Internal problem ID [6075]
Internal file name [OUTPUT/5323_Sunday_June_05_2022_03_34_10_PM_89530852/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(a).
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`]]
\[ \boxed {y^{\prime }-\frac {x -y+2}{x +y-1}=0} \]
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 \(a_1=1, b_1=-1, c_1 =2, a_2=1, b_2=1, c_2=-1\). 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} -y_{0} +2 &= 0 \\ x_{0} +y_{0} -1 &= 0 \\ \end {align*}
Solving for \(x_0,y_0\) from the above gives \begin {align*} x_0 &= -{\frac {1}{2}} \\ y_0 &= {\frac {3}{2}} \end {align*}
Therefore the transformation becomes \begin {align*} X &=x+{\frac {1}{2}} \\ Y &=y-{\frac {3}{2}} \end {align*}
Using this transformation in \(y^{\prime }-\frac {x -y+2}{x +y-1} = 0\) result in \begin {align*} \frac {dY}{dX} &= \frac {X -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 +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 \(M=X -Y\) 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 {-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 \[ X \left (u \left (X \right )+1\right ) \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}+2 u \left (X \right )-1 = 0 \] 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}{X \left (u +1\right )} \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 \[ \sqrt {\frac {-\left (x +\frac {1}{2}\right )^{2}+2 \left (y-\frac {3}{2}\right ) \left (x +\frac {1}{2}\right )+\left (y-\frac {3}{2}\right )^{2}}{\left (x +\frac {1}{2}\right )^{2}}} = \frac {c_{4} {\mathrm e}^{c_{3}}}{x +\frac {1}{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {\frac {-\left (x +\frac {1}{2}\right )^{2}+2 \left (y-\frac {3}{2}\right ) \left (x +\frac {1}{2}\right )+\left (y-\frac {3}{2}\right )^{2}}{\left (x +\frac {1}{2}\right )^{2}}} &= \frac {c_{4} {\mathrm e}^{c_{3}}}{x +\frac {1}{2}} \\ \end{align*}
Verification of solutions
\[ \sqrt {\frac {-\left (x +\frac {1}{2}\right )^{2}+2 \left (y-\frac {3}{2}\right ) \left (x +\frac {1}{2}\right )+\left (y-\frac {3}{2}\right )^{2}}{\left (x +\frac {1}{2}\right )^{2}}} = \frac {c_{4} {\mathrm e}^{c_{3}}}{x +\frac {1}{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {x -y+2}{x +y-1}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x -y+2}{x +y-1} \end {array} \]
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: 0.391 (sec). Leaf size: 33
dsolve(diff(y(x),x)=(x-y(x)+2)/(x+y(x)-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\sqrt {1+8 \left (x +\frac {1}{2}\right )^{2} c_{1}^{2}}+\left (-2 x +2\right ) c_{1}}{2 c_{1}} \]
✓ Solution by Mathematica
Time used: 0.154 (sec). Leaf size: 53
DSolve[y'[x]==(x-y[x]+2)/(x+y[x]-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {2 x^2+2 x+1+c_1}-x+1 \\ y(x)\to \sqrt {2 x^2+2 x+1+c_1}-x+1 \\ \end{align*}