Internal problem ID [5772]
Internal file name [OUTPUT/5020_Sunday_June_05_2022_03_17_39_PM_50656827/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: 24.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program :
Maple gives the following as the ode type
[[_homogeneous, `class A`], _dAlembert]
\[ \boxed {\left (y^{\prime } x +y\right )^{2}-y^{\prime } y^{2}=0} \]
Solving for \(y^{\prime }\) gives \begin {align*} y^{\prime }&=\frac {\left (-2 x +y+\sqrt {y^{2}-4 y x}\right ) y}{2 x^{2}}\tag {1} \\ y^{\prime }&=\frac {\left (-2 x +y-\sqrt {y^{2}-4 y x}\right ) y}{2 x^{2}}\tag {2} \end {align*}
Now ODE (1) is solved In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\left (-2 x +y +\sqrt {-4 x y +y^{2}}\right ) y}{2 x^{2}}\tag {1} \end {align*}
An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if \[ f(t^n x, t^n y)= t^n f(x,y) \] In this case, it can be seen that both \(M=\left (-2 x +y +\sqrt {-4 x y +y^{2}}\right ) y\) and \(N=2 x^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence \[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \] Applying the transformation \(y=ux\) to the above ODE in (1) gives \begin {align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= \frac {u \left (\sqrt {u \left (u -4\right )}+u -2\right )}{2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {u \left (x \right ) \left (\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}+u \left (x \right )-2\right )}{2}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right ) \left (\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}+u \left (x \right )-2\right )}{2}-u \left (x \right )}{x} = 0 \] Or \[ 2 u^{\prime }\left (x \right ) x -u \left (x \right )^{2}-u \left (x \right ) \sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}+4 u \left (x \right ) = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (\sqrt {u \left (u -4\right )}+u -4\right )}{2 x} \end {align*}
Where \(f(x)=\frac {1}{2 x}\) and \(g(u)=\left (\sqrt {u \left (u -4\right )}+u -4\right ) u\). Integrating both sides gives \begin{align*} \frac {1}{\left (\sqrt {u \left (u -4\right )}+u -4\right ) u} \,du &= \frac {1}{2 x} \,d x \\ \int { \frac {1}{\left (\sqrt {u \left (u -4\right )}+u -4\right ) u} \,du} &= \int {\frac {1}{2 x} \,d x} \\ \frac {\ln \left (\frac {u -2+\sqrt {u^{2}-4 u}}{u}\right )}{4}&=\frac {\ln \left (x \right )}{2}+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\left (\frac {u -2+\sqrt {u^{2}-4 u}}{u}\right )}^{\frac {1}{4}} &= {\mathrm e}^{\frac {\ln \left (x \right )}{2}+c_{2}} \end {align*}
Which simplifies to \begin {align*} \left (\frac {\sqrt {u \left (u -4\right )}+u -2}{u}\right )^{\frac {1}{4}} &= c_{3} \sqrt {x} \end {align*}
Which simplifies to \[ \left (\frac {\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}+u \left (x \right )-2}{u \left (x \right )}\right )^{\frac {1}{4}} = c_{3} {\mathrm e}^{c_{2}} \sqrt {x} \] The solution is \[ \left (\frac {\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}+u \left (x \right )-2}{u \left (x \right )}\right )^{\frac {1}{4}} = c_{3} {\mathrm e}^{c_{2}} \sqrt {x} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \left (\frac {\left (\sqrt {\frac {y \left (\frac {y}{x}-4\right )}{x}}+\frac {y}{x}-2\right ) x}{y}\right )^{\frac {1}{4}} = c_{3} {\mathrm e}^{c_{2}} \sqrt {x} \] Now ODE (2) is solved In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\left (-2 x +y -\sqrt {-4 x y +y^{2}}\right ) y}{2 x^{2}}\tag {1} \end {align*}
An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if \[ f(t^n x, t^n y)= t^n f(x,y) \] In this case, it can be seen that both \(M=-\left (2 x -y +\sqrt {-4 x y +y^{2}}\right ) y\) and \(N=2 x^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence \[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \] Applying the transformation \(y=ux\) to the above ODE in (1) gives \begin {align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= -\frac {u \left (\sqrt {u \left (u -4\right )}-u +2\right )}{2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-\frac {u \left (x \right ) \left (\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}-u \left (x \right )+2\right )}{2}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {-\frac {u \left (x \right ) \left (\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}-u \left (x \right )+2\right )}{2}-u \left (x \right )}{x} = 0 \] Or \[ 2 u^{\prime }\left (x \right ) x -u \left (x \right )^{2}+u \left (x \right ) \sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}+4 u \left (x \right ) = 0 \] Which is now solved as separable in \(u \left (x \right )\). Which is now solved in \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (-\sqrt {u \left (u -4\right )}+u -4\right )}{2 x} \end {align*}
