4.21 problem Problem 29(b)

Internal problem ID [2685]
Internal file name [OUTPUT/2177_Sunday_June_05_2022_02_52_11_AM_56933570/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 29(b).
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"

Maple gives the following as the ode type

[[_homogeneous, `class A`], _rational, [_Abel, `2nd type`, `class A`]]

\[ \boxed {y^{\prime }-\frac {x +\frac {y}{2}}{\frac {x}{2}-y}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1] \end {align*}

4.21.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -\frac {y +2 x}{2 y -x} \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=1\) is \[ \{x <2\boldsymbol {\lor }2

The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=1\) is \[ \{x <2\boldsymbol {\lor }2

4.21.2 Solving as homogeneous ode

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

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

Where \(f(x)=-\frac {2}{x}\) and \(g(u)=\frac {u^{2}+1}{2 u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+1}{2 u -1}} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}+1}{2 u -1}} \,du} &= \int {-\frac {2}{x} \,d x} \\ \ln \left (u^{2}+1\right )-\arctan \left (u \right )&=-2 \ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \ln \left (u \left (x \right )^{2}+1\right )-\arctan \left (u \left (x \right )\right )+2 \ln \left (x \right )-c_{2} = 0 \] Now \(u\) in the above solution is replaced back by \(y\) using \(u=\frac {y}{x}\) which results in the solution \[ \ln \left (\frac {y^{2}}{x^{2}}+1\right )-\arctan \left (\frac {y}{x}\right )+2 \ln \left (x \right )-c_{2} = 0 \] Substituting initial conditions and solving for \(c_{2}\) gives \(c_{2} = \ln \left (2\right )-\frac {\pi }{4}\). Hence the solution becomes

Summary

The solution(s) found are the following \begin{align*} \tag{1} \ln \left (\frac {y^{2}}{x^{2}}+1\right )-\arctan \left (\frac {y}{x}\right )+2 \ln \left (x \right )-\ln \left (2\right )+\frac {\pi }{4} &= 0 \\ \end{align*}

Verification of solutions

\[ \ln \left (\frac {y^{2}}{x^{2}}+1\right )-\arctan \left (\frac {y}{x}\right )+2 \ln \left (x \right )-\ln \left (2\right )+\frac {\pi }{4} = 0 \] Verified OK.

4.21.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {x +\frac {y}{2}}{\frac {x}{2}-y}=0, y \left (1\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x +\frac {y}{2}}{\frac {x}{2}-y} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]

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`
 

✓ Solution by Maple

Time used: 0.188 (sec). Leaf size: 30

dsolve([diff(y(x),x)=(x+1/2*y(x))/(1/2*x-y(x)),y(1) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = \tan \left (\operatorname {RootOf}\left (4 \textit {\_Z} -4 \ln \left (\sec \left (\textit {\_Z} \right )^{2}\right )-8 \ln \left (x \right )+4 \ln \left (2\right )-\pi \right )\right ) x \]

✓ Solution by Mathematica

Time used: 0.046 (sec). Leaf size: 42

DSolve[{y'[x]==(x+1/2*y[x])/(1/2*x-y[x]),{y[1]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\log \left (\frac {y(x)^2}{x^2}+1\right )-\arctan \left (\frac {y(x)}{x}\right )=\frac {1}{4} (4 \log (2)-\pi )-2 \log (x),y(x)\right ] \]