Internal problem ID [5779]
Internal file name [OUTPUT/5027_Sunday_June_05_2022_03_18_00_PM_4330689/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 3.
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`]]
\[ \boxed {y^{\prime }-\frac {x +y-2}{y-x -4}=0} \]
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 \(a_1=-1, b_1=-1, c_1 =2, a_2=1, b_2=-1, c_2=4\). 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} -y_{0} +2 &= 0 \\ x_{0} -y_{0} +4 &= 0 \\ \end {align*}
Solving for \(x_0,y_0\) from the above gives \begin {align*} x_0 &= -1 \\ y_0 &= 3 \end {align*}
Therefore the transformation becomes \begin {align*} X &=x+1 \\ Y &=y-3 \end {align*}
Using this transformation in \(y^{\prime }-\frac {x +y-2}{y-x -4} = 0\) result in \begin {align*} \frac {dY}{dX} &= \frac {-X -Y}{-Y +X} \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 +Y}{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 \(M=-X -Y\) and \(N=-Y +X\) 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 +1}{u -1}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {u \left (X \right )+1}{u \left (X \right )-1}-u \left (X \right )}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {u \left (X \right )+1}{u \left (X \right )-1}-u \left (X \right )}{X} = 0 \] 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 \[ \left (u \left (X \right )-1\right ) X \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}-2 u \left (X \right )-1 = 0 \] 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}-2 u -1}{\left (u -1\right ) X} \end {align*}
Where \(f(X)=-\frac {1}{X}\) and \(g(u)=\frac {u^{2}-2 u -1}{u -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-2 u -1}{u -1}} \,du &= -\frac {1}{X} \,d X \\ \int { \frac {1}{\frac {u^{2}-2 u -1}{u -1}} \,du} &= \int {-\frac {1}{X} \,d X} \\ \frac {\ln \left (u^{2}-2 u -1\right )}{2}&=-\ln \left (X \right )+c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}-2 u -1} &= {\mathrm e}^{-\ln \left (X \right )+c_{3}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u^{2}-2 u -1} &= \frac {c_{4}}{X} \end {align*}
Which simplifies to \[ \sqrt {u \left (X \right )^{2}-2 u \left (X \right )-1} = \frac {c_{4} {\mathrm e}^{c_{3}}}{X} \] The solution is \[ \sqrt {u \left (X \right )^{2}-2 u \left (X \right )-1} = \frac {c_{4} {\mathrm e}^{c_{3}}}{X} \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \sqrt {\frac {Y \left (X \right )^{2}}{X^{2}}-\frac {2 Y \left (X \right )}{X}-1} = \frac {c_{4} {\mathrm e}^{c_{3}}}{X} \] The solution is implicit \(\sqrt {\frac {Y \left (X \right )^{2}-2 Y \left (X \right ) X -X^{2}}{X^{2}}} = \frac {c_{4} {\mathrm e}^{c_{3}}}{X}\). Replacing \(Y=y-y_0, X=x-x_0\) gives \[ \sqrt {\frac {-\left (1+x \right )^{2}-2 \left (y-3\right ) \left (1+x \right )+\left (y-3\right )^{2}}{\left (1+x \right )^{2}}} = \frac {c_{4} {\mathrm e}^{c_{3}}}{1+x} \]
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {-\left (1+x \right )^{2}-2 \left (y-3\right ) \left (1+x \right )+\left (y-3\right )^{2}}{\left (1+x \right )^{2}}} &= \frac {c_{4} {\mathrm e}^{c_{3}}}{1+x} \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {-\left (1+x \right )^{2}-2 \left (y-3\right ) \left (1+x \right )+\left (y-3\right )^{2}}{\left (1+x \right )^{2}}} = \frac {c_{4} {\mathrm e}^{c_{3}}}{1+x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {x +y-2}{y-x -4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x +y-2}{y-x -4} \end {array} \]
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: 0.265 (sec). Leaf size: 30
dsolve(diff(y(x),x)=(x+y(x)-2)/(y(x)-x-4),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\sqrt {2 \left (x +1\right )^{2} c_{1}^{2}+1}+\left (x +4\right ) c_{1}}{c_{1}} \]
✓ Solution by Mathematica
Time used: 0.807 (sec). Leaf size: 59
DSolve[y'[x]==(x+y[x]-2)/(y[x]-x-4),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -i \sqrt {-2 x^2-4 x-16-c_1}+x+4 \\ y(x)\to i \sqrt {-2 x^2-4 x-16-c_1}+x+4 \\ \end{align*}