problem 754

Internal problem ID [9088]
Internal ﬁle name [OUTPUT/8023_Monday_June_06_2022_01_19_08_AM_61030337/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear ﬁrst order
Problem number: 754.
ODE order: 1.
ODE degree: 1.

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

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

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

Veriﬁcation of solutions

\[ \int _{}^{\frac {y}{x}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}+1}d \textit {\_a} = x +c_{2} \] Veriﬁed OK.

2.178.2 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 {y}{x}+x\tag {1} \end {align*}

Therefore \begin {align*} f_0(x) &= x\\ f_1(x) &= \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 {u \left (x \right )}{3}+\frac {29 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 u \left (x \right )^{3} x^{3}-9 u \left (x \right ) x^{3}-27 \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}+29 x^{3}+27 u \left (x \right ) x^{2} = 0 \end {align*}

Integrating both sides gives \begin {align*} \int _{}^{u \left (x \right )}\frac {1}{\textit {\_a}^{3}-\frac {1}{3} \textit {\_a} +\frac {29}{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 {1}{3} \textit {\_a} +\frac {29}{27}}d \textit {\_a} = x +c_{4}\\ 27 \left (\int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{27 \textit {\_a}^{3}-9 \textit {\_a} +29}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}-9 \textit {\_a} +29}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}-9 \textit {\_a} +29}d \textit {\_a} \right ) &= x +c_{4} \\ \end{align*}

Veriﬁcation of solutions

\[ 27 \left (\int _{}^{\frac {y-\frac {x}{3}}{x}}\frac {1}{27 \textit {\_a}^{3}-9 \textit {\_a} +29}d \textit {\_a} \right ) = x +c_{4} \] Veriﬁed OK.

2.178.3 Maple step by step solution

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

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`

\[ y \left (x \right ) = \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a}^{3}+\textit {\_a}^{2}+1}d \textit {\_a} \right )+x +c_{1} \right ) x \]

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

2.178 problem 754

2.178.1 Solving as homogeneousTypeD2 ode

2.178.2 Solving as abelFirstKind ode

2.178.3 Maple step by step solution