problem 20

Internal problem ID [5768]
Internal ﬁle name [OUTPUT/5016_Sunday_June_05_2022_03_17_28_PM_96250479/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: 20.
ODE order: 1.
ODE degree: 1.

2.20.1 Solving as homogeneous ode

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

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

Where \(f(x)=\frac {1}{x}\) and \(g(u)=\sqrt {u^{2}-1}\). Integrating both sides gives \begin{align*} \frac {1}{\sqrt {u^{2}-1}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\sqrt {u^{2}-1}} \,du} &= \int {\frac {1}{x} \,d x} \\ \ln \left (u +\sqrt {u^{2}-1}\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} u +\sqrt {u^{2}-1} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} u +\sqrt {u^{2}-1} &= c_{3} x \end {align*}

Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ y = \frac {\left ({\mathrm e}^{2 c_{2}} x^{2} c_{3}^{2}+1\right ) {\mathrm e}^{-c_{2}}}{2 c_{3}} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left ({\mathrm e}^{2 c_{2}} x^{2} c_{3}^{2}+1\right ) {\mathrm e}^{-c_{2}}}{2 c_{3}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {\left ({\mathrm e}^{2 c_{2}} x^{2} c_{3}^{2}+1\right ) {\mathrm e}^{-c_{2}}}{2 c_{3}} \] Veriﬁed OK. {0 < x}

2.20.2 Maple step by step solution

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

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`

\[ \frac {-c_{1} x^{2}+y \left (x \right )+\sqrt {y \left (x \right )^{2}-x^{2}}}{x^{2}} = 0 \]

2.20 problem 20

2.20.1 Solving as homogeneous ode

2.20.2 Maple step by step solution