Where \(f(x)=\frac {1}{2 x}\) and \(g(u)=u \left (-\sqrt {u \left (u -4\right )}+u -4\right )\). Integrating both sides gives \begin{align*} \frac {1}{u \left (-\sqrt {u \left (u -4\right )}+u -4\right )} \,du &= \frac {1}{2 x} \,d x \\ \int { \frac {1}{u \left (-\sqrt {u \left (u -4\right )}+u -4\right )} \,du} &= \int {\frac {1}{2 x} \,d x} \\ \frac {\ln \left (\frac {\sqrt {u^{2}-4 u}+2-u}{u}\right )}{4}&=\frac {\ln \left (x \right )}{2}+c_{5} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\left (\frac {\sqrt {u^{2}-4 u}+2-u}{u}\right )}^{\frac {1}{4}} &= {\mathrm e}^{\frac {\ln \left (x \right )}{2}+c_{5}} \end {align*}
Which simplifies to \begin {align*} \left (\frac {\sqrt {u \left (u -4\right )}-u +2}{u}\right )^{\frac {1}{4}} &= c_{6} \sqrt {x} \end {align*}
Which simplifies to \[ \left (\frac {\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}-u \left (x \right )+2}{u \left (x \right )}\right )^{\frac {1}{4}} = c_{6} \sqrt {x}\, {\mathrm e}^{c_{5}} \] The solution is \[ \left (\frac {\sqrt {u \left (x \right ) \left (u \left (x \right )-4\right )}-u \left (x \right )+2}{u \left (x \right )}\right )^{\frac {1}{4}} = c_{6} \sqrt {x}\, {\mathrm e}^{c_{5}} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \left (\frac {\left (\sqrt {\frac {y \left (\frac {y}{x}-4\right )}{x}}-\frac {y}{x}+2\right ) x}{y}\right )^{\frac {1}{4}} = c_{6} \sqrt {x}\, {\mathrm e}^{c_{5}} \]
The solution(s) found are the following \begin{align*} \tag{1} {\left (\frac {\sqrt {\frac {y^{2}-4 y x}{x^{2}}}\, x +y-2 x}{y}\right )}^{\frac {1}{4}} &= c_{3} {\mathrm e}^{c_{2}} \sqrt {x} \\ \tag{2} {\left (\frac {\sqrt {\frac {y^{2}-4 y x}{x^{2}}}\, x +2 x -y}{y}\right )}^{\frac {1}{4}} &= c_{6} \sqrt {x}\, {\mathrm e}^{c_{5}} \\ \end{align*}
Verification of solutions
\[
{\left (\frac {\sqrt {\frac {y^{2}-4 y x}{x^{2}}}\, x +y-2 x}{y}\right )}^{\frac {1}{4}} = c_{3} {\mathrm e}^{c_{2}} \sqrt {x}
\] Verified OK. {0 < x}
\[
{\left (\frac {\sqrt {\frac {y^{2}-4 y x}{x^{2}}}\, x +2 x -y}{y}\right )}^{\frac {1}{4}} = c_{6} \sqrt {x}\, {\mathrm e}^{c_{5}}
\] Verified OK. {0 < x}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y^{\prime } x +y\right )^{2}-y^{\prime } y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {\left (-2 x +y-\sqrt {y^{2}-4 y x}\right ) y}{2 x^{2}}, y^{\prime }=\frac {\left (-2 x +y+\sqrt {y^{2}-4 y x}\right ) y}{2 x^{2}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (-2 x +y-\sqrt {y^{2}-4 y x}\right ) y}{2 x^{2}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (-2 x +y+\sqrt {y^{2}-4 y x}\right ) y}{2 x^{2}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: *** Sublevel 2 *** Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations <- symmetries for implicit equations successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 124
dsolve((x*diff(y(x),x)+y(x))^2=y(x)^2*diff(y(x),x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 4 x \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= -\frac {2 c_{1}^{2} \left (-\sqrt {2}\, c_{1} +x \right )}{-2 c_{1}^{2}+x^{2}} \\ y \left (x \right ) &= -\frac {2 c_{1}^{2} \left (\sqrt {2}\, c_{1} +x \right )}{-2 c_{1}^{2}+x^{2}} \\ y \left (x \right ) &= \frac {c_{1}^{3} \sqrt {2}-2 c_{1}^{2} x}{-2 c_{1}^{2}+4 x^{2}} \\ y \left (x \right ) &= \frac {c_{1}^{2} \left (\sqrt {2}\, c_{1} +2 x \right )}{2 c_{1}^{2}-4 x^{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.501 (sec). Leaf size: 62
DSolve[(x*y'[x]+y[x])^2==y[x]^2*y'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {4 e^{-2 c_1}}{2+e^{2 c_1} x} \\ y(x)\to -\frac {e^{-2 c_1}}{2+4 e^{2 c_1} x} \\ y(x)\to 0 \\ y(x)\to 4 x \\ \end{align*}