Internal problem ID [4293]
Internal file name [OUTPUT/3786_Sunday_June_05_2022_10_56_54_AM_77427436/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 36
Problem number: 1076.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "exact", "quadrature", "differentialType", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[_quadrature]
\[ \boxed {\left (2 y+x \right ) {y^{\prime }}^{3}+3 \left (x +y\right ) {y^{\prime }}^{2}+\left (2 x +y\right ) y^{\prime }=0} \] The ode \begin {align*} \left (2 y+x \right ) {y^{\prime }}^{3}+3 \left (x +y\right ) {y^{\prime }}^{2}+\left (2 x +y\right ) y^{\prime } = 0 \end {align*}
is factored to \begin {align*} y^{\prime } \left (y^{\prime }+1\right ) \left (2 y^{\prime } y+x y^{\prime }+y+2 x \right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{\prime } = 0\tag {1} \\ y^{\prime }+1 = 0\tag {2} \\ 2 y^{\prime } y+x y^{\prime }+y+2 x = 0\tag {3} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} y &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \] Verified OK.
Solving ODE (2) Integrating both sides gives \begin {align*} y &= \int { -1\,\mathop {\mathrm {d}x}}\\ &= -x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = -x +c_{2} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -x +c_{2} \\ \end{align*}
Verification of solutions
\[ y = -x +c_{2} \] Verified OK.
Solving ODE (3) Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} 2 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right ) x +x \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+u \left (x \right ) x = -2 x \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {2 \left (u^{2}+u +1\right )}{x \left (2 u +1\right )} \end {align*}
Where \(f(x)=-\frac {2}{x}\) and \(g(u)=\frac {u^{2}+u +1}{2 u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+u +1}{2 u +1}} \,du &= -\frac {2}{x} \,d x \\ \int { \frac {1}{\frac {u^{2}+u +1}{2 u +1}} \,du} &= \int {-\frac {2}{x} \,d x} \\ \ln \left (u^{2}+u +1\right )&=-2 \ln \left (x \right )+c_{4} \\ \end{align*} Raising both side to exponential gives \begin {align*} u^{2}+u +1 &= {\mathrm e}^{-2 \ln \left (x \right )+c_{4}} \end {align*}
Which simplifies to \begin {align*} u^{2}+u +1 &= \frac {c_{5}}{x^{2}} \end {align*}
Which simplifies to \[ u \left (x \right )^{2}+u \left (x \right )+1 = \frac {c_{5} {\mathrm e}^{c_{4}}}{x^{2}} \] The solution is \[ u \left (x \right )^{2}+u \left (x \right )+1 = \frac {c_{5} {\mathrm e}^{c_{4}}}{x^{2}} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y^{2}}{x^{2}}+\frac {y}{x}+1 = \frac {c_{5} {\mathrm e}^{c_{4}}}{x^{2}}\\ \frac {y^{2}}{x^{2}}+\frac {y}{x}+1 = \frac {c_{5} {\mathrm e}^{c_{4}}}{x^{2}} \end {align*}
Which simplifies to \begin {align*} y^{2}+x y+x^{2} = c_{5} {\mathrm e}^{c_{4}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y^{2}+x y+x^{2} &= c_{5} {\mathrm e}^{c_{4}} \\ \end{align*}
Verification of solutions
\[ y^{2}+x y+x^{2} = c_{5} {\mathrm e}^{c_{4}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y^{2}+x y+x^{2} &= c_{5} {\mathrm e}^{c_{4}} \\ \end{align*}
Verification of solutions
\[ y^{2}+x y+x^{2} = c_{5} {\mathrm e}^{c_{4}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (2 y+x \right ) {y^{\prime }}^{3}+3 \left (x +y\right ) {y^{\prime }}^{2}+\left (2 x +y\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=0, y^{\prime }=-1, y^{\prime }=-\frac {2 x +y}{2 y+x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=0 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int 0d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=c_{1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int y^{\prime }d x =\int \left (-1\right )d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & y=-x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-x +c_{1} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {2 x +y}{2 y+x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , y=c_{1} , y=-x +c_{1} \right \} \end {array} \]
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 trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous 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: 0.031 (sec). Leaf size: 65
dsolve((x+2*y(x))*diff(y(x),x)^3+3*(x+y(x))*diff(y(x),x)^2+(2*x+y(x))*diff(y(x),x) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} -x \\ y \left (x \right ) &= \frac {-c_{1} x -\sqrt {-3 c_{1}^{2} x^{2}+4}}{2 c_{1}} \\ y \left (x \right ) &= \frac {-c_{1} x +\sqrt {-3 c_{1}^{2} x^{2}+4}}{2 c_{1}} \\ y \left (x \right ) &= c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.471 (sec). Leaf size: 130
DSolve[(x+2 y[x])(y'[x])^3+3 (x+y[x]) (y'[x])^2+ (2 x+y[x]) y'[x] ==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} \left (-x-\sqrt {-3 x^2+4 e^{c_1}}\right ) \\ y(x)\to \frac {1}{2} \left (-x+\sqrt {-3 x^2+4 e^{c_1}}\right ) \\ y(x)\to c_1 \\ y(x)\to -x+c_1 \\ y(x)\to \frac {1}{2} \left (-\sqrt {3} \sqrt {-x^2}-x\right ) \\ y(x)\to \frac {1}{2} \left (\sqrt {3} \sqrt {-x^2}-x\right ) \\ \end{align*}