Internal problem ID [5785]
Internal file name [OUTPUT/5033_Sunday_June_05_2022_03_18_17_PM_72205197/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: 35.
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`]]
\[ \boxed {y-\left (4 x +2 y-3\right ) y^{\prime }=-2 x -1} \]
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 \(a_1=2, b_1=1, c_1 =1, a_2=4, b_2=2, c_2=-3\). 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 \(\frac {a_1}{b_1}=\frac {2}{1}=2\) and \(\frac {a_2}{b_2}=\frac {4}{2}=2\). Hence this is case two, where the lines are parallel. Let \(U(x)=2 x +y\). Solving for \(y\) gives \[ y=-2 x +U \left (x \right ) \] Taking derivative w.r.t \(x\) gives \[ y^{\prime }=-2+U^{\prime }\left (x \right ) \] Substituting the above into the ODE results in the ODE \[ -2 x +U \left (x \right )-\left (2 U \left (x \right )-3\right ) \left (-2+U^{\prime }\left (x \right )\right ) = -2 x -1 \] Or \[ \left (-2 U \left (x \right )+3\right ) U^{\prime }\left (x \right )-2 x +5 U \left (x \right )-6 = -2 x -1 \] Or \[ U^{\prime }\left (x \right )=\frac {5 U \left (x \right )-5}{2 U \left (x \right )-3} \] 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 {5 U -5}{2 U -3} \end {align*}
Where \(f(x)=1\) and \(g(U)=\frac {5 U -5}{2 U -3}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {5 U -5}{2 U -3}} \,dU &= 1 \,d x \\ \int { \frac {1}{\frac {5 U -5}{2 U -3}} \,dU} &= \int {1 \,d x} \\ \frac {2 U}{5}-\frac {\ln \left (U -1\right )}{5}&=x +c_{2} \\ \end{align*} The solution is \[ \frac {2 U \left (x \right )}{5}-\frac {\ln \left (U \left (x \right )-1\right )}{5}-x -c_{2} = 0 \] The solution \(\frac {2 U \left (x \right )}{5}-\frac {\ln \left (U \left (x \right )-1\right )}{5}-x -c_{2} = 0\) is converted to \(y\) using \(U \left (x \right ) = 2 x +y\). Which gives \[ -\frac {x}{5}+\frac {2 y}{5}-\frac {\ln \left (2 x +y-1\right )}{5}-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -\frac {x}{5}+\frac {2 y}{5}-\frac {\ln \left (2 x +y-1\right )}{5}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {x}{5}+\frac {2 y}{5}-\frac {\ln \left (2 x +y-1\right )}{5}-c_{2} = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y-\left (4 x +2 y-3\right ) y^{\prime }=-2 x -1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {-2 x -y-1}{4 x +2 y-3} \end {array} \]
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: 0.031 (sec). Leaf size: 23
dsolve((2*x+y(x)+1)-(4*x+2*y(x)-3)*diff(y(x),x)=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\operatorname {LambertW}\left (-2 \,{\mathrm e}^{-5 x +2+5 c_{1}}\right )}{2}-2 x +1 \]
✓ Solution by Mathematica
Time used: 11.239 (sec). Leaf size: 35
DSolve[(2*x+y[x]+1)-(4*x+2*y[x]-3)*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {1}{2} W\left (-e^{-5 x-1+c_1}\right )-2 x+1 \\ y(x)\to 1-2 x \\ \end{align*}