Internal problem ID [6782]
Internal file name [OUTPUT/6029_Tuesday_July_26_2022_05_04_40_AM_96206336/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 94. Factoring the left member. EXERCISES
Page 309
Problem number: 16.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, [_Abel, `2nd type`, `class A`]]
\[ \boxed {\left (x +y\right )^{2} {y^{\prime }}^{2}+\left (2 y^{2}+y x -x^{2}\right ) y^{\prime }+y \left (y-x \right )=0} \] The ode \begin {align*} \left (x +y\right )^{2} {y^{\prime }}^{2}+\left (2 y^{2}+y x -x^{2}\right ) y^{\prime }+y \left (y-x \right ) = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime } y+y^{\prime } x +y-x \right ) \left (y^{\prime } y+y^{\prime } x +y\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } y+y^{\prime } x +y-x = 0\tag {1} \\ y^{\prime } y+y^{\prime } x +y = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right ) x +\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x +u \left (x \right ) x = x \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u^{2}+2 u -1}{x \left (u +1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{2}+2 u -1}{u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+2 u -1}{u +1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}+2 u -1}{u +1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \frac {\ln \left (u^{2}+2 u -1\right )}{2}&=-\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}+2 u -1} &= {\mathrm e}^{-\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u^{2}+2 u -1} &= \frac {c_{3}}{x} \end {align*}
Which simplifies to \[ \sqrt {u \left (x \right )^{2}+2 u \left (x \right )-1} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] The solution is \[ \sqrt {u \left (x \right )^{2}+2 u \left (x \right )-1} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \sqrt {\frac {y^{2}}{x^{2}}+\frac {2 y}{x}-1} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x}\\ \sqrt {\frac {y^{2}+2 y x -x^{2}}{x^{2}}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {y^{2}+2 y x -x^{2}}{x^{2}}} &= \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {y^{2}+2 y x -x^{2}}{x^{2}}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {y^{2}+2 y x -x^{2}}{x^{2}}} &= \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {y^{2}+2 y x -x^{2}}{x^{2}}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{x} \] Verified OK.
Solving ODE (2) Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right ) x +\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x +u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {u \left (u +2\right )}{x \left (u +1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u \left (u +2\right )}{u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u \left (u +2\right )}{u +1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u \left (u +2\right )}{u +1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \frac {\ln \left (u \left (u +2\right )\right )}{2}&=-\ln \left (x \right )+c_{5} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u \left (u +2\right )} &= {\mathrm e}^{-\ln \left (x \right )+c_{5}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u \left (u +2\right )} &= \frac {c_{6}}{x} \end {align*}
Which simplifies to \[ \sqrt {u \left (x \right ) \left (u \left (x \right )+2\right )} = \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \] The solution is \[ \sqrt {u \left (x \right ) \left (u \left (x \right )+2\right )} = \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \sqrt {\frac {y \left (\frac {y}{x}+2\right )}{x}} = \frac {c_{6} {\mathrm e}^{c_{5}}}{x}\\ \sqrt {\frac {y \left (y+2 x \right )}{x^{2}}} = \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {y \left (y+2 x \right )}{x^{2}}} &= \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {y \left (y+2 x \right )}{x^{2}}} = \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {y \left (y+2 x \right )}{x^{2}}} &= \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {y \left (y+2 x \right )}{x^{2}}} = \frac {c_{6} {\mathrm e}^{c_{5}}}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x +y\right )^{2} {y^{\prime }}^{2}+\left (2 y^{2}+y x -x^{2}\right ) y^{\prime }+y \left (y-x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {y}{x +y}, y^{\prime }=-\frac {y-x}{x +y}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y}{x +y} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y-x}{x +y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear <- 1st order linear successful <- inverse linear successful 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 D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 85
dsolve((y(x)+x)^2*diff(y(x),x)^2+(2*y(x)^2+x*y(x)-x^2)*diff(y(x),x)+y(x)*(y(x)-x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -x -\sqrt {x^{2}+2 c_{1}} \\ y \left (x \right ) &= -x +\sqrt {x^{2}+2 c_{1}} \\ y \left (x \right ) &= \frac {-c_{1} x -\sqrt {2 x^{2} c_{1}^{2}+1}}{c_{1}} \\ y \left (x \right ) &= \frac {-c_{1} x +\sqrt {2 x^{2} c_{1}^{2}+1}}{c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.492 (sec). Leaf size: 172
DSolve[(y[x]+x)^2*(y'[x])^2+(2*y[x]^2+x*y[x]-x^2)*y'[x]+y[x]*(y[x]-x)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -x-\sqrt {x^2+e^{2 c_1}} \\ y(x)\to -x+\sqrt {x^2+e^{2 c_1}} \\ y(x)\to -x-\sqrt {2 x^2+e^{2 c_1}} \\ y(x)\to -x+\sqrt {2 x^2+e^{2 c_1}} \\ y(x)\to -\sqrt {x^2}-x \\ y(x)\to \sqrt {x^2}-x \\ y(x)\to -\sqrt {2} \sqrt {x^2}-x \\ y(x)\to \sqrt {2} \sqrt {x^2}-x \\ \end{align*}