2.32 problem Example 4

2.32.1 Solving as polynomial ode
2.32.2 Maple step by step solution

Internal problem ID [5780]
Internal file name [OUTPUT/5028_Sunday_June_05_2022_03_18_03_PM_69075019/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: Example 4.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program :

Maple gives the following as the ode type

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

\[ \boxed {-4 y+\left (x +y-2\right ) y^{\prime }=-2 x -6} \]

2.32.1 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 \(a_1=-2, b_1=4, c_1 =-6, a_2=1, b_2=1, c_2=-2\). 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} +4 y_{0} -6 &= 0 \\ x_{0} +y_{0} -2 &= 0 \\ \end {align*}

Solving for \(x_0,y_0\) from the above gives \begin {align*} x_0 &= {\frac {1}{3}} \\ y_0 &= {\frac {5}{3}} \end {align*}

Therefore the transformation becomes \begin {align*} X &=x-{\frac {1}{3}} \\ Y &=y-{\frac {5}{3}} \end {align*}

Using this transformation in \(-4 y+\left (x +y-2\right ) y^{\prime } = -2 x -6\) result in \begin {align*} \frac {dY}{dX} &= \frac {-2 X +4 Y}{X +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 +4 Y}{X +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 \(M=-2 X +4 Y\) and \(N=X +Y\) 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 {4 u -2}{u +1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {4 u \left (X \right )-2}{u \left (X \right )+1}-u \left (X \right )}{X} \end {align*}

Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {4 u \left (X \right )-2}{u \left (X \right )+1}-u \left (X \right )}{X} = 0 \] Or \[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )+\left (\frac {d}{d X}u \left (X \right )\right ) X +u \left (X \right )^{2}-3 u \left (X \right )+2 = 0 \] Or \[ \left (u \left (X \right )+1\right ) X \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}-3 u \left (X \right )+2 = 0 \] 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 {u^{2}-3 u +2}{\left (u +1\right ) X} \end {align*}

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

Which simplifies to \begin {align*} \frac {\left (u -2\right )^{3}}{\left (u -1\right )^{2}} &= \frac {c_{4}}{X} \end {align*}

The solution is \[ \frac {\left (u \left (X \right )-2\right )^{3}}{\left (u \left (X \right )-1\right )^{2}} = \frac {c_{4}}{X} \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \frac {\left (\frac {Y \left (X \right )}{X}-2\right )^{3}}{\left (\frac {Y \left (X \right )}{X}-1\right )^{2}} = \frac {c_{4}}{X} \] Which simplifies to \begin {align*} -\frac {\left (-Y \left (X \right )+2 X \right )^{3}}{\left (-Y \left (X \right )+X \right )^{2}} = c_{4} \end {align*}

The solution is implicit \(-\frac {\left (-Y \left (X \right )+2 X \right )^{3}}{\left (-Y \left (X \right )+X \right )^{2}} = c_{4}\). Replacing \(Y=y-y_0, X=x-x_0\) gives \[ -\frac {\left (2 x -y+1\right )^{3}}{\left (x +\frac {4}{3}-y\right )^{2}} = c_{4} \]

Summary

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

Figure 66: Slope field plot

Verification of solutions

\[ -\frac {\left (2 x -y+1\right )^{3}}{\left (x +\frac {4}{3}-y\right )^{2}} = c_{4} \] Verified OK.

2.32.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -4 y+\left (x +y-2\right ) y^{\prime }=-2 x -6 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-2 x +4 y-6}{x +y-2} \end {array} \]

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

Solution by Maple

Time used: 0.172 (sec). Leaf size: 198

dsolve((2*x-4*y(x)+6)+(x+y(x)-2)*diff(y(x),x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = -\frac {2 \left (\left (\frac {i \sqrt {3}}{72}-\frac {1}{72}\right ) \left (36 \sqrt {3}\, \left (x -\frac {1}{3}\right ) c_{1}^{2} \sqrt {\frac {243 \left (x -\frac {1}{3}\right )^{2} c_{1} -12 x +4}{c_{1}}}+8+972 \left (x -\frac {1}{3}\right )^{2} c_{1}^{2}+\left (-216 x +72\right ) c_{1} \right )^{\frac {2}{3}}+\left (\frac {1}{18}+\left (-x -\frac {1}{2}\right ) c_{1} \right ) \left (36 \sqrt {3}\, \left (x -\frac {1}{3}\right ) c_{1}^{2} \sqrt {\frac {243 \left (x -\frac {1}{3}\right )^{2} c_{1} -12 x +4}{c_{1}}}+8+972 \left (x -\frac {1}{3}\right )^{2} c_{1}^{2}+\left (-216 x +72\right ) c_{1} \right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) \left (-\frac {1}{18}+\left (x -\frac {1}{3}\right ) c_{1} \right )\right )}{\left (36 \sqrt {3}\, \left (x -\frac {1}{3}\right ) c_{1}^{2} \sqrt {\frac {243 \left (x -\frac {1}{3}\right )^{2} c_{1} -12 x +4}{c_{1}}}+8+972 \left (x -\frac {1}{3}\right )^{2} c_{1}^{2}+\left (-216 x +72\right ) c_{1} \right )^{\frac {1}{3}} c_{1}} \]

Solution by Mathematica

Time used: 60.144 (sec). Leaf size: 2563

DSolve[(2*x-4*y[x]+6)+(x+y[x]-2)*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

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