Internal problem ID [5783]
Internal file name [OUTPUT/5031_Sunday_June_05_2022_03_18_12_PM_39213340/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: 33.
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`]]
\[ \boxed {y^{\prime }-\frac {x +3 y-5}{x -y-1}=0} \]
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 \(a_1=1, b_1=3, c_1 =-5, a_2=1, b_2=-1, c_2=-1\). 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*} x_{0} +3 y_{0} -5 &= 0 \\ x_{0} -y_{0} -1 &= 0 \\ \end {align*}
Solving for \(x_0,y_0\) from the above gives \begin {align*} x_0 &= 2 \\ y_0 &= 1 \end {align*}
Therefore the transformation becomes \begin {align*} X &=x-2 \\ Y &=y-1 \end {align*}
Using this transformation in \(y^{\prime }-\frac {x +3 y-5}{x -y-1} = 0\) result in \begin {align*} \frac {dY}{dX} &= \frac {X +3 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 {X +3 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 \(M=X +3 Y\) 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 {-3 u -1}{u -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {-3 u \left (X \right )-1}{u \left (X \right )-1}-u \left (X \right )}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {-3 u \left (X \right )-1}{u \left (X \right )-1}-u \left (X \right )}{X} = 0 \] Or \[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )-\left (\frac {d}{d X}u \left (X \right )\right ) X +u \left (X \right )^{2}+2 u \left (X \right )+1 = 0 \] Or \[ X \left (u \left (X \right )-1\right ) \left (\frac {d}{d X}u \left (X \right )\right )+\left (1+u \left (X \right )\right )^{2} = 0 \] 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 {\left (1+u \right )^{2}}{X \left (u -1\right )} \end {align*}
Where \(f(X)=-\frac {1}{X}\) and \(g(u)=\frac {\left (1+u \right )^{2}}{u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (1+u \right )^{2}}{u -1}} \,du &= -\frac {1}{X} \,d X \\ \int { \frac {1}{\frac {\left (1+u \right )^{2}}{u -1}} \,du} &= \int {-\frac {1}{X} \,d X} \\ \frac {2}{1+u}+\ln \left (1+u \right )&=-\ln \left (X \right )+c_{3} \\ \end{align*} The solution is \[ \frac {2}{1+u \left (X \right )}+\ln \left (1+u \left (X \right )\right )+\ln \left (X \right )-c_{3} = 0 \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \frac {2}{1+\frac {Y \left (X \right )}{X}}+\ln \left (1+\frac {Y \left (X \right )}{X}\right )+\ln \left (X \right )-c_{3} = 0 \] The solution is implicit \(\frac {2 X}{X +Y \left (X \right )}+\ln \left (\frac {X +Y \left (X \right )}{X}\right )+\ln \left (X \right )-c_{3} = 0\). Replacing \(Y=y-y_0, X=x-x_0\) gives \[ \frac {2 x -4}{x +y-3}+\ln \left (\frac {x +y-3}{x -2}\right )+\ln \left (x -2\right )-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {2 x -4}{x +y-3}+\ln \left (\frac {x +y-3}{x -2}\right )+\ln \left (x -2\right )-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {2 x -4}{x +y-3}+\ln \left (\frac {x +y-3}{x -2}\right )+\ln \left (x -2\right )-c_{3} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {x +3 y-5}{x -y-1}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x +3 y-5}{x -y-1} \end {array} \]
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: 0.188 (sec). Leaf size: 32
dsolve(diff(y(x),x)=(x+3*y(x)-5)/(x-y(x)-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-x +3\right ) \operatorname {LambertW}\left (2 c_{1} \left (-2+x \right )\right )-2 x +4}{\operatorname {LambertW}\left (2 c_{1} \left (-2+x \right )\right )} \]
✓ Solution by Mathematica
Time used: 1.041 (sec). Leaf size: 148
DSolve[y'[x]==(x+3*y[x]-5)/(x-y[x]-1),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {2^{2/3} \left (x \log \left (-\frac {y(x)+x-3}{-y(x)+x-1}\right )-(x-3) \log \left (\frac {x-2}{-y(x)+x-1}\right )-3 \log \left (-\frac {y(x)+x-3}{-y(x)+x-1}\right )-y(x) \left (\log \left (\frac {x-2}{-y(x)+x-1}\right )-\log \left (-\frac {y(x)+x-3}{-y(x)+x-1}\right )+1+\log (2)\right )+x-x \log (6)+x \log (3)-1+\log (8)\right )}{9 (y(x)+x-3)}=\frac {1}{9} 2^{2/3} \log (x-2)+c_1,y(x)\right ] \]