Internal
problem
ID
[9580]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
Additional
non-linear
first
order
Problem
number
:
601
Date
solved
:
Thursday, October 17, 2024 at 08:47:25 PM
CAS
classification
:
[[_1st_order, `_with_symmetry_[F(x),G(y)]`]]
\begin{align*} y^{\prime }&=\frac {F \left (-\left (x -y\right ) \left (x +y\right )\right ) x}{y} \end{align*}
Unknown ode type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {F \left (-\left (x -y \left (x \right )\right ) \left (y \left (x \right )+x \right )\right ) x}{y \left (x \right )} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {F \left (-\left (x -y \left (x \right )\right ) \left (y \left (x \right )+x \right )\right ) x}{y \left (x \right )} \end {array} \]
2.25.3 Maple dsolve solution
Solving time : 0.033
(sec)
Leaf size : 77
dsolve(diff(y(x),x) = F(-(x-y(x))*(x+y(x)))*x/y(x),
y(x),singsol=all)
\begin{align*}
y &= \operatorname {RootOf}\left (F \left (\textit {\_Z}^{2}-x^{2}\right )-1\right ) \\
y &= \sqrt {x^{2}+\operatorname {RootOf}\left (-x^{2}+\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )-1}d \textit {\_a} +2 c_{1} \right )} \\
y &= -\sqrt {x^{2}+\operatorname {RootOf}\left (-x^{2}+\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )-1}d \textit {\_a} +2 c_{1} \right )} \\
\end{align*}
2.25.4 Mathematica DSolve solution
Solving time : 0.23
(sec)
Leaf size : 182
DSolve[{D[y[x],x] == (x*F[(-x + y[x])*(x + y[x])])/y[x],{}},
y[x],x,IncludeSingularSolutions->True]
\[
\text {Solve}\left [\int _1^{y(x)}\left (\frac {K[2]}{F((K[2]-x) (x+K[2]))-1}-\int _1^x\left (\frac {2 F((K[2]-K[1]) (K[1]+K[2])) K[1] K[2] F'((K[2]-K[1]) (K[1]+K[2]))}{(F((K[2]-K[1]) (K[1]+K[2]))-1)^2}-\frac {2 K[1] K[2] F'((K[2]-K[1]) (K[1]+K[2]))}{F((K[2]-K[1]) (K[1]+K[2]))-1}\right )dK[1]\right )dK[2]+\int _1^x-\frac {F((y(x)-K[1]) (K[1]+y(x))) K[1]}{F((y(x)-K[1]) (K[1]+y(x)))-1}dK[1]=c_1,y(x)\right ]
\]