problem 5(b)

Internal problem ID [6076]
Internal ﬁle name [OUTPUT/5324_Sunday_June_05_2022_03_34_13_PM_13344468/index.tex]

Book: An introduction to Ordinary Diﬀerential Equations. Earl A. Coddington. Dover. NY 1961
Section: Chapter 5. Existence and uniqueness of solutions to ﬁrst order equations. Page 190
Problem number: 5(b).
ODE order: 1.
ODE degree: 1.

21.14.1 Solving as polynomial ode

This is ODE of type polynomial. Where the RHS of the ode is ratio of equations of two lines. Writing the ODE in the form \[ y^{\prime }= \frac {a_1 x + b_1 y + c_1}{ a_2 x + b_2 y + c_3 } \] Where \(a_1=2, b_1=3, c_1 =1, a_2=1, b_2=-2, c_2=-1\). There are now two possible solution methods. The ﬁrst case is when the two lines \(a_1 x + b_1 y + c_1\),\( a_2 x + b_2 y + c_3\) are not parallel and the second case is if they are parallel. If they are not parallel, then the transformation \(X=x-x_0\), \(Y=y-y_0\) converts the ODE to a homogeneous ODE. The values \( x_0,y_0\) have to be determined. If they are parallel then a transformation \(U(x)=a_1 x + b_1 y\) converts the given ODE in \(y\) to a separable ODE in \(U(x)\). The ﬁrst case is when \(\frac {a_1}{b_1} \neq \frac {a_2}{b_2}\) and the second case when \(\frac {a_1}{b_1} = \frac {a_2}{b_2}\). From the above we see that \(\frac {a_1}{b_1}\neq \frac {a_2}{b_2}\). Hence this is case one where lines are not parallel. Using the transformation \begin {align*} X &=x-x_0 \\ Y &=y-y_0 \end {align*}

Where the constants \(x_0,y_0\) are obtained by solving the following two linear algebraic equations \begin {align*} a_1 x_0 + b_1 y_0 + c_1 &= 0\\ a_2 x_0 + b_2 y_0 + c_2 &= 0 \end {align*}

Substituting the values for \(a_1,b_1,c_1,a_2,b_2,c_2\) gives \begin {align*} 2 x_{0} +3 y_{0} +1 &= 0 \\ x_{0} -2 y_{0} -1 &= 0 \\ \end {align*}

Solving for \(x_0,y_0\) from the above gives \begin {align*} x_0 &= {\frac {1}{7}} \\ y_0 &= -{\frac {3}{7}} \end {align*}

Therefore the transformation becomes \begin {align*} X &=x-{\frac {1}{7}} \\ Y &=y+{\frac {3}{7}} \end {align*}

Using this transformation in \(y^{\prime }-\frac {2 x +3 y+1}{-1+x -2 y} = 0\) result in \begin {align*} \frac {dY}{dX} &= \frac {2 X +3 Y}{X -2 Y} \end {align*}

This is now a homogeneous ODE which will now be solved for \(Y(X)\). In canonical form, the ODE is \begin {align*} Y' &= F(X,Y)\\ &= -\frac {2 X +3 Y}{-X +2 Y}\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 \(M=2 X +3 Y\) 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 {-3 u -2}{2 u -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {-3 u \left (X \right )-2}{2 u \left (X \right )-1}-u \left (X \right )}{X} \end {align*}

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

Where \(f(X)=-\frac {2}{X}\) and \(g(u)=\frac {u^{2}+u +1}{2 u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+u +1}{2 u -1}} \,du &= -\frac {2}{X} \,d X \\ \int { \frac {1}{\frac {u^{2}+u +1}{2 u -1}} \,du} &= \int {-\frac {2}{X} \,d X} \\ \ln \left (u^{2}+u +1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2 u +1\right ) \sqrt {3}}{3}\right )}{3}&=-2 \ln \left (X \right )+c_{3} \\ \end{align*} The solution is \[ \ln \left (u \left (X \right )^{2}+u \left (X \right )+1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2 u \left (X \right )+1\right ) \sqrt {3}}{3}\right )}{3}+2 \ln \left (X \right )-c_{3} = 0 \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \ln \left (\frac {Y \left (X \right )^{2}}{X^{2}}+\frac {Y \left (X \right )}{X}+1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (\frac {2 Y \left (X \right )}{X}+1\right ) \sqrt {3}}{3}\right )}{3}+2 \ln \left (X \right )-c_{3} = 0 \] The solution is implicit \(\ln \left (\frac {Y \left (X \right )^{2}}{X^{2}}+\frac {Y \left (X \right )}{X}+1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2 Y \left (X \right )+X \right ) \sqrt {3}}{3 X}\right )}{3}+2 \ln \left (X \right )-c_{3} = 0\). Replacing \(Y=y-y_0, X=x-x_0\) gives \[ \ln \left (\frac {\left (y+\frac {3}{7}\right )^{2}}{\left (x -\frac {1}{7}\right )^{2}}+\frac {y+\frac {3}{7}}{x -\frac {1}{7}}+1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2 y+\frac {5}{7}+x \right ) \sqrt {3}}{3 x -\frac {3}{7}}\right )}{3}+2 \ln \left (x -\frac {1}{7}\right )-c_{3} = 0 \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \ln \left (\frac {\left (y+\frac {3}{7}\right )^{2}}{\left (x -\frac {1}{7}\right )^{2}}+\frac {y+\frac {3}{7}}{x -\frac {1}{7}}+1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2 y+\frac {5}{7}+x \right ) \sqrt {3}}{3 x -\frac {3}{7}}\right )}{3}+2 \ln \left (x -\frac {1}{7}\right )-c_{3} &= 0 \\ \end{align*}

Veriﬁcation of solutions

\[ \ln \left (\frac {\left (y+\frac {3}{7}\right )^{2}}{\left (x -\frac {1}{7}\right )^{2}}+\frac {y+\frac {3}{7}}{x -\frac {1}{7}}+1\right )-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2 y+\frac {5}{7}+x \right ) \sqrt {3}}{3 x -\frac {3}{7}}\right )}{3}+2 \ln \left (x -\frac {1}{7}\right )-c_{3} = 0 \] Veriﬁed OK.

21.14.2 Maple step by step solution

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

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 C 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful 
<- homogeneous successful`

\[ y \left (x \right ) = -\frac {5}{14}-\frac {x}{2}+\frac {\sqrt {3}\, \left (7 x -1\right ) \tan \left (\operatorname {RootOf}\left (-2 \sqrt {3}\, \ln \left (2\right )+\sqrt {3}\, \ln \left (\sec \left (\textit {\_Z} \right )^{2} \left (7 x -1\right )^{2}\right )+\sqrt {3}\, \ln \left (3\right )+2 \sqrt {3}\, c_{1} -4 \textit {\_Z} \right )\right )}{14} \]

\[ \text {Solve}\left [32 \sqrt {3} \arctan \left (\frac {4 y(x)+5 x+1}{\sqrt {3} (-2 y(x)+x-1)}\right )=3 \left (8 \log \left (\frac {4 \left (7 x^2+7 y(x)^2+(7 x+5) y(x)+x+1\right )}{(1-7 x)^2}\right )+16 \log (7 x-1)+7 c_1\right ),y(x)\right ] \]

21.14 problem 5(b)

21.14.1 Solving as polynomial ode

21.14.2 Maple step by step solution