Internal problem ID [5761]
Internal file name [OUTPUT/5009_Sunday_June_05_2022_03_17_11_PM_61223389/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: 13.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _dAlembert]
\[ \boxed {2 y x -y^{2}+\left (y^{2}+2 y x -x^{2}\right ) y^{\prime }=-x^{2}} \] With initial conditions \begin {align*} [y \left (1\right ) = -1] \end {align*}
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-x^{2}-2 x y +y^{2}}{-x^{2}+2 x y +y^{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 \(M=x^{2}+2 x y -y^{2}\) and \(N=x^{2}-2 x y -y^{2}\) 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^{2}-2 u -1}{u^{2}+2 u -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {u \left (x \right )^{2}-2 u \left (x \right )-1}{u \left (x \right )^{2}+2 u \left (x \right )-1}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right )^{2}-2 u \left (x \right )-1}{u \left (x \right )^{2}+2 u \left (x \right )-1}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) u \left (x \right )^{2} x +2 u^{\prime }\left (x \right ) u \left (x \right ) x +u \left (x \right )^{3}-u^{\prime }\left (x \right ) x +u \left (x \right )^{2}+u \left (x \right )+1 = 0 \] Or \[ x \left (u \left (x \right )^{2}+2 u \left (x \right )-1\right ) u^{\prime }\left (x \right )+\left (u \left (x \right )+1\right ) \left (u \left (x \right )^{2}+1\right ) = 0 \] 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 {\left (u +1\right ) \left (u^{2}+1\right )}{x \left (u^{2}+2 u -1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {\left (u +1\right ) \left (u^{2}+1\right )}{u^{2}+2 u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (u +1\right ) \left (u^{2}+1\right )}{u^{2}+2 u -1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {\left (u +1\right ) \left (u^{2}+1\right )}{u^{2}+2 u -1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ -\ln \left (u +1\right )+\ln \left (u^{2}+1\right )&=-\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (u +1\right )+\ln \left (u^{2}+1\right )} &= {\mathrm e}^{-\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {u^{2}+1}{u +1} &= \frac {c_{3}}{x} \end {align*}
The solution is \[ \frac {u \left (x \right )^{2}+1}{u \left (x \right )+1} = \frac {c_{3}}{x} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {\frac {y^{2}}{x^{2}}+1}{\frac {y}{x}+1} = \frac {c_{3}}{x} \] Which simplifies to \begin {align*} \frac {x^{2}+y^{2}}{x +y} = c_{3} \end {align*}
Writing the solution as \begin {align*} c_{1} \left (x^{2}+y^{2}\right ) = x +y \end {align*}
Where \(c_{1} = \frac {1}{c_{3}}\) and solving for \(c_{1}\) after applying initial conditions gives \(c_{1}=0\). Hence the above solution becomes \begin {align*} 0 = x +y \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} 0 &= x +y \\ \end{align*}
Verification of solutions
\[ 0 = x +y \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 y x -y^{2}+\left (y^{2}+2 y x -x^{2}\right ) y^{\prime }=-x^{2}, y \left (1\right )=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-x^{2}-2 y x +y^{2}}{y^{2}+2 y x -x^{2}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=-1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
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 D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.203 (sec). Leaf size: 7
dsolve([(x^2+2*x*y(x)-y(x)^2)+(y(x)^2+2*x*y(x)-x^2)*diff(y(x),x)=0,y(1) = -1],y(x), singsol=all)
\[ y \left (x \right ) = -x \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[{(x^2+2*x*y[x]-y[x]^2)+(y[x]^2+2*x*y[x]-x^2)*y'[x]==0,{y[1]==-1}},y[x],x,IncludeSingularSolutions -> True]
{}