1.162 problem 163

1.162.1 Solving as riccati ode
1.162.2 Maple step by step solution

Internal problem ID [8499]
Internal file name [OUTPUT/7432_Sunday_June_05_2022_10_55_04_PM_80503644/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 163.
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]

\[ \boxed {2 x^{2} y^{\prime }-2 y^{2}-y x=-2 a^{2} x} \]

1.162.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-2 a^{2} x +x y +2 y^{2}}{2 x^{2}} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {a^{2}}{x}+\frac {y}{2 x}+\frac {y^{2}}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {a^{2}}{x}\), \(f_1(x)=\frac {1}{2 x}\) and \(f_2(x)=\frac {1}{x^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x^{2}}} \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 {2}{x^{3}}\\ f_1 f_2 &=\frac {1}{2 x^{3}}\\ f_2^2 f_0 &=-\frac {a^{2}}{x^{5}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x^{2}}+\frac {3 u^{\prime }\left (x \right )}{2 x^{3}}-\frac {a^{2} u \left (x \right )}{x^{5}} &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {a \left (-c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )-c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )\right )}{x^{{3}/{2}}} \] Using the above in (1) gives the solution \[ y = -\frac {a \sqrt {x}\, \left (-c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )-c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )\right )}{c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {\left (c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )\right ) a \sqrt {x}}{c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )\right ) a \sqrt {x}}{c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right )} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )\right ) a \sqrt {x}}{c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right )} \] Verified OK.

1.162.2 Maple step by step solution

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

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 Chini 
<- Chini successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 25

dsolve(2*x^2*diff(y(x),x) - 2*y(x)^2 - x*y(x) + 2*a^2*x=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \tanh \left (\frac {i c_{1} \sqrt {x}+2 a}{\sqrt {x}}\right ) \sqrt {x}\, a \]

Solution by Mathematica

Time used: 0.374 (sec). Leaf size: 43

DSolve[2*x^2*y'[x] - 2*y[x]^2 - x*y[x] + 2*a^2*x==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -\sqrt {-a^2} \sqrt {x} \tan \left (\frac {2 \sqrt {-a^2}}{\sqrt {x}}-c_1\right ) \]