sqrt {x^{2} u \left (x \right )} \] Or \[ u^{\prime }\left (x \right ) = \frac {-4 x u \left (x \right )+4 \sqrt {x^{2} u \left (x \right )}-x}{2 x^{2}} \] Simplifying the above ode, assuming \(x>0\) gives \[ u^{\prime }\left (x \right ) = \frac {-4 u \left (x \right )+4 \sqrt {u \left (x \right )}-1}{2 x} \] 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 {-2 u +2 \sqrt {u}-\frac {1}{2}}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=-2 u +2 \sqrt {u}-\frac {1}{2}\). Integrating both sides gives \begin{align*} \frac {1}{-2 u +2 \sqrt {u}-\frac {1}{2}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{-2 u +2 \sqrt {u}-\frac {1}{2}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {1}{2 \sqrt {u}-1}-\ln \left (2 \sqrt {u}-1\right )&=\ln \left (x \right )+c_{1} \\ \end{align*} The solution is \[ \frac {1}{2 \sqrt {u \left (x \right )}-1}-\ln \left (2 \sqrt {u \left (x \right )}-1\right )-\ln \left (x \right )-c_{1} = 0 \] 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 \[ \frac {1}{2 \sqrt {\frac {y}{x^{2}}}-1}-\ln \left (2 \sqrt {\frac {y}{x^{2}}}-1\right )-\ln \left (x \right )-c_{1} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {1}{2 \sqrt {\frac {y}{x^{2}}}-1}-\ln \left (2 \sqrt {\frac {y}{x^{2}}}-1\right )-\ln \left (x \right )-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[
\frac {1}{2 \sqrt {\frac {y}{x^{2}}}-1}-\ln \left (2 \sqrt {\frac {y}{x^{2}}}-1\right )-\ln \left (x \right )-c_{1} = 0
\] Verified OK. {0 < x}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime }-4 \sqrt {y}=-x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x}{2}+2 \sqrt {y} \end {array} \]
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 Chini <- Chini successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 100
dsolve(2*diff(y(x),x)+x=4*sqrt(y(x)),y(x), singsol=all)
\[ \frac {\left (-x^{2}+4 y \left (x \right )\right ) \ln \left (\frac {x^{2}-4 y \left (x \right )}{x^{2}}\right )+2 i \left (x^{2}-4 y \left (x \right )\right ) \arctan \left (2 \sqrt {-\frac {y \left (x \right )}{x^{2}}}\right )-4 i \sqrt {-\frac {y \left (x \right )}{x^{2}}}\, x^{2}+4 \left (-c_{1} +2 \ln \left (x \right )\right ) y \left (x \right )+x^{2} \left (c_{1} -2 \ln \left (x \right )-2\right )}{x^{2}-4 y \left (x \right )} = 0 \]
✓ Solution by Mathematica
Time used: 0.104 (sec). Leaf size: 49
DSolve[2*y'[x]+x==4*Sqrt[y[x]],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [4 \left (\frac {4}{4 \sqrt {\frac {y(x)}{x^2}}+2}+2 \log \left (4 \sqrt {\frac {y(x)}{x^2}}+2\right )\right )=-8 \log (x)+c_1,y(x)\right ] \]