Internal problem ID [3635]
Internal file name [OUTPUT/3128_Sunday_June_05_2022_08_53_01_AM_13705603/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 13
Problem number: 379.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "homogeneousTypeD2"
Maple gives the following as the ode type
[[_homogeneous, `class D`], _rational, _Riccati]
\[ \boxed {x \left (-x^{4}+1\right ) y^{\prime }-2 x \left (x^{2}-y^{2}\right )-\left (-x^{4}+1\right ) y=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} x \left (-x^{4}+1\right ) \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )-2 x \left (x^{2}-u \left (x \right )^{2} x^{2}\right )-\left (-x^{4}+1\right ) u \left (x \right ) x = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {x \left (2 u^{2}-2\right )}{x^{4}-1} \end {align*}
Where \(f(x)=\frac {x}{x^{4}-1}\) and \(g(u)=2 u^{2}-2\). Integrating both sides gives \begin{align*} \frac {1}{2 u^{2}-2} \,du &= \frac {x}{x^{4}-1} \,d x \\ \int { \frac {1}{2 u^{2}-2} \,du} &= \int {\frac {x}{x^{4}-1} \,d x} \\ -\frac {\operatorname {arctanh}\left (u \right )}{2}&=\frac {\ln \left (x -1\right )}{4}+\frac {\ln \left (x +1\right )}{4}-\frac {\ln \left (x^{2}+1\right )}{4}+c_{2} \\ \end{align*} The solution is \[ -\frac {\operatorname {arctanh}\left (u \left (x \right )\right )}{2}-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} -\frac {\operatorname {arctanh}\left (\frac {y}{x}\right )}{2}-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}-c_{2} = 0\\ -\frac {\operatorname {arctanh}\left (\frac {y}{x}\right )}{2}-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}-c_{2} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\operatorname {arctanh}\left (\frac {y}{x}\right )}{2}-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ -\frac {\operatorname {arctanh}\left (\frac {y}{x}\right )}{2}-\frac {\ln \left (x -1\right )}{4}-\frac {\ln \left (x +1\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}-c_{2} = 0 \] Verified OK.
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{4} y -2 x^{3}+2 x \,y^{2}-y}{x \left (x^{4}-1\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{3} y}{x^{4}-1}-\frac {2 x^{2}}{x^{4}-1}+\frac {2 y^{2}}{x^{4}-1}-\frac {y}{x \left (x^{4}-1\right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {2 x^{2}}{x^{4}-1}\), \(f_1(x)=\frac {1}{x}\) and \(f_2(x)=\frac {2}{x^{4}-1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {2 u}{x^{4}-1}} \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 {8 x^{3}}{\left (x^{4}-1\right )^{2}}\\ f_1 f_2 &=\frac {2}{x \left (x^{4}-1\right )}\\ f_2^2 f_0 &=-\frac {8 x^{2}}{\left (x^{4}-1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {2 u^{\prime \prime }\left (x \right )}{x^{4}-1}-\left (-\frac {8 x^{3}}{\left (x^{4}-1\right )^{2}}+\frac {2}{x \left (x^{4}-1\right )}\right ) u^{\prime }\left (x \right )-\frac {8 x^{2} u \left (x \right )}{\left (x^{4}-1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+c_{2} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {2 x \left (c_{1} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+c_{2} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )\right )}{x^{4}-1} \] Using the above in (1) gives the solution \[ y = -\frac {\left (c_{1} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+c_{2} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )\right ) x}{c_{1} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+c_{2} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (c_{3} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+\sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )\right ) x}{c_{3} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+\cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (c_{3} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+\sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )\right ) x}{c_{3} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+\cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (c_{3} \cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+\sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )\right ) x}{c_{3} \sinh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )+\cosh \left (\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (-x^{4}+1\right ) y^{\prime }-2 x \left (x^{2}-y^{2}\right )-\left (-x^{4}+1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 x \left (x^{2}-y^{2}\right )+\left (-x^{4}+1\right ) y}{x \left (-x^{4}+1\right )} \end {array} \]
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 homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 32
dsolve(x*(-x^4+1)*diff(y(x),x) = 2*x*(x^2-y(x)^2)+(-x^4+1)*y(x),y(x), singsol=all)
\[ y \left (x \right ) = -\tanh \left (\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}+\frac {\ln \left (x -1\right )}{2}+2 c_{1} \right ) x \]
✓ Solution by Mathematica
Time used: 0.329 (sec). Leaf size: 58
DSolve[x(1-x^4)y'[x]==2 x(x^2-y[x]^2)+(1-x^4) y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x \left (x^2+e^{2 c_1} \left (x^2-1\right )+1\right )}{-x^2+e^{2 c_1} \left (x^2-1\right )-1} \\ y(x)\to -x \\ y(x)\to x \\ \end{align*}