Internal problem ID [8853]
Internal file name [OUTPUT/7788_Monday_June_06_2022_12_20_49_AM_32022248/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 519.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "first_order_nonlinear_p_but_separable"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
\[ \boxed {{y^{\prime }}^{3}-f \left (x \right ) \left (y^{2} a +b y+c \right )^{2}=0} \]
The ode has the form \begin {align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end {align*}
Where \(n=3, m=1, f=f \left (x \right ) , g=\left (y^{2} a +y b +c \right )^{2}\). Hence the ode is \begin {align*} (y')^{3} &= f \left (x \right ) \left (y^{2} a +y b +c \right )^{2} \end {align*}
Solving for \(y^{\prime }\) from (1) gives \begin {align*} y^{\prime } &=\left (f g \right )^{\frac {1}{3}}\\ y^{\prime } &=-\frac {\left (f g \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (f g \right )^{\frac {1}{3}}}{2}\\ y^{\prime } &=-\frac {\left (f g \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (f g \right )^{\frac {1}{3}}}{2} \end {align*}
To be able to solve as separable ode, we have to now assume that \(f>0,g>0\). \begin {align*} f \left (x \right ) &> 0\\ \left (y^{2} a +y b +c \right )^{2} &> 0 \end {align*}
Under the above assumption the differential equations become separable and can be written as \begin {align*} y^{\prime } &=f^{\frac {1}{3}} g^{\frac {1}{3}}\\ y^{\prime } &=\frac {f^{\frac {1}{3}} g^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}{2}\\ y^{\prime } &=-\frac {f^{\frac {1}{3}} g^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{2} \end {align*}
Therefore \begin {align*} \frac {1}{g^{\frac {1}{3}}} \, dy &= \left (f^{\frac {1}{3}}\right )\,dx\\ \frac {2}{g^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )} \, dy &= \left (f^{\frac {1}{3}}\right )\,dx\\ -\frac {2}{g^{\frac {1}{3}} \left (1+i \sqrt {3}\right )} \, dy &= \left (f^{\frac {1}{3}}\right )\,dx \end {align*}
Replacing \(f(x),g(y)\) by their values gives \begin {align*} \frac {1}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}}} \, dy &= \left (f \left (x \right )^{\frac {1}{3}}\right )\,dx\\ \frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )} \, dy &= \left (f \left (x \right )^{\frac {1}{3}}\right )\,dx\\ -\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )} \, dy &= \left (f \left (x \right )^{\frac {1}{3}}\right )\,dx \end {align*}
Integrating now gives the solutions. \begin {align*} \int \frac {1}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}}}d y &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}\\ \int \frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}d y &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}\\ \int -\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d y &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \end {align*}
Integrating gives \begin {align*} \int _{}^{y}\frac {1}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}\\ \int _{}^{y}\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}\\ \int _{}^{y}-\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \end {align*}
Therefore \begin{align*} \int _{}^{y}\frac {1}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \\ \int _{}^{y}\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \\ \int _{}^{y}-\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \\ \end{align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \\ \tag{2} \int _{}^{y}\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \\ \tag{3} \int _{}^{y}-\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} &= \int f \left (x \right )^{\frac {1}{3}}d x +c_{1} \\ \end{align*}
Verification of solutions
\[
\int _{}^{y}\frac {1}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}
\] Verified OK. {0 < (a*y^2+b*y+c)^2, 0 < f(x)}
\[
\int _{}^{y}\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}
\] Verified OK. {0 < (a*y^2+b*y+c)^2, 0 < f(x)}
\[
\int _{}^{y}-\frac {2}{{\left (\left (y^{2} a +y b +c \right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} = \int f \left (x \right )^{\frac {1}{3}}d x +c_{1}
\] Verified OK. {0 < (a*y^2+b*y+c)^2, 0 < f(x)}
