Internal problem ID [8777]
Internal file name [OUTPUT/7711_Sunday_June_05_2022_11_45_33_PM_46554956/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 442.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "exact", "linear", "separable", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_linear]
\[ \boxed {x^{2} {y^{\prime }}^{2}+\left (y x^{2}-2 y x +x^{3}\right ) y^{\prime }+\left (-y x^{2}+y^{2}\right ) \left (1-x \right )=0} \] The ode \begin {align*} x^{2} {y^{\prime }}^{2}+\left (y x^{2}-2 y x +x^{3}\right ) y^{\prime }+\left (-y x^{2}+y^{2}\right ) \left (1-x \right ) = 0 \end {align*}
is factored to \begin {align*} \left (y^{\prime } x +y x -y\right ) \left (-y^{\prime } x -x^{2}+y\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } x +y x -y = 0\tag {1} \\ -y^{\prime } x -x^{2}+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 \left (x -1\right )}{x} \end {align*}
Where \(f(x)=-\frac {x -1}{x}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= -\frac {x -1}{x} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {-\frac {x -1}{x} \,d x}\\ \ln \left (y \right )&=-x +\ln \left (x \right )+c_{1}\\ y&={\mathrm e}^{-x +\ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-x +\ln \left (x \right )} \end {align*}
Which simplifies to \[ y = c_{1} {\mathrm e}^{-x} x \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x} x \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x} x \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x} x \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-x} x \] Verified OK.
Solving ODE (2)
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{x}\\ q(x) &=-x \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{x} = -x \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {1}{x}d x} \\ &= \frac {1}{x} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (-x\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{x}\right ) &= \left (\frac {1}{x}\right ) \left (-x\right )\\ \mathrm {d} \left (\frac {y}{x}\right ) &= -1\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \frac {y}{x} &= \int {-1\,\mathrm {d} x}\\ \frac {y}{x} &= -x + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {1}{x}\) results in \begin {align*} y &= c_{2} x -x^{2} \end {align*}
which simplifies to \begin {align*} y &= x \left (-x +c_{2} \right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \left (-x +c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = x \left (-x +c_{2} \right ) \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x \left (-x +c_{2} \right ) \\ \end{align*}
Verification of solutions
\[ y = x \left (-x +c_{2} \right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} {y^{\prime }}^{2}+\left (y x^{2}-2 y x +x^{3}\right ) y^{\prime }+\left (-y x^{2}+y^{2}\right ) \left (1-x \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 {-x^{2}+y}{x}, y^{\prime }=-\frac {y \left (x -1\right )}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-x^{2}+y}{x} \\ {} & \circ & \textrm {Collect w.r.t.}\hspace {3pt} y\hspace {3pt}\textrm {and simplify}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y}{x}-x \\ {} & \circ & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y}{x}=-x \\ {} & \circ & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{x}\right )=-\mu \left (x \right ) x \\ {} & \circ & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{x}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=-\frac {\mu \left (x \right )}{x} \\ {} & \circ & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int -\mu \left (x \right ) x d x +c_{1} \\ {} & \circ & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int -\mu \left (x \right ) x d x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int -\mu \left (x \right ) x d x +c_{1}}{\mu \left (x \right )} \\ {} & \circ & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{x} \\ {} & {} & y=x \left (\int \left (-1\right )d x +c_{1} \right ) \\ {} & \circ & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=x \left (-x +c_{1} \right ) \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {y \left (x -1\right )}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=-\frac {x -1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int -\frac {x -1}{x}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=-x +\ln \left (x \right )+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x}{{\mathrm e}^{x -c_{1}}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=x \left (-x +c_{1} \right ), y=\frac {x}{{\mathrm e}^{x -c_{1}}}\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: 0.016 (sec). Leaf size: 21
dsolve(x^2*diff(y(x),x)^2+(x^2*y(x)-2*x*y(x)+x^3)*diff(y(x),x)+(y(x)^2-x^2*y(x))*(1-x) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \left (-x +c_{1} \right ) x \\ y \left (x \right ) &= c_{1} {\mathrm e}^{-x} x \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.047 (sec). Leaf size: 26
DSolve[(1 - x)*(-(x^2*y[x]) + y[x]^2) + (x^3 - 2*x*y[x] + x^2*y[x])*y'[x] + x^2*y'[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_1 e^{-x} x \\ y(x)\to x (-x+c_1) \\ \end{align*}