Internal problem ID [5762]
Internal file name [OUTPUT/5010_Sunday_June_05_2022_03_17_14_PM_18960310/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: 14.
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, [_Abel, `2nd type`, `class A`]]
\[ \boxed {x y^{\prime }-y-y^{\prime } y=0} \]
In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y}{-x +y}\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 \(M=y\) and \(N=x -y\) are both homogeneous and of the same order \(n=1\). 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}{u -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-\frac {u \left (x \right )}{u \left (x \right )-1}-u \left (x \right )}{x} \end {align*}
Or \[ u^{\prime }\left (x \right )-\frac {-\frac {u \left (x \right )}{u \left (x \right )-1}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x u \left (x \right )-u^{\prime }\left (x \right ) x +u \left (x \right )^{2} = 0 \] Or \[ x \left (u \left (x \right )-1\right ) u^{\prime }\left (x \right )+u \left (x \right )^{2} = 0 \] 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^{2}}{x \left (u -1\right )} \end {align*}
Where \(f(x)=-\frac {1}{x}\) and \(g(u)=\frac {u^{2}}{u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}}{u -1}} \,du &= -\frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}}{u -1}} \,du} &= \int {-\frac {1}{x} \,d x} \\ \ln \left (u \right )+\frac {1}{u}&=-\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \ln \left (u \left (x \right )\right )+\frac {1}{u \left (x \right )}+\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \ln \left (\frac {y}{x}\right )+\frac {x}{y}+\ln \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (\frac {y}{x}\right )+\frac {x}{y}+\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ \ln \left (\frac {y}{x}\right )+\frac {x}{y}+\ln \left (x \right )-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-y-y^{\prime } y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{x -y} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear <- 1st order linear successful <- inverse linear successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 17
dsolve(x*diff(y(x),x)-y(x)=y(x)*diff(y(x),x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x}{\operatorname {LambertW}\left (-x \,{\mathrm e}^{-c_{1}}\right )} \]
✓ Solution by Mathematica
Time used: 3.949 (sec). Leaf size: 25
DSolve[x*y'[x]-y[x]==y[x]*y'[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x}{W\left (-e^{-c_1} x\right )} \\ y(x)\to 0 \\ \end{align*}