problem 978

Internal problem ID [9311]
Internal ﬁle name [OUTPUT/8247_Monday_June_06_2022_02_30_18_AM_61420509/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear ﬁrst order
Problem number: 978.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "abelFirstKind", "homogeneousTypeD2"

\[ \boxed {y^{\prime }-\frac {y \left (y^{2}+y x +x^{2}+x \right )}{x^{2}}=0} \]

2.401.1 Solving as homogeneousTypeD2 ode

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*} x^{2} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )-u \left (x \right )^{3} x^{3}-u \left (x \right )^{2} x^{3}-u \left (x \right ) x^{3}-u \left (x \right ) x^{2} = 0 \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{u \left (u^{2}+u +1\right )}d u &= \int {dx}\\ \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a} \left (\textit {\_a}^{2}+\textit {\_a} +1\right )}d \textit {\_a}&= x +c_{2} \end {align*}

Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\textit {\_a}^{2}+\textit {\_a} +1\right )}d \textit {\_a} = x +c_{2}\\ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\textit {\_a}^{2}+\textit {\_a} +1\right )}d \textit {\_a} = x +c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\textit {\_a}^{2}+\textit {\_a} +1\right )}d \textit {\_a} &= x +c_{2} \\ \end{align*}

Veriﬁcation of solutions

\[ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a} \left (\textit {\_a}^{2}+\textit {\_a} +1\right )}d \textit {\_a} = x +c_{2} \] Veriﬁed OK.

2.401.2 Solving as abelFirstKind ode

This is Abel ﬁrst kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=\frac {y^{3}}{x^{2}}+\frac {y^{2}}{x}+\frac {\left (x^{2}+x \right ) y}{x^{2}}\tag {1} \end {align*}

Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 1+\frac {1}{x}\\ f_2(x) &= \frac {1}{x}\\ f_3(x) &= \frac {1}{x^{2}} \end {align*}

Since \(f_2(x)=\frac {1}{x}\) is not zero, then the ﬁrst step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {\frac {1}{x}}{\frac {3}{x^{2}}} \right ) \\ &= u \left (x \right )-\frac {x}{3} \end {align*}

The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {u \left (x \right )^{3}}{x^{2}}+\frac {2 u \left (x \right )}{3}-\frac {7 x}{27}+\frac {u \left (x \right )}{x}\tag {2} \end {align*}

Using the change of variables \(u \left (x \right ) = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} 27 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}-27 u \left (x \right )^{3} x^{3}-18 u \left (x \right ) x^{3}+7 x^{3}-27 u \left (x \right ) x^{2} = 0 \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{u^{3}+\frac {2}{3} u -\frac {7}{27}}d u &= \int {dx}\\ \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {7}{27}}d \textit {\_a}&= x +c_{4} \end {align*}

Replacing \(u(x)\) in the above solution by \(\frac {u \left (x \right )}{x}\) results in the solution for \(u \left (x \right )\) in implicit form \begin {align*} \int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{\textit {\_a}^{3}+\frac {2}{3} \textit {\_a} -\frac {7}{27}}d \textit {\_a} = x +c_{4}\\ 27 \left (\int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{27 \textit {\_a}^{3}+18 \textit {\_a} -7}d \textit {\_a} \right ) = x +c_{4} \end {align*}

Substituting \(u=y+\frac {x}{3}\) in the above solution gives \begin {align*} 27 \left (\int _{}^{\frac {y+\frac {x}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}+18 \textit {\_a} -7}d \textit {\_a} \right ) = x +c_{4} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} 27 \left (\int _{}^{\frac {y+\frac {x}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}+18 \textit {\_a} -7}d \textit {\_a} \right ) &= x +c_{4} \\ \end{align*}

Veriﬁcation of solutions

\[ 27 \left (\int _{}^{\frac {y+\frac {x}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}+18 \textit {\_a} -7}d \textit {\_a} \right ) = x +c_{4} \] Veriﬁed OK.

2.401.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-y^{3}-y^{2} x -y x^{2}-y x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3}+y^{2} x +y x^{2}+y x}{x^{2}} \end {array} \]

Maple trace

`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`

\[ y \left (x \right ) = \frac {x \left (\sqrt {3}\, \tan \left (\operatorname {RootOf}\left (2 \sqrt {3}\, \ln \left (3\right )-\sqrt {3}\, \ln \left (\cos \left (\textit {\_Z} \right )^{2}\right )-2 \sqrt {3}\, \ln \left (-\sqrt {3}+3 \tan \left (\textit {\_Z} \right )\right )+2 \sqrt {3}\, c_{1} +2 \sqrt {3}\, x +2 \textit {\_Z} \right )\right )-1\right )}{2} \]

\[ \text {Solve}\left [-\frac {\arctan \left (\frac {\frac {2 y(x)}{x}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\frac {y(x)^2}{x^2}+\frac {y(x)}{x}+1\right )+\log \left (\frac {y(x)}{x}\right )=x+c_1,y(x)\right ] \]

2.401 problem 978

2.401.1 Solving as homogeneousTypeD2 ode

2.401.2 Solving as abelFirstKind ode

2.401.3 Maple step by step solution