Internal problem ID [1016]
Internal file name [OUTPUT/1017_Sunday_June_05_2022_01_56_56_AM_54647075/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 2, First order equations. Transformation of Nonlinear Equations into Separable
Equations. Section 2.4 Page 68
Problem number: 41.
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`]]
\[ \boxed {y^{\prime }-\frac {-6 x +y-3}{2 x -y-1}=0} \]
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 \(a_1=-6, b_1=1, c_1 =-3, a_2=2, b_2=-1, c_2=-1\). 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*} -6 x_{0} +y_{0} -3 &= 0 \\ 2 x_{0} -y_{0} -1 &= 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 \(y^{\prime }-\frac {-6 x +y-3}{2 x -y-1} = 0\) result in \begin {align*} \frac {dY}{dX} &= \frac {-6 X +Y}{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 {-6 X +Y}{-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 \(M=-6 X +Y\) 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 {-u +6}{u -2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {-u \left (X \right )+6}{u \left (X \right )-2}-u \left (X \right )}{X} \end {align*}
Or \[ \frac {d}{d X}u \left (X \right )-\frac {\frac {-u \left (X \right )+6}{u \left (X \right )-2}-u \left (X \right )}{X} = 0 \] Or \[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )-2 \left (\frac {d}{d X}u \left (X \right )\right ) X +u \left (X \right )^{2}-u \left (X \right )-6 = 0 \] Or \[ X \left (u \left (X \right )-2\right ) \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}-u \left (X \right )-6 = 0 \] 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}-u -6}{X \left (u -2\right )} \end {align*}
Where \(f(X)=-\frac {1}{X}\) and \(g(u)=\frac {u^{2}-u -6}{u -2}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-u -6}{u -2}} \,du &= -\frac {1}{X} \,d X \\ \int { \frac {1}{\frac {u^{2}-u -6}{u -2}} \,du} &= \int {-\frac {1}{X} \,d X} \\ \frac {\ln \left (u -3\right )}{5}+\frac {4 \ln \left (u +2\right )}{5}&=-\ln \left (X \right )+c_{3} \\ \end{align*} The above can be written as \begin {align*} \frac {\ln \left (u -3\right )+4 \ln \left (u +2\right )}{5} &= -\ln \left (X \right )+c_{3}\\ \ln \left (u -3\right )+4 \ln \left (u +2\right ) &= \left (5\right ) \left (-\ln \left (X \right )+c_{3}\right ) \\ &= -5 \ln \left (X \right )+5 c_{3} \end {align*}
Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (u -3\right )+4 \ln \left (u +2\right )} &= {\mathrm e}^{-5 \ln \left (X \right )+5 c_{3}} \end {align*}
Which simplifies to \begin {align*} \left (u -3\right ) \left (u +2\right )^{4} &= \frac {5 c_{3}}{X^{5}}\\ &= \frac {c_{4}}{X^{5}} \end {align*}
