Internal problem ID [7080]
Internal file name [OUTPUT/6066_Sunday_June_05_2022_04_17_16_PM_84389658/index.tex]
Book: Own collection of miscellaneous problems
Section: section 1.0
Problem number: 37.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {x y^{\prime }-2 y+b y^{2}=c \,x^{4}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {-c \,x^{4}+b \,y^{2}-2 y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{3} c -\frac {b \,y^{2}}{x}+\frac {2 y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{3} c\), \(f_1(x)=\frac {2}{x}\) and \(f_2(x)=-\frac {b}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {b u}{x}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=\frac {b}{x^{2}}\\ f_1 f_2 &=-\frac {2 b}{x^{2}}\\ f_2^2 f_0 &=b^{2} x c \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {b u^{\prime \prime }\left (x \right )}{x}+\frac {b u^{\prime }\left (x \right )}{x^{2}}+b^{2} x c u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+c_{2} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x \sqrt {b}\, \sqrt {c}\, \left (c_{1} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+c_{2} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {x^{2} \sqrt {c}\, \left (c_{1} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+c_{2} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )}{\sqrt {b}\, \left (c_{1} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+c_{2} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{2} \sqrt {c}\, \left (c_{3} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+\sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )}{\sqrt {b}\, \left (c_{3} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+\cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2} \sqrt {c}\, \left (c_{3} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+\sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )}{\sqrt {b}\, \left (c_{3} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+\cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2} \sqrt {c}\, \left (c_{3} \cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+\sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )}{\sqrt {b}\, \left (c_{3} \sinh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )+\cosh \left (\frac {x^{2} \sqrt {b}\, \sqrt {c}}{2}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-2 y+b y^{2}=c \,x^{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y-b y^{2}+c \,x^{4}}{x} \end {array} \]
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 Chini <- Chini successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 31
dsolve(x*diff(y(x),x)-2*y(x)+b*y(x)^2=c*x^4,y(x), singsol=all)
\[ y \left (x \right ) = \frac {i \tan \left (-\frac {i x^{2} \sqrt {b}\, \sqrt {c}}{2}+c_{1} \right ) x^{2} \sqrt {c}}{\sqrt {b}} \]
✓ Solution by Mathematica
Time used: 0.251 (sec). Leaf size: 153
DSolve[x*y'[x]-2*y[x]+b*y[x]^2==c*x^4,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {c} x^2 \left (-\cos \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )+c_1 \sin \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )\right )}{\sqrt {-b} \left (\sin \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )+c_1 \cos \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )\right )} \\ y(x)\to \frac {\sqrt {c} x^2 \tan \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )}{\sqrt {-b}} \\ \end{align*}