Internal problem ID [3313]
Internal file name [OUTPUT/2805_Sunday_June_05_2022_08_40_48_AM_60658320/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 2
Problem number: 49.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }-\left (\sin \left (2 x \right )+y\right ) y=\cos \left (2 x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y \sin \left (2 x \right )+y^{2}+\cos \left (2 x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = 2 \cos \left (x \right ) \sin \left (x \right ) y +y^{2}+2 \cos \left (x \right )^{2}-1 \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\cos \left (2 x \right )\), \(f_1(x)=\sin \left (2 x \right )\) 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 &=\sin \left (2 x \right )\\ f_2^2 f_0 &=\cos \left (2 x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\sin \left (2 x \right ) u^{\prime }\left (x \right )+\cos \left (2 x \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+c_{2} \cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -2 \left (c_{2} \cos \left (x \right )^{2} \operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+c_{1} \cos \left (x \right ) \operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\frac {\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) c_{2}}{2}\right ) \sin \left (x \right ) \] Using the above in (1) gives the solution \[ y = \frac {2 \left (c_{2} \cos \left (x \right )^{2} \operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+c_{1} \cos \left (x \right ) \operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\frac {\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) c_{2}}{2}\right ) \sin \left (x \right )}{c_{1} \operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+c_{2} \cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (2 \operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) \cos \left (x \right )^{2}+2 c_{3} \cos \left (x \right ) \operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )\right ) \sin \left (x \right )}{c_{3} \operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 \operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) \cos \left (x \right )^{2}+2 c_{3} \cos \left (x \right ) \operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )\right ) \sin \left (x \right )}{c_{3} \operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (2 \operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) \cos \left (x \right )^{2}+2 c_{3} \cos \left (x \right ) \operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )\right ) \sin \left (x \right )}{c_{3} \operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (\sin \left (2 x \right )+y\right ) y=\cos \left (2 x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\cos \left (2 x \right )+\left (\sin \left (2 x \right )+y\right ) y \end {array} \]
Maple trace Kovacic algorithm successful
`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: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = 2*(diff(y(x), x))*cos(x)*sin(x)+(1-2*cos(x)^2)*y(x), y(x)` *** Su Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 <- Kovacics algorithm successful Change of variables used: [x = 1/2*arccos(t)] Linear ODE actually solved: t*u(t)+(-2*t^2-4*t+2)*diff(u(t),t)+(-4*t^2+4)*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 96
dsolve(diff(y(x),x) = cos(2*x)+(sin(2*x)+y(x))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sin \left (x \right ) \left (\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) c_{1} +2 \cos \left (x \right ) \left (\cos \left (x \right ) \operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) c_{1} +\operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )\right )\right )}{c_{1} \cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )} \]
✓ Solution by Mathematica
Time used: 2.305 (sec). Leaf size: 111
DSolve[y'[x]==Cos[2 x]+(Sin[2 x]+y[x])y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sec (x) \left (\sin (x) \int _1^{\cos (x)}\frac {e^{-K[1]^2}}{K[1]^2 \sqrt {K[1]^2-1}}dK[1]+c_1 \sin (x)+\frac {e^{-\cos ^2(x)} \tan (x)}{\sqrt {-\sin ^2(x)}}\right )}{\int _1^{\cos (x)}\frac {e^{-K[1]^2}}{K[1]^2 \sqrt {K[1]^2-1}}dK[1]+c_1} \\ y(x)\to \tan (x) \\ \end{align*}