Internal problem ID [4132]
Internal file name [OUTPUT/3625_Sunday_June_05_2022_09_50_15_AM_30951881/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 30
Problem number: 896.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "differentialType", "homogeneousTypeD2", "homogeneousTypeMapleC", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_separable]
\[ \boxed {x^{2} {y^{\prime }}^{2}-x y^{\prime }+y \left (1-y\right )=0} \] The ode \begin {align*} x^{2} {y^{\prime }}^{2}-x y^{\prime }+y \left (1-y\right ) = 0 \end {align*}
is factored to \begin {align*} \left (x y^{\prime }+y-1\right ) \left (-x y^{\prime }+y\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} x y^{\prime }+y-1 = 0\tag {1} \\ -x y^{\prime }+y = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-y +1}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=-y +1\). Integrating both sides gives \begin{align*} \frac {1}{-y +1} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{-y +1} \,dy} &= \int {\frac {1}{x} \,d x} \\ -\ln \left (y -1\right )&=\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \frac {1}{y -1} &= {\mathrm e}^{\ln \left (x \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {1}{y -1} &= c_{2} x \end {align*}
Which simplifies to \[ y = \frac {\left (c_{2} {\mathrm e}^{c_{1}} x +1\right ) {\mathrm e}^{-c_{1}}}{c_{2} x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{2} {\mathrm e}^{c_{1}} x +1\right ) {\mathrm e}^{-c_{1}}}{c_{2} x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{2} {\mathrm e}^{c_{1}} x +1\right ) {\mathrm e}^{-c_{1}}}{c_{2} x} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{2} {\mathrm e}^{c_{1}} x +1\right ) {\mathrm e}^{-c_{1}}}{c_{2} x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{2} {\mathrm e}^{c_{1}} x +1\right ) {\mathrm e}^{-c_{1}}}{c_{2} x} \] Verified OK.
Solving ODE (2) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {1}{x} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {1}{x} \,d x}\\ \ln \left (y \right )&=\ln \left (x \right )+c_{3}\\ y&={\mathrm e}^{\ln \left (x \right )+c_{3}}\\ &=c_{3} x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x \\ \end{align*}
Verification of solutions
\[ y = c_{3} x \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x \\ \end{align*}
Verification of solutions
\[ y = c_{3} x \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} {y^{\prime }}^{2}-x y^{\prime }+y \left (1-y\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {y}{x}, y^{\prime }=-\frac {y-1}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {y}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int \frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{c_{1}} x \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y-1}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y-1}=-\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y-1}d x =\int -\frac {1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y-1\right )=-\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{c_{1}}+x}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=\frac {{\mathrm e}^{c_{1}}+x}{x}, y={\mathrm e}^{c_{1}} x \right \} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear 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: 0.0 (sec). Leaf size: 17
dsolve(x^2*diff(y(x),x)^2-x*diff(y(x),x)+y(x)*(1-y(x)) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} x \\ y \left (x \right ) &= \frac {c_{1} +x}{x} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.044 (sec). Leaf size: 31
DSolve[x^2 (y'[x])^2-x y'[x]+y[x](1-y[x])==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 x \\ y(x)\to \frac {x+c_1}{x} \\ y(x)\to 0 \\ y(x)\to 1 \\ \end{align*}