Internal problem ID [6077]
Internal file name [OUTPUT/5325_Sunday_June_05_2022_03_34_16_PM_75701382/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(c).
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`]]
\[ \boxed {y^{\prime }-\frac {x +y+1}{2 x +2 y-1}=0} \]
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 \(a_1=1, b_1=1, c_1 =1, a_2=2, b_2=2, c_2=-1\). 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 \(\frac {a_1}{b_1}=\frac {1}{1}=1\) and \(\frac {a_2}{b_2}=\frac {2}{2}=1\). Hence this is case two, where the lines are parallel. Let \(U(x)=x +y\). Solving for \(y\) gives \[ y=-x +U \left (x \right ) \] Taking derivative w.r.t \(x\) gives \[ y^{\prime }=-1+U^{\prime }\left (x \right ) \] Substituting the above into the ODE results in the ODE \[ -1+U^{\prime }\left (x \right )-\frac {U \left (x \right )+1}{2 U \left (x \right )-1} = 0 \] Or \[ -1+U^{\prime }\left (x \right )+\frac {-U \left (x \right )-1}{2 U \left (x \right )-1} = 0 \] Or \[ U^{\prime }\left (x \right )=\frac {3 U \left (x \right )}{2 U \left (x \right )-1} \] Which is now solved as separable in \(U \left (x \right )\). In canonical form the ODE is \begin {align*} U' &= F(x,U)\\ &= f( x) g(U)\\ &= \frac {3 U}{2 U -1} \end {align*}
Where \(f(x)=1\) and \(g(U)=\frac {3 U}{2 U -1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {3 U}{2 U -1}} \,dU &= 1 \,d x \\ \int { \frac {1}{\frac {3 U}{2 U -1}} \,dU} &= \int {1 \,d x} \\ \frac {2 U}{3}-\frac {\ln \left (U \right )}{3}&=x +c_{2} \\ \end{align*} The solution is \[ \frac {2 U \left (x \right )}{3}-\frac {\ln \left (U \left (x \right )\right )}{3}-x -c_{2} = 0 \] The solution \(\frac {2 U \left (x \right )}{3}-\frac {\ln \left (U \left (x \right )\right )}{3}-x -c_{2} = 0\) is converted to \(y\) using \(U \left (x \right ) = x +y\). Which gives \[ -\frac {x}{3}+\frac {2 y}{3}-\frac {\ln \left (x +y\right )}{3}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -\frac {x}{3}+\frac {2 y}{3}-\frac {\ln \left (x +y\right )}{3}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {x}{3}+\frac {2 y}{3}-\frac {\ln \left (x +y\right )}{3}-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {x +y+1}{2 x +2 y-1}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x +y+1}{2 x +2 y-1} \end {array} \]
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 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 21
dsolve(diff(y(x),x)=(x+y(x)+1)/(2*x+2*y(x)-1),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\operatorname {LambertW}\left (-2 \,{\mathrm e}^{-3 x +3 c_{1}}\right )}{2}-x \]
✓ Solution by Mathematica
Time used: 4.2 (sec). Leaf size: 32
DSolve[y'[x]==(x+y[x]+1)/(2*x+2*y[x]-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -x-\frac {1}{2} W\left (-e^{-3 x-1+c_1}\right ) \\ y(x)\to -x \\ \end{align*}