Internal problem ID [11238]
Internal file name [OUTPUT/10224_Sunday_December_11_2022_01_20_22_AM_84656961/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter V, Singular solutions. Article 33. Page 73
Problem number: Ex 1.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "quadrature"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {{y^{\prime }}^{2} x^{2}=\left (x -1\right )^{2}} \] The ode \begin {align*} {y^{\prime }}^{2} x^{2} = \left (x -1\right )^{2} \end {align*}
is factored to \begin {align*} \left (y^{\prime } x +x -1\right ) \left (y^{\prime } x -x +1\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } x +x -1 = 0\tag {1} \\ y^{\prime } x -x +1 = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} y &= \int { -\frac {x -1}{x}\,\mathop {\mathrm {d}x}}\\ &= -x +\ln \left (x \right )+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +\ln \left (x \right )+c_{1} \\ \end{align*}
Verification of solutions
\[ y = -x +\ln \left (x \right )+c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +\ln \left (x \right )+c_{1} \\ \end{align*}
Verification of solutions
\[ y = -x +\ln \left (x \right )+c_{1} \] Verified OK.
Solving ODE (2) Integrating both sides gives \begin {align*} y &= \int { \frac {x -1}{x}\,\mathop {\mathrm {d}x}}\\ &= x -\ln \left (x \right )+c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x -\ln \left (x \right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = x -\ln \left (x \right )+c_{2} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x -\ln \left (x \right )+c_{2} \\ \end{align*}
Verification of solutions
\[ y = x -\ln \left (x \right )+c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2} x^{2}=\left (x -1\right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {x -1}{x}, y^{\prime }=-\frac {x -1}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {x -1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \frac {x -1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=x -\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=x -\ln \left (x \right )+c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {x -1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int -\frac {x -1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-x +\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-x +\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=-x +\ln \left (x \right )+c_{1} , y=x -\ln \left (x \right )+c_{1} \right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 21
dsolve(x^2*diff(y(x),x)^2-(x-1)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= x -\ln \left (x \right )+c_{1} \\ y \left (x \right ) &= -x +\ln \left (x \right )+c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 25
DSolve[x^2*(y'[x])^2-(x-1)^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to x-\log (x)+c_1 \\ y(x)\to -x+\log (x)+c_1 \\ \end{align*}