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{3}-f \left (x \right ) \left (y^{2} a +b y+c \right )^{2}=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 }=\left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}, y^{\prime }=-\frac {\left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2}-\frac {\mathrm {I} \sqrt {3}\, \left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2}, y^{\prime }=-\frac {\left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2}+\frac {\mathrm {I} \sqrt {3}\, \left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2}-\frac {\mathrm {I} \sqrt {3}\, \left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2}+\frac {\mathrm {I} \sqrt {3}\, \left (f \left (x \right ) y^{4} a^{2}+2 f \left (x \right ) y^{3} a b +2 f \left (x \right ) y^{2} a c +f \left (x \right ) y^{2} b^{2}+2 f \left (x \right ) y b c +f \left (x \right ) c^{2}\right )^{\frac {1}{3}}}{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations Successful isolation of dy/dx: 3 solutions were found. Trying to solve each resulting ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying exact <- exact successful ------------------- * Tackling next ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying exact <- exact successful ------------------- * Tackling next ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying exact <- exact successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 197
dsolve(diff(y(x),x)^3-f(x)*(a*y(x)^2+b*y(x)+c)^2=0,y(x), singsol=all)
\begin{align*} \int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} a +\textit {\_a} b +c \right )^{\frac {2}{3}}}d \textit {\_a} -\frac {\int _{}^{x}{\left (f \left (\textit {\_a} \right ) \left (a y \left (x \right )^{2}+b y \left (x \right )+c \right )^{2}\right )}^{\frac {1}{3}}d \textit {\_a}}{\left (a y \left (x \right )^{2}+b y \left (x \right )+c \right )^{\frac {2}{3}}}+c_{1} &= 0 \\ \int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} a +\textit {\_a} b +c \right )^{\frac {2}{3}}}d \textit {\_a} +\frac {\left (1+i \sqrt {3}\right ) \left (\int _{}^{x}{\left (f \left (\textit {\_a} \right ) \left (a y \left (x \right )^{2}+b y \left (x \right )+c \right )^{2}\right )}^{\frac {1}{3}}d \textit {\_a} \right )}{2 \left (a y \left (x \right )^{2}+b y \left (x \right )+c \right )^{\frac {2}{3}}}+c_{1} &= 0 \\ \int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} a +\textit {\_a} b +c \right )^{\frac {2}{3}}}d \textit {\_a} -\frac {\left (i \sqrt {3}-1\right ) \left (\int _{}^{x}{\left (f \left (\textit {\_a} \right ) \left (a y \left (x \right )^{2}+b y \left (x \right )+c \right )^{2}\right )}^{\frac {1}{3}}d \textit {\_a} \right )}{2 \left (a y \left (x \right )^{2}+b y \left (x \right )+c \right )^{\frac {2}{3}}}+c_{1} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 21.249 (sec). Leaf size: 405
DSolve[-(f[x]*(c + b*y[x] + a*y[x]^2)^2) + y'[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [\frac {\sqrt [3]{2} (2 \text {$\#$1} a+b) \left (\frac {a (\text {$\#$1} (\text {$\#$1} a+b)+c)}{4 a c-b^2}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 a \text {$\#$1})^2}{b^2-4 a c}\right )}{a (\text {$\#$1} (\text {$\#$1} a+b)+c)^{2/3}}\&\right ]\left [\int _1^x\sqrt [3]{f(K[1])}dK[1]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt [3]{2} (2 \text {$\#$1} a+b) \left (\frac {a (\text {$\#$1} (\text {$\#$1} a+b)+c)}{4 a c-b^2}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 a \text {$\#$1})^2}{b^2-4 a c}\right )}{a (\text {$\#$1} (\text {$\#$1} a+b)+c)^{2/3}}\&\right ]\left [\int _1^x-\sqrt [3]{-1} \sqrt [3]{f(K[2])}dK[2]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt [3]{2} (2 \text {$\#$1} a+b) \left (\frac {a (\text {$\#$1} (\text {$\#$1} a+b)+c)}{4 a c-b^2}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(b+2 a \text {$\#$1})^2}{b^2-4 a c}\right )}{a (\text {$\#$1} (\text {$\#$1} a+b)+c)^{2/3}}\&\right ]\left [\int _1^x(-1)^{2/3} \sqrt [3]{f(K[3])}dK[3]+c_1\right ] \\ y(x)\to -\frac {\sqrt {b^2-4 a c}+b}{2 a} \\ y(x)\to \frac {\sqrt {b^2-4 a c}-b}{2 a} \\ \end{align*}