Which simplifies to \[ u \left (X \right ) = \operatorname {RootOf}\left (\textit {\_Z}^{5}+5 \textit {\_Z}^{4}-\frac {c_{4} {\mathrm e}^{5 c_{3}}}{X^{5}}-40 \textit {\_Z}^{2}-80 \textit {\_Z} -48\right ) \] Now \(u\) in the above solution is replaced back by \(Y\) using \(u=\frac {Y}{X}\) which results in the solution \[ Y \left (X \right ) = X \operatorname {RootOf}\left (\textit {\_Z}^{5} X^{5}+5 \textit {\_Z}^{4} X^{5}-40 \textit {\_Z}^{2} X^{5}-c_{4} {\mathrm e}^{5 c_{3}}-80 \textit {\_Z} \,X^{5}-48 X^{5}\right ) \] The solution is \[ Y \left (X \right ) = X \operatorname {RootOf}\left (\textit {\_Z}^{5} X^{5}+5 \textit {\_Z}^{4} X^{5}-40 \textit {\_Z}^{2} X^{5}-c_{4} {\mathrm e}^{5 c_{3}}-80 \textit {\_Z} \,X^{5}-48 X^{5}\right ) \] Replacing \(Y=y-y_0, X=x-x_0\) gives \[ y+3 = \left (x +1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{5} \left (x +1\right )^{5}+5 \textit {\_Z}^{4} \left (x +1\right )^{5}-40 \textit {\_Z}^{2} \left (x +1\right )^{5}-c_{4} {\mathrm e}^{5 c_{3}}-80 \textit {\_Z} \left (x +1\right )^{5}-48 \left (x +1\right )^{5}\right ) \] Or \[ y = \left (x +1\right ) \operatorname {RootOf}\left (\textit {\_Z}^{5} \left (x +1\right )^{5}+5 \textit {\_Z}^{4} \left (x +1\right )^{5}-40 \textit {\_Z}^{2} \left (x +1\right )^{5}-c_{4} {\mathrm e}^{5 c_{3}}-80 \textit {\_Z} \left (x +1\right )^{5}-48 \left (x +1\right )^{5}\right )-3 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (x +1\right ) \operatorname {RootOf}\left (\left (x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+5 x +1\right ) \textit {\_Z}^{5}+\left (5 x^{5}+25 x^{4}+50 x^{3}+50 x^{2}+25 x +5\right ) \textit {\_Z}^{4}+\left (-40 x^{5}-200 x^{4}-400 x^{3}-400 x^{2}-200 x -40\right ) \textit {\_Z}^{2}+\left (-80 x^{5}-400 x^{4}-800 x^{3}-800 x^{2}-400 x -80\right ) \textit {\_Z} -48 x^{5}-240 x^{4}-480 x^{3}-c_{4} {\mathrm e}^{5 c_{3}}-480 x^{2}-240 x -48\right )-3 \\ \end{align*}
Verification of solutions
\[ y = \left (x +1\right ) \operatorname {RootOf}\left (\left (x^{5}+5 x^{4}+10 x^{3}+10 x^{2}+5 x +1\right ) \textit {\_Z}^{5}+\left (5 x^{5}+25 x^{4}+50 x^{3}+50 x^{2}+25 x +5\right ) \textit {\_Z}^{4}+\left (-40 x^{5}-200 x^{4}-400 x^{3}-400 x^{2}-200 x -40\right ) \textit {\_Z}^{2}+\left (-80 x^{5}-400 x^{4}-800 x^{3}-800 x^{2}-400 x -80\right ) \textit {\_Z} -48 x^{5}-240 x^{4}-480 x^{3}-c_{4} {\mathrm e}^{5 c_{3}}-480 x^{2}-240 x -48\right )-3 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {-6 x +y-3}{2 x -y-1}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-6 x +y-3}{2 x -y-1} \end {array} \]
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: 1.422 (sec). Leaf size: 99
dsolve(diff(y(x),x)=(-6*x+y(x)-3)/(2*x-y(x)-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\operatorname {RootOf}\left (\textit {\_Z}^{25}+\left (-5 c_{1} x^{5}-25 c_{1} x^{4}-50 c_{1} x^{3}-50 c_{1} x^{2}-25 c_{1} x -5 c_{1} \right ) \textit {\_Z}^{5}-c_{1} x^{5}-5 c_{1} x^{4}-10 c_{1} x^{3}-10 c_{1} x^{2}-5 c_{1} x -c_{1} \right )^{20}+3 c_{1} x \left (x +1\right )^{4}}{c_{1} \left (x +1\right )^{4}} \]
✓ Solution by Mathematica
Time used: 60.095 (sec). Leaf size: 3011
DSolve[y'[x]==(-6*x+y[x]-3)/(2*x-y[x]-1),y[x],x,IncludeSingularSolutions -> True]
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