Internal problem ID [13055]
Internal file name [OUTPUT/11708_Wednesday_November_08_2023_03_29_03_AM_89585763/index.tex]
Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th
edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Review Exercises for chapter 1. page
136
Problem number: 43.
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]
\[ \boxed {y^{\prime }-\left (y-2\right ) \left (y+1-\cos \left (t \right )\right )=0} \]
In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= -\left (y -2\right ) \left (-y +\cos \left (t \right )-1\right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\cos \left (t \right ) y +y^{2}+2 \cos \left (t \right )-y -2 \] With Riccati ODE standard form \[ y' = f_0(t)+ f_1(t)y+f_2(t)y^{2} \] Shows that \(f_0(t)=2 \cos \left (t \right )-2\), \(f_1(t)=-\cos \left (t \right )-1\) and \(f_2(t)=1\). 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''(t) -\left ( f_2' + f_1 f_2 \right ) u'(t) + f_2^2 f_0 u(t) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=-\cos \left (t \right )-1\\ f_2^2 f_0 &=2 \cos \left (t \right )-2 \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (t \right )-\left (-\cos \left (t \right )-1\right ) u^{\prime }\left (t \right )+\left (2 \cos \left (t \right )-2\right ) u \left (t \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (t \right ) = c_{1} {\mathrm e}^{-2 t +\pi }+i c_{2} {\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right ) \] The above shows that \[ u^{\prime }\left (t \right ) = -2 c_{1} {\mathrm e}^{-2 t +\pi }-2 i c_{2} {\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )+i c_{2} {\mathrm e}^{t -\frac {3 \pi }{2}-\sin \left (t \right )} \] Using the above in (1) gives the solution \[ y = -\frac {-2 c_{1} {\mathrm e}^{-2 t +\pi }-2 i c_{2} {\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )+i c_{2} {\mathrm e}^{t -\frac {3 \pi }{2}-\sin \left (t \right )}}{c_{1} {\mathrm e}^{-2 t +\pi }+i c_{2} {\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {2 i c_{3} {\mathrm e}^{-2 t +\pi }-2 \,{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )+{\mathrm e}^{t -\frac {3 \pi }{2}-\sin \left (t \right )}}{i c_{3} {\mathrm e}^{-2 t +\pi }-{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 i c_{3} {\mathrm e}^{-2 t +\pi }-2 \,{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )+{\mathrm e}^{t -\frac {3 \pi }{2}-\sin \left (t \right )}}{i c_{3} {\mathrm e}^{-2 t +\pi }-{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 i c_{3} {\mathrm e}^{-2 t +\pi }-2 \,{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )+{\mathrm e}^{t -\frac {3 \pi }{2}-\sin \left (t \right )}}{i c_{3} {\mathrm e}^{-2 t +\pi }-{\mathrm e}^{-2 t} \left (\int {\mathrm e}^{3 t -\frac {3 \pi }{2}-\sin \left (t \right )}d t \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (y-2\right ) \left (y+1-\cos \left (t \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (y-2\right ) \left (y+1-\cos \left (t \right )\right ) \end {array} \]
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 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) = (-cos(x)-1)*(diff(y(x), x))+(2-2*cos(x))*y(x), y(x)` *** Sublevel 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 symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> 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: 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients <- with_periodic_functions successful Change of variables used: [x = arccos(t)] Linear ODE actually solved: (2*t-2)*u(t)+(-(-t^2+1)^(1/2)*t-(-t^2+1)^(1/2)-t)*diff(u(t),t)+(-t^2+1)*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: 81
dsolve(diff(y(t),t)=(y(t)-2)*(y(t)+1-cos(t)),y(t), singsol=all)
\[ y \left (t \right ) = \frac {-2 c_{1} {\mathrm e}^{-2 t} \left (\int {\mathrm e}^{-\frac {3 \pi }{2}+3 t -\sin \left (t \right )}d t \right )+c_{1} {\mathrm e}^{t -\frac {3 \pi }{2}-\sin \left (t \right )}+2 i {\mathrm e}^{-2 t +\pi }}{-c_{1} {\mathrm e}^{-2 t} \left (\int {\mathrm e}^{-\frac {3 \pi }{2}+3 t -\sin \left (t \right )}d t \right )+i {\mathrm e}^{-2 t +\pi }} \]
✓ Solution by Mathematica
Time used: 3.379 (sec). Leaf size: 224
DSolve[y'[t]==(y[t]-2)*(y[t]+1-Cos[t]),y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to -\frac {-2 \int _1^{e^{i t}}e^{\frac {i \left (K[1]^2-1\right )}{2 K[1]}} K[1]^{-1-3 i}dK[1]+i e^{\frac {1}{2} i e^{-i t} \left (-1+e^{2 i t}\right )} \left (e^{i t}\right )^{-3 i}-2 c_1}{\int _1^{e^{i t}}e^{\frac {i \left (K[1]^2-1\right )}{2 K[1]}} K[1]^{-1-3 i}dK[1]+c_1} \\ y(t)\to 2 \\ y(t)\to 2-\frac {i e^{\frac {1}{2} i e^{-i t} \left (-1+e^{2 i t}\right )} \left (e^{i t}\right )^{-3 i}}{\int _1^{e^{i t}}e^{\frac {i \left (K[1]^2-1\right )}{2 K[1]}} K[1]^{-1-3 i}dK[1]} \\ \end{align*}