Internal problem ID [5753]
Internal file name [OUTPUT/5001_Sunday_June_05_2022_03_16_48_PM_72453845/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: 5.
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 A`], _rational, _dAlembert]
\[ \boxed {\left (x^{2}+y^{2}\right ) y^{\prime }-2 y x=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 y x}{x^{2}+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 \(M=2 y x\) and \(N=x^{2}+y^{2}\) 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 {2 u}{u^{2}+1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {2 u \left (x \right )}{u \left (x \right )^{2}+1}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {\frac {2 u \left (x \right )}{u \left (x \right )^{2}+1}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) u \left (x \right )^{2} x +u \left (x \right )^{3}+u^{\prime }\left (x \right ) x -u \left (x \right ) = 0 \] Or \[ \left (u \left (x \right )^{2}+1\right ) x u^{\prime }\left (x \right )+u \left (x \right )^{3}-u \left (x \right ) = 0 \] 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^{3}-u}{x \left (u^{2}+1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{3}-u}{u^{2}+1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{3}-u}{u^{2}+1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{3}-u}{u^{2}+1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ -\ln \left (u \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 \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} &= \frac {c_{3}}{x} \end {align*}
The solution is \[ \frac {u \left (x \right )^{2}-1}{u \left (x \right )} = \frac {c_{3}}{x} \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {x \left (\frac {y^{2}}{x^{2}}-1\right )}{y} = \frac {c_{3}}{x} \] Which simplifies to \begin {align*} -\frac {\left (x -y\right ) \left (y+x \right )}{y} = c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (x -y\right ) \left (y+x \right )}{y} &= c_{3} \\ \end{align*}
Verification of solutions
\[ -\frac {\left (x -y\right ) \left (y+x \right )}{y} = c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+y^{2}\right ) y^{\prime }-2 y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y x}{x^{2}+y^{2}} \end {array} \]
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 D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 47
dsolve((x^2+y(x)^2)*diff(y(x),x)=2*x*y(x),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {1-\sqrt {4 c_{1}^{2} x^{2}+1}}{2 c_{1}} \\ y \left (x \right ) &= \frac {1+\sqrt {4 c_{1}^{2} x^{2}+1}}{2 c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.931 (sec). Leaf size: 70
DSolve[(x^2+y[x]^2)*y'[x]==2*x*y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} \left (-\sqrt {4 x^2+e^{2 c_1}}-e^{c_1}\right ) \\ y(x)\to \frac {1}{2} \left (\sqrt {4 x^2+e^{2 c_1}}-e^{c_1}\right ) \\ y(x)\to 0 \\ \end{align*}