Internal problem ID [3026]
Internal file name [OUTPUT/2518_Sunday_June_05_2022_03_17_55_AM_70956231/index.tex]
Book: Theory and solutions of Ordinary Differential equations, Donald Greenspan,
1960
Section: Exercises, page 14
Problem number: 2(o).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\left (2-x +2 y\right ) y^{\prime }-x y \left (y^{\prime }-1\right )=-x} \]
Integrating both sides gives \begin {align*} y &= \int { \frac {x}{-2+x}\,\mathop {\mathrm {d}x}}\\ &= x +2 \ln \left (-2+x \right )+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x +2 \ln \left (-2+x \right )+c_{1} \\ \end{align*}
Verification of solutions
\[ y = x +2 \ln \left (-2+x \right )+c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2-x +2 y\right ) y^{\prime }-x y \left (y^{\prime }-1\right )=-x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y x +x}{-y x +2 y-x +2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x}{-2+x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {x}{-2+x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=x +2 \ln \left (-2+x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=x +2 \ln \left (-2+x \right )+c_{1} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 17
dsolve(x+(2-x+2*y(x))*diff(y(x),x)=x*y(x)*(diff(y(x),x)-1),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -1 \\ y \left (x \right ) &= x +2 \ln \left (-2+x \right )+c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 20
DSolve[x+(2-x+2*y[x])*y'[x]==x*y[x]*(y'[x]-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -1 \\ y(x)\to x+2 \log (x-2)+c_1 \\ \end{align*}