problem Problem 23

Internal problem ID [2679]
Internal ﬁle name [OUTPUT/2171_Sunday_June_05_2022_02_51_40_AM_38121038/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 1, First-Order Diﬀerential Equations. Section 1.8, Change of Variables. page 79
Problem number: Problem 23.
ODE order: 1.
ODE degree: 1.

4.15.1 Solving as homogeneous ode

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

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

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

Summary

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

Veriﬁcation of solutions

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

4.15.2 Maple step by step solution

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

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`

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

\begin{align*} y(x)\to -x \sqrt {\log ^2(x)+2 c_1 \log (x)-1+c_1{}^2} \\ y(x)\to x \sqrt {\log ^2(x)+2 c_1 \log (x)-1+c_1{}^2} \\ \end{align*}

4.15 problem Problem 23

4.15.1 Solving as homogeneous ode

4.15.2 Maple step by step solution