problem 1(c)

Internal problem ID [6205]
Internal ﬁle name [OUTPUT/5453_Sunday_June_05_2022_03_39_02_PM_43511898/index.tex]

Book: Diﬀerential Equations: Theory, Technique, and Practice by George Simmons, Steven Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 1. What is a diﬀerential equation. Section 1.7. Homogeneous Equations. Page 28
Problem number: 1(c).
ODE order: 1.
ODE degree: 1.

\[ \boxed {x^{2} y^{\prime }-3 \left (x^{2}+y^{2}\right ) \arctan \left (\frac {y}{x}\right )-y x=0} \]

5.3.1 Solving as homogeneous ode

In canonical form, the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {3 \arctan \left (\frac {y}{x}\right ) x^{2}+3 \arctan \left (\frac {y}{x}\right ) y^{2}+x y}{x^{2}}\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 \(M=3 \arctan \left (\frac {y}{x}\right ) x^{2}+3 \arctan \left (\frac {y}{x}\right ) y^{2}+x y\) and \(N=x^{2}\) 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 &= 3 \arctan \left (u \right )+3 \arctan \left (u \right ) u^{2}+u\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {3 \arctan \left (u \left (x \right )\right )+3 \arctan \left (u \left (x \right )\right ) u \left (x \right )^{2}}{x} \end {align*}

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

Where \(f(x)=\frac {3}{x}\) and \(g(u)=\left (u^{2}+1\right ) \arctan \left (u \right )\). Integrating both sides gives \begin{align*} \frac {1}{\left (u^{2}+1\right ) \arctan \left (u \right )} \,du &= \frac {3}{x} \,d x \\ \int { \frac {1}{\left (u^{2}+1\right ) \arctan \left (u \right )} \,du} &= \int {\frac {3}{x} \,d x} \\ \ln \left (\arctan \left (u \right )\right )&=3 \ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \arctan \left (u \right ) &= {\mathrm e}^{3 \ln \left (x \right )+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} \arctan \left (u \right ) &= c_{3} x^{3} \end {align*}

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

Summary

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

Veriﬁcation of solutions

\[ y = x \tan \left (c_{3} {\mathrm e}^{c_{2}} x^{3}\right ) \] Veriﬁed OK.

5.3.2 Maple step by step solution

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

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`

\begin{align*} y(x)\to x \tan \left (x^3 (\cosh (3 c_1)-\sinh (3 c_1))\right ) \\ y(x)\to 0 \\ \end{align*}

5.3 problem 1(c)

5.3.1 Solving as homogeneous ode

5.3.2 Maple step by step solution