problem 15

Internal problem ID [5763]
Internal ﬁle name [OUTPUT/5011_Sunday_June_05_2022_03_17_15_PM_86048617/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: 15.
ODE order: 1.
ODE degree: 1.

2.15.1 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y^{2}}{x \left (-x +y \right )}\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 \(M=-y^{2}\) and \(N=x \left (x -y \right )\) 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}}{u -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {u \left (x \right )^{2}}{u \left (x \right )-1}-u \left (x \right )}{x} \end {align*}

Or \[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right )^{2}}{u \left (x \right )-1}-u \left (x \right )}{x} = 0 \] Or \[ u^{\prime }\left (x \right ) x u \left (x \right )-u^{\prime }\left (x \right ) x -u \left (x \right ) = 0 \] Or \[ x \left (u \left (x \right )-1\right ) u^{\prime }\left (x \right )-u \left (x \right ) = 0 \] 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}{x \left (u -1\right )} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {u}{u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u}{u -1}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {u}{u -1}} \,du} &= \int {\frac {1}{x} \,d x} \\ u -\ln \left (u \right )&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ u \left (x \right )-\ln \left (u \left (x \right )\right )-\ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \frac {y}{x}-\ln \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {y}{x}-\ln \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ \frac {y}{x}-\ln \left (\frac {y}{x}\right )-\ln \left (x \right )-c_{2} = 0 \] Veriﬁed OK.

2.15.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2}+\left (x^{2}-x y\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2}}{x^{2}-x y} \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful`

\[ y \left (x \right ) = -x \operatorname {LambertW}\left (-\frac {{\mathrm e}^{-c_{1}}}{x}\right ) \]

\begin{align*} y(x)\to -x W\left (-\frac {e^{-c_1}}{x}\right ) \\ y(x)\to 0 \\ \end{align*}

2.15 problem 15

2.15.1 Solving as homogeneous ode

2.15.2 Maple step by step solution