problem 21

Internal problem ID [5769]
Internal ﬁle name [OUTPUT/5017_Sunday_June_05_2022_03_17_30_PM_64848144/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: 21.
ODE order: 1.
ODE degree: 1.

2.21.1 Solving as homogeneous ode

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

Or \[ u^{\prime }\left (x \right )-\frac {\frac {u \left (x \right )}{-2 \sqrt {u \left (x \right )}+1}-u \left (x \right )}{x} = 0 \] Or \[ 2 u^{\prime }\left (x \right ) x \sqrt {u \left (x \right )}-u^{\prime }\left (x \right ) x +2 u \left (x \right )^{\frac {3}{2}} = 0 \] 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 {2 u^{\frac {3}{2}}}{x \left (2 \sqrt {u}-1\right )} \end {align*}

Where \(f(x)=-\frac {2}{x}\) and \(g(u)=\frac {u^{\frac {3}{2}}}{2 \sqrt {u}-1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{\frac {3}{2}}}{2 \sqrt {u}-1}} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{\frac {3}{2}}}{2 \sqrt {u}-1}} \,du} &= \int {-\frac {2}{x} \,d x} \\ 2 \ln \left (u \right )+\frac {2}{\sqrt {u}}&=-2 \ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ 2 \ln \left (u \left (x \right )\right )+\frac {2}{\sqrt {u \left (x \right )}}+2 \ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ 2 \ln \left (\frac {y}{x}\right )+\frac {2}{\sqrt {\frac {y}{x}}}+2 \ln \left (x \right )-c_{2} = 0 \]

Summary

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

Veriﬁcation of solutions

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

2.21.2 Maple step by step solution

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

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying homogeneous types: 
trying homogeneous G 
1st order, trying the canonical coordinates of the invariance group 
<- 1st order, canonical coordinates successful 
<- homogeneous successful`

\[ \ln \left (y \left (x \right )\right )+\frac {x}{\sqrt {x y \left (x \right )}}-c_{1} = 0 \]

\[ \text {Solve}\left [\frac {2}{\sqrt {\frac {y(x)}{x}}}+2 \log \left (\frac {y(x)}{x}\right )=-2 \log (x)+c_1,y(x)\right ] \]

2.21 problem 21

2.21.1 Solving as homogeneous ode

2.21.2 Maple step by step solution