Internal problem ID [5324]
Internal file name [OUTPUT/4815_Friday_February_02_2024_05_14_02_AM_34816384/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 9. Equations of first order and higher degree. Supplemetary problems. Page
65
Problem number: 18.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "quadrature", "separable", "differentialType", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {x {y^{\prime }}^{2}+\left (y-1-x^{2}\right ) y^{\prime }-\left (y-1\right ) x=0} \] The ode \begin {align*} x {y^{\prime }}^{2}+\left (y-1-x^{2}\right ) y^{\prime }-\left (y-1\right ) x = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime }-x \right ) \left (y^{\prime } x +y-1\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime }-x = 0\tag {1} \\ y^{\prime } x +y-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 { x\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{2}}{2}+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2}}{2}+c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2}}{2}+c_{1} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {1-y}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=1-y\). Integrating both sides gives \begin{align*} \frac {1}{1-y} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{1-y} \,dy} &= \int {\frac {1}{x} \,d x} \\ -\ln \left (y -1\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{y -1} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {1}{y -1} &= c_{3} x \end {align*}
Which simplifies to \[ y = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} {\mathrm e}^{c_{2}} x +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} {\mathrm e}^{c_{2}} x +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x +1\right ) {\mathrm e}^{-c_{2}}}{c_{3} x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x {y^{\prime }}^{2}+\left (y-1-x^{2}\right ) y^{\prime }-\left (y-1\right ) x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=x , y^{\prime }=-\frac {y-1}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x^{2}}{2}+c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y-1}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y-1}=-\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y-1}d x =\int -\frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y-1\right )=-\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x +{\mathrm e}^{c_{1}}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {x +{\mathrm e}^{c_{1}}}{x}, y=\frac {x^{2}}{2}+c_{1} \right \} \end {array} \]
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 trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 21
dsolve(x*diff(y(x),x)^2+(y(x)-1-x^2)*diff(y(x),x)-x*(y(x)-1)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x^{2}}{2}+c_{1} \\ y \left (x \right ) &= \frac {x +c_{1}}{x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 32
DSolve[x*(y'[x])^2+(y[x]-1-x^2)*y'[x]-x*(y[x]-1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^2}{2}+c_1 \\ y(x)\to \frac {x+c_1}{x} \\ y(x)\to 1 \\ \end{align*}