2.1.36 problem 37

Internal problem ID [8174]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 37
Date solved : Tuesday, November 12, 2024 at 11:07:20 PM
CAS classification : [_rational, _Riccati]

Solve

\begin{align*} x y^{\prime }-2 y+b y^{2}&=c \,x^{4} \end{align*}

Solved as first order ode of type Riccati

Time used: 0.520 (sec)

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

\[ y' = c \,x^{3}-\frac {b \,y^{2}}{x}+\frac {2 y}{x} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=c \,x^{3}\), \(f_1(x)=\frac {2}{x}\) and \(f_2(x)=-\frac {b}{x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u b}{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*}

In normal form the ode

\begin{align*} -\frac {b \left (\frac {d^{2}u}{d x^{2}}\right )}{x}+\frac {b \left (\frac {d u}{d x}\right )}{x^{2}}+b^{2} x c u&=0 \tag {1} \end{align*}

Becomes

\begin{align*} \frac {d^{2}u}{d x^{2}}+p \left (x \right ) \left (\frac {d u}{d x}\right )+q \left (x \right ) u&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {1}{x}\\ q \left (x \right )&=-b c \,x^{2} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }u \left (\tau \right )\right )+q_{1} u \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\frac {d^{2}}{d x^{2}}\tau \left (x \right )+p \left (x \right ) \left (\frac {d}{d x}\tau \left (x \right )\right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \frac {d^{2}}{d x^{2}}\tau \left (x \right )+p \left (x \right ) \left (\frac {d}{d x}\tau \left (x \right )\right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int -\frac {1}{x}d x \right )}d x\\ &= \int e^{\ln \left (x \right )} \,dx\\ &= \int x d x\\ &= \frac {x^{2}}{2}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\\ &= \frac {-b c \,x^{2}}{x^{2}}\\ &= -b c\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+q_{1} u \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )-b c u \left (\tau \right )&=0 \end{align*}

The above ode is now solved for \(u \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A u''(\tau ) + B u'(\tau ) + C u(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=-b c\). Let the solution be \(u \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }-b c \,{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ -b c +\lambda ^{2} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=-b c\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-b c\right )}\\ &= \pm \sqrt {b c} \end{align*}

Hence

\begin{align*} \lambda _1 &= + \sqrt {b c} \\ \lambda _2 &= - \sqrt {b c} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= \sqrt {b c} \\ \lambda _2 &= -\sqrt {b c} \\ \end{align*}

Since roots are real and distinct, then the solution is

\begin{align*} u \left (\tau \right ) &= c_1 e^{\lambda _1 \tau } + c_2 e^{\lambda _2 \tau } \\ u \left (\tau \right ) &= c_1 e^{\left (\sqrt {b c}\right )\tau } +c_2 e^{\left (-\sqrt {b c}\right )\tau } \\ \end{align*}

Or

\[ u \left (\tau \right ) =c_1 \,{\mathrm e}^{\tau \sqrt {b c}}+c_2 \,{\mathrm e}^{-\tau \sqrt {b c}} \]

Will add steps showing solving for IC soon.

The above solution is now transformed back to \(u\) using (6) which results in

\[ u = c_1 \,{\mathrm e}^{\frac {x^{2} \sqrt {b c}}{2}}+c_2 \,{\mathrm e}^{-\frac {x^{2} \sqrt {b c}}{2}} \]

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = c_1 x \sqrt {b c}\, {\mathrm e}^{\frac {x^{2} \sqrt {b c}}{2}}-c_2 x \sqrt {b c}\, {\mathrm e}^{-\frac {x^{2} \sqrt {b c}}{2}} \]

Doing change of constants, the solution becomes

\[ y = \frac {\left (c_3 x \sqrt {b c}\, {\mathrm e}^{\frac {x^{2} \sqrt {b c}}{2}}-x \sqrt {b c}\, {\mathrm e}^{-\frac {x^{2} \sqrt {b c}}{2}}\right ) x}{b \left (c_3 \,{\mathrm e}^{\frac {x^{2} \sqrt {b c}}{2}}+{\mathrm e}^{-\frac {x^{2} \sqrt {b c}}{2}}\right )} \]

Summary of solutions found

\begin{align*} y &= \frac {x^{2} \sqrt {b c}\, \left (c_3 \,{\mathrm e}^{x^{2} \sqrt {b c}}-1\right )}{b \left (c_3 \,{\mathrm e}^{x^{2} \sqrt {b c}}+1\right )} \\ \end{align*}

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y \left (x \right )\right )-2 y \left (x \right )+b y \left (x \right )^{2}=c \,x^{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {2 y \left (x \right )-b y \left (x \right )^{2}+c \,x^{4}}{x} \end {array} \]

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`
 

Maple dsolve solution

Solving time : 0.009 (sec)
Leaf size : 29

dsolve(diff(y(x),x)*x-2*y(x)+b*y(x)^2 = c*x^4, 
       y(x),singsol=all)
 

\[ y = \frac {i \tan \left (-\frac {i x^{2} \sqrt {b}\, \sqrt {c}}{2}+c_{1} \right ) x^{2} \sqrt {c}}{\sqrt {b}} \]

Mathematica DSolve solution

Solving time : 0.235 (sec)
Leaf size : 153

DSolve[{x*D[y[x],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*}