Internal problem ID [5781]
Internal file name [OUTPUT/5029_Sunday_June_05_2022_03_18_06_PM_70744627/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: 31.
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`]]
\[ \boxed {y^{\prime }-\frac {2 y-x +5}{2 x -y-4}=0} \]
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 \(a_1=-1, b_1=2, c_1 =5, a_2=2, b_2=-1, c_2=-4\). 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} +2 y_{0} +5 &= 0 \\ 2 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 &= -2 \end {align*}
Therefore the transformation becomes \begin {align*} X &=x-1 \\ Y &=y+2 \end {align*}
Using this transformation in \(y^{\prime }-\frac {2 y-x +5}{2 x -y-4} = 0\) result in \begin {align*} \frac {dY}{dX} &= \frac {2 Y -X}{2 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 Y -X}{-2 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 \(M=2 Y -X\) and \(N=2 X -Y\) 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 {-2 u +1}{u -2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {-2 u \left (X \right )+1}{u \left (X \right )-2}-u \left (X \right )}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {-2 u \left (X \right )+1}{u \left (X \right )-2}-u \left (X \right )}{X} = 0 \] Or \[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )-2 \left (\frac {d}{d X}u \left (X \right )\right ) X +u \left (X \right )^{2}-1 = 0 \] Or \[ X \left (u \left (X \right )-2\right ) \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}-1 = 0 \] 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}-1}{X \left (u -2\right )} \end {align*}
Where \(f(X)=-\frac {1}{X}\) and \(g(u)=\frac {u^{2}-1}{u -2}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-1}{u -2}} \,du &= -\frac {1}{X} \,d X \\ \int { \frac {1}{\frac {u^{2}-1}{u -2}} \,du} &= \int {-\frac {1}{X} \,d X} \\ -\frac {\ln \left (u -1\right )}{2}+\frac {3 \ln \left (u +1\right )}{2}&=-\ln \left (X \right )+c_{3} \\ \end{align*} The above can be written as \begin {align*} \frac {-\ln \left (u -1\right )+3 \ln \left (u +1\right )}{2} &= -\ln \left (X \right )+c_{3}\\ -\ln \left (u -1\right )+3 \ln \left (u +1\right ) &= \left (2\right ) \left (-\ln \left (X \right )+c_{3}\right ) \\ &= -2 \ln \left (X \right )+2 c_{3} \end {align*}
Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (u -1\right )+3 \ln \left (u +1\right )} &= {\mathrm e}^{-2 \ln \left (X \right )+2 c_{3}} \end {align*}
Which simplifies to \begin {align*} \frac {\left (u +1\right )^{3}}{u -1} &= \frac {2 c_{3}}{X^{2}}\\ &= \frac {c_{4}}{X^{2}} \end {align*}
Which simplifies to \[ \frac {\left (u \left (X \right )+1\right )^{3}}{u \left (X \right )-1} = \frac {c_{4} {\mathrm e}^{2 c_{3}}}{X^{2}} \] The solution is \[ \frac {\left (u \left (X \right )+1\right )^{3}}{u \left (X \right )-1} = \frac {c_{4} {\mathrm e}^{2 c_{3}}}{X^{2}} \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ \frac {\left (\frac {Y \left (X \right )}{X}+1\right )^{3}}{\frac {Y \left (X \right )}{X}-1} = \frac {c_{4} {\mathrm e}^{2 c_{3}}}{X^{2}} \] Which simplifies to \begin {align*} -\frac {\left (Y \left (X \right )+X \right )^{3}}{-Y \left (X \right )+X} = c_{4} {\mathrm e}^{2 c_{3}} \end {align*}
The solution is implicit \(-\frac {\left (Y \left (X \right )+X \right )^{3}}{-Y \left (X \right )+X} = c_{4} {\mathrm e}^{2 c_{3}}\). Replacing \(Y=y-y_0, X=x-x_0\) gives \[ -\frac {\left (y+1+x \right )^{3}}{-y-3+x} = c_{4} {\mathrm e}^{2 c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (y+1+x \right )^{3}}{-y-3+x} &= c_{4} {\mathrm e}^{2 c_{3}} \\ \end{align*}
Verification of solutions
\[ -\frac {\left (y+1+x \right )^{3}}{-y-3+x} = c_{4} {\mathrm e}^{2 c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {2 y-x +5}{2 x -y-4}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y-x +5}{2 x -y-4} \end {array} \]
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: 0.391 (sec). Leaf size: 117
dsolve(diff(y(x),x)=(2*y(x)-x+5)/(2*x-y(x)-4),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (i \sqrt {3}-1\right ) \left (27 c_{1} \left (x -1\right )+3 \sqrt {3}\, \sqrt {27 \left (x -1\right )^{2} c_{1}^{2}-1}\right )^{\frac {2}{3}}-3 i \sqrt {3}-3+6 \left (3 \sqrt {3}\, \sqrt {27 \left (x -1\right )^{2} c_{1}^{2}-1}+27 c_{1} x -27 c_{1} \right )^{\frac {1}{3}} \left (x +1\right ) c_{1}}{6 \left (27 c_{1} \left (x -1\right )+3 \sqrt {3}\, \sqrt {27 \left (x -1\right )^{2} c_{1}^{2}-1}\right )^{\frac {1}{3}} c_{1}} \]
✓ Solution by Mathematica
Time used: 60.196 (sec). Leaf size: 1601
DSolve[y'[x]==(2*y[x]-x+5)/(2*x-y[x]-4),y[x],x,IncludeSingularSolutions -> True]
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