Internal problem ID [4518]
Internal file name [OUTPUT/4011_Sunday_June_05_2022_12_07_58_PM_94540826/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 2. Special types of differential equations of the first kind. Lesson 11, Bernoulli
Equations
Problem number: Exercise 11.26, page 97.
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]
\[ \boxed {y^{\prime }-\frac {2 y}{x}+\frac {y^{2}}{x}=x^{3}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {-x^{4}+y^{2}-2 y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{3}+\frac {2 y}{x}-\frac {y^{2}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{3}\), \(f_1(x)=\frac {2}{x}\) and \(f_2(x)=-\frac {1}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {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 {1}{x^{2}}\\ f_1 f_2 &=-\frac {2}{x^{2}}\\ f_2^2 f_0 &=x \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {u^{\prime \prime }\left (x \right )}{x}+\frac {u^{\prime }\left (x \right )}{x^{2}}+x u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sinh \left (\frac {x^{2}}{2}\right )+c_{2} \cosh \left (\frac {x^{2}}{2}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x \left (c_{1} \cosh \left (\frac {x^{2}}{2}\right )+c_{2} \sinh \left (\frac {x^{2}}{2}\right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {x^{2} \left (c_{1} \cosh \left (\frac {x^{2}}{2}\right )+c_{2} \sinh \left (\frac {x^{2}}{2}\right )\right )}{c_{1} \sinh \left (\frac {x^{2}}{2}\right )+c_{2} \cosh \left (\frac {x^{2}}{2}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{2} \left (c_{3} \cosh \left (\frac {x^{2}}{2}\right )+\sinh \left (\frac {x^{2}}{2}\right )\right )}{c_{3} \sinh \left (\frac {x^{2}}{2}\right )+\cosh \left (\frac {x^{2}}{2}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{2} \left (c_{3} \cosh \left (\frac {x^{2}}{2}\right )+\sinh \left (\frac {x^{2}}{2}\right )\right )}{c_{3} \sinh \left (\frac {x^{2}}{2}\right )+\cosh \left (\frac {x^{2}}{2}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{2} \left (c_{3} \cosh \left (\frac {x^{2}}{2}\right )+\sinh \left (\frac {x^{2}}{2}\right )\right )}{c_{3} \sinh \left (\frac {x^{2}}{2}\right )+\cosh \left (\frac {x^{2}}{2}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -x^{4}+y^{2}+x y^{\prime }-2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{4}-y^{2}+2 y}{x} \end {array} \]
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: 0.0 (sec). Leaf size: 19
dsolve(diff(y(x),x)=x^3+2/x*y(x)-1/x*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = i \tan \left (-\frac {i x^{2}}{2}+c_{1} \right ) x^{2} \]
✓ Solution by Mathematica
Time used: 0.162 (sec). Leaf size: 75
DSolve[y'[x]==x^3+2/x*y[x]-1/x*y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^2 \left (i \cosh \left (\frac {x^2}{2}\right )+c_1 \sinh \left (\frac {x^2}{2}\right )\right )}{i \sinh \left (\frac {x^2}{2}\right )+c_1 \cosh \left (\frac {x^2}{2}\right )} \\ y(x)\to x^2 \tanh \left (\frac {x^2}{2}\right ) \\ \end{align*}