Internal problem ID [1153]
Internal file name [OUTPUT/1154_Sunday_June_05_2022_02_03_40_AM_94871224/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 5 linear second order equations. Section 5.6 Reduction or order. Page
253
Problem number: 39 part(e).
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, [_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {x^{2} \left (y^{\prime }+y^{2}\right )+y x=\frac {1}{4}-x^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {4 x^{2} y^{2}+4 x^{2}+4 y x -1}{4 x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y^{2}-1-\frac {y}{x}+\frac {1}{4 x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {4 x^{2}-1}{4 x^{2}}\), \(f_1(x)=-\frac {1}{x}\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \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' &=0\\ f_1 f_2 &=\frac {1}{x}\\ f_2^2 f_0 &=-\frac {4 x^{2}-1}{4 x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-\frac {u^{\prime }\left (x \right )}{x}-\frac {\left (4 x^{2}-1\right ) u \left (x \right )}{4 x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {c_{1} \sin \left (x \right )+c_{2} \cos \left (x \right )}{\sqrt {x}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (c_{1} x -\frac {c_{2}}{2}\right ) \cos \left (x \right )-\left (c_{2} x +\frac {c_{1}}{2}\right ) \sin \left (x \right )}{x^{\frac {3}{2}}} \] Using the above in (1) gives the solution \[ y = \frac {\left (c_{1} x -\frac {c_{2}}{2}\right ) \cos \left (x \right )-\left (c_{2} x +\frac {c_{1}}{2}\right ) \sin \left (x \right )}{x \left (c_{1} \sin \left (x \right )+c_{2} \cos \left (x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (2 c_{3} x -1\right ) \cos \left (x \right )-2 \left (x +\frac {c_{3}}{2}\right ) \sin \left (x \right )}{2 \left (c_{3} \sin \left (x \right )+\cos \left (x \right )\right ) x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{3} x -1\right ) \cos \left (x \right )-2 \left (x +\frac {c_{3}}{2}\right ) \sin \left (x \right )}{2 \left (c_{3} \sin \left (x \right )+\cos \left (x \right )\right ) x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (2 c_{3} x -1\right ) \cos \left (x \right )-2 \left (x +\frac {c_{3}}{2}\right ) \sin \left (x \right )}{2 \left (c_{3} \sin \left (x \right )+\cos \left (x \right )\right ) x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (y^{\prime }+y^{2}\right )+y x =\frac {1}{4}-x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {4 x^{2} y^{2}+4 y x +4 x^{2}-1}{4 x^{2}} \end {array} \]
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 differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: <- Riccati particular polynomial solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 46
dsolve(x^2*(diff(y(x),x)+y(x)^2)+x*y(x)+x^2-1/4=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-4 c_{1} x -{\mathrm e}^{-2 i x}-2 i {\mathrm e}^{-2 i x} x -2 i c_{1}}{2 x \left ({\mathrm e}^{-2 i x}+2 i c_{1} \right )} \]
✓ Solution by Mathematica
Time used: 0.377 (sec). Leaf size: 22
DSolve[x^2*(y'[x]+y[x]^2)+x*y[x]+x^2-1/4==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {1}{2 x}-\tan (x-c_1) \]