35.19 problem 1052

Internal problem ID [4273]
Internal file name [OUTPUT/3766_Sunday_June_05_2022_10_48_55_AM_57294556/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 35
Problem number: 1052.
ODE order: 1.
ODE degree: 3.

The type(s) of ODE detected by this program : "exact", "riccati", "quadrature", "separable", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[_quadrature]

\[ \boxed {{y^{\prime }}^{3}-\left (x^{2}+x y^{2}+y^{4}\right ) {y^{\prime }}^{2}+x y^{2} \left (x^{2}+x y^{2}+y^{4}\right ) y^{\prime }-x^{3} y^{6}=0} \] The ode \begin {align*} {y^{\prime }}^{3}-\left (x^{2}+x y^{2}+y^{4}\right ) {y^{\prime }}^{2}+x y^{2} \left (x^{2}+x y^{2}+y^{4}\right ) y^{\prime }-x^{3} y^{6} = 0 \end {align*}

is factored to \begin {align*} \left (-x^{2}+y^{\prime }\right ) \left (y^{4}-y^{\prime }\right ) \left (x y^{2}-y^{\prime }\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} -x^{2}+y^{\prime } = 0\tag {1} \\ y^{4}-y^{\prime } = 0\tag {2} \\ x y^{2}-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^{4}}d y &= x +c_{2}\\ -\frac {1}{3 y^{3}}&=x +c_{2} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{3 c_{2} +3 x}\\ y_2&=-\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}-\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}\\ y_3&=-\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}+\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 x +6 c_{2}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{3 c_{2} +3 x} \\ \tag{2} y &= -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}-\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )} \\ \tag{3} y &= -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}+\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 x +6 c_{2}} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{3 c_{2} +3 x} \] Verified OK.

\[ y = -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}-\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )} \] Verified OK.

\[ y = -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}+\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 x +6 c_{2}} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{3 c_{2} +3 x} \\ \tag{2} y &= -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}-\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )} \\ \tag{3} y &= -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}+\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 x +6 c_{2}} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{3 c_{2} +3 x} \] Verified OK.

\[ y = -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}-\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )} \] Verified OK.

\[ y = -\frac {\left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 \left (x +c_{2} \right )}+\frac {i \sqrt {3}\, \left (-9 \left (x +c_{2} \right )^{2}\right )^{\frac {1}{3}}}{6 x +6 c_{2}} \] Verified OK.

Solving ODE (3) In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= x \,y^{2} \end {align*}

Where \(f(x)=x\) and \(g(y)=y^{2}\). Integrating both sides gives \begin{align*} \frac {1}{y^{2}} \,dy &= x \,d x \\ \int { \frac {1}{y^{2}} \,dy} &= \int {x \,d x} \\ -\frac {1}{y}&=\frac {x^{2}}{2}+c_{3} \\ \end{align*} Which results in \begin{align*} y &= -\frac {2}{x^{2}+2 c_{3}} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2}{x^{2}+2 c_{3}} \\ \end{align*}

Verification of solutions

\[ y = -\frac {2}{x^{2}+2 c_{3}} \] Verified OK.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2}{x^{2}+2 c_{3}} \\ \end{align*}

Verification of solutions

\[ y = -\frac {2}{x^{2}+2 c_{3}} \] Verified OK.

35.19.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{3}-\left (x^{2}+x y^{2}+y^{4}\right ) {y^{\prime }}^{2}+x y^{2} \left (x^{2}+x y^{2}+y^{4}\right ) y^{\prime }-x^{3} y^{6}=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^{4}, y^{\prime }=x y^{2}\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^{4} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{4}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{4}}d x =\int 1d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{3 y^{3}}=x +c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\left (-9 \left (x +c_{1} \right )^{2}\right )^{\frac {1}{3}}}{3 \left (x +c_{1} \right )} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=x y^{2} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y^{2}}=x \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y^{2}}d x =\int x d x +c_{1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{y}=\frac {x^{2}}{2}+c_{1} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {2}{x^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{y=-\frac {2}{x^{2}+2 c_{1}}, y=\frac {\left (-9 \left (x +c_{1} \right )^{2}\right )^{\frac {1}{3}}}{3 \left (x +c_{1} \right )}, y=\frac {x^{3}}{3}+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 
trying Bernoulli 
<- Bernoulli successful`
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 75

dsolve(diff(y(x),x)^3-(x^2+x*y(x)^2+y(x)^4)*diff(y(x),x)^2+x*y(x)^2*(x^2+x*y(x)^2+y(x)^4)*diff(y(x),x)-x^3*y(x)^6 = 0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= \frac {x^{3}}{3}+c_{1} \\ y \left (x \right ) &= \frac {1}{\left (-3 x +c_{1} \right )^{\frac {1}{3}}} \\ y \left (x \right ) &= -\frac {1+i \sqrt {3}}{2 \left (-3 x +c_{1} \right )^{\frac {1}{3}}} \\ y \left (x \right ) &= \frac {-1+i \sqrt {3}}{2 \left (-3 x +c_{1} \right )^{\frac {1}{3}}} \\ y \left (x \right ) &= -\frac {2}{x^{2}-2 c_{1}} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.241 (sec). Leaf size: 110

DSolve[(y'[x])^3 -(x^2+x y[x]^2+ y[x]^4) (y'[x])^2 +x y[x]^2(x^2 +x y[x]^2+ y[x]^4) y'[x]-x^3 y[x]^6==0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -\frac {\sqrt [3]{-\frac {1}{3}}}{\sqrt [3]{-x-c_1}} \\ y(x)\to \frac {1}{\sqrt [3]{3} \sqrt [3]{-x-c_1}} \\ y(x)\to \frac {(-1)^{2/3}}{\sqrt [3]{3} \sqrt [3]{-x-c_1}} \\ y(x)\to \frac {x^3}{3}+c_1 \\ y(x)\to -\frac {2}{x^2+2 c_1} \\ y(x)\to 0 \\ \end{align*}