Internal problem ID [8860]
Internal file name [OUTPUT/7795_Monday_June_06_2022_12_21_47_AM_13628467/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 526.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "exact", "linear", "quadrature", "separable", "homogeneousTypeD2", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {{y^{\prime }}^{3}-\left (y^{2}+x y+x^{2}\right ) {y^{\prime }}^{2}+\left (x y^{3}+y^{2} x^{2}+x^{3} y\right ) y^{\prime }-y^{3} x^{3}=0} \] The ode \begin {align*} {y^{\prime }}^{3}-\left (y^{2}+x y+x^{2}\right ) {y^{\prime }}^{2}+\left (x y^{3}+y^{2} x^{2}+x^{3} y\right ) y^{\prime }-y^{3} x^{3} = 0 \end {align*}
is factored to \begin {align*} \left (-x^{2}+y^{\prime }\right ) \left (y^{2}-y^{\prime }\right ) \left (x y-y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} -x^{2}+y^{\prime } = 0\tag {1} \\ y^{2}-y^{\prime } = 0\tag {2} \\ x y-y^{\prime } = 0\tag {3} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} y &= \int { x^{2}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{3}}{3}+c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{3}}{3}+c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{3}}{3}+c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{3}}{3}+c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{3}}{3}+c_{1} \] Verified OK.
Solving ODE (2) Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}}d y &= x +c_{2}\\ -\frac {1}{y}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {1}{x +c_{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{x +c_{2}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {1}{x +c_{2}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{x +c_{2}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {1}{x +c_{2}} \] Verified OK.
Solving ODE (3) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= x y \end {align*}
Where \(f(x)=x\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= x \,d x\\ \int { \frac {1}{y} \,dy} &= \int {x \,d x}\\ \ln \left (y \right )&=\frac {x^{2}}{2}+c_{3}\\ y&={\mathrm e}^{\frac {x^{2}}{2}+c_{3}}\\ &=c_{3} {\mathrm e}^{\frac {x^{2}}{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{\frac {x^{2}}{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{\frac {x^{2}}{2}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{\frac {x^{2}}{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{\frac {x^{2}}{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{3}-\left (y^{2}+x y+x^{2}\right ) {y^{\prime }}^{2}+\left (x y^{3}+y^{2} x^{2}+x^{3} y\right ) y^{\prime }-y^{3} x^{3}=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^{2}, y^{\prime }=y^{2}, y^{\prime }=x y\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x^{2} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int x^{2}d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=\frac {x^{3}}{3}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {x^{3}}{3}+c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=y^{2} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}}d x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y}=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {1}{x +c_{1}} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x y \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\frac {x^{2}}{2}+c_{1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=-\frac {1}{x +c_{1}}, y=\frac {x^{3}}{3}+c_{1} , y={\mathrm e}^{\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 trying Bernoulli <- Bernoulli 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: 32
dsolve(diff(y(x),x)^3-(y(x)^2+x*y(x)+x^2)*diff(y(x),x)^2+(x*y(x)^3+x^2*y(x)^2+x^3*y(x))*diff(y(x),x)-x^3*y(x)^3=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x^{3}}{3}+c_{1} \\ y \left (x \right ) &= \frac {1}{-x +c_{1}} \\ y \left (x \right ) &= {\mathrm e}^{\frac {x^{2}}{2}} c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.137 (sec). Leaf size: 48
DSolve[-(x^3*y[x]^3) + (x^3*y[x] + x^2*y[x]^2 + x*y[x]^3)*y'[x] - (x^2 + x*y[x] + y[x]^2)*y'[x]^2 + y'[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {1}{x+c_1} \\ y(x)\to c_1 e^{\frac {x^2}{2}} \\ y(x)\to \frac {x^3}{3}+c_1 \\ y(x)\to 0 \\ \end{align*}