Internal problem ID [10520]
Internal file name [OUTPUT/9468_Monday_June_06_2022_02_45_15_PM_21042046/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.6-2. Equations with
cosine.
Problem number: 23.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-a \cos \left (\lambda x +\mu \right )^{k} \left (y-b \,x^{n}-c \right )^{2}=b n \,x^{n -1}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \cos \left (\lambda x +\mu \right )^{k} x^{2 n} a \,b^{2}+2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b c -2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b y +\cos \left (\lambda x +\mu \right )^{k} a \,c^{2}-2 \cos \left (\lambda x +\mu \right )^{k} a c y +\cos \left (\lambda x +\mu \right )^{k} a \,y^{2}+b n \,x^{n -1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{k} x^{2 n} a \,b^{2}+2 \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{k} x^{n} a b c -2 \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{k} x^{n} a b y +\left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{k} a \,c^{2}-2 \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{k} a c y +\left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{k} a \,y^{2}+\frac {b n \,x^{n}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\cos \left (\lambda x +\mu \right )^{k} x^{2 n} a \,b^{2}+2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b c +\cos \left (\lambda x +\mu \right )^{k} a \,c^{2}+b n \,x^{n -1}\), \(f_1(x)=-2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b -2 a c \cos \left (\lambda x +\mu \right )^{k}\) and \(f_2(x)=a \cos \left (\lambda x +\mu \right )^{k}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \cos \left (\lambda x +\mu \right )^{k} 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' &=-\frac {a \cos \left (\lambda x +\mu \right )^{k} k \lambda \sin \left (\lambda x +\mu \right )}{\cos \left (\lambda x +\mu \right )}\\ f_1 f_2 &=\left (-2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b -2 a c \cos \left (\lambda x +\mu \right )^{k}\right ) a \cos \left (\lambda x +\mu \right )^{k}\\ f_2^2 f_0 &=a^{2} \cos \left (\lambda x +\mu \right )^{2 k} \left (\cos \left (\lambda x +\mu \right )^{k} x^{2 n} a \,b^{2}+2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b c +\cos \left (\lambda x +\mu \right )^{k} a \,c^{2}+b n \,x^{n -1}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \cos \left (\lambda x +\mu \right )^{k} u^{\prime \prime }\left (x \right )-\left (-\frac {a \cos \left (\lambda x +\mu \right )^{k} k \lambda \sin \left (\lambda x +\mu \right )}{\cos \left (\lambda x +\mu \right )}+\left (-2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b -2 a c \cos \left (\lambda x +\mu \right )^{k}\right ) a \cos \left (\lambda x +\mu \right )^{k}\right ) u^{\prime }\left (x \right )+a^{2} \cos \left (\lambda x +\mu \right )^{2 k} \left (\cos \left (\lambda x +\mu \right )^{k} x^{2 n} a \,b^{2}+2 \cos \left (\lambda x +\mu \right )^{k} x^{n} a b c +\cos \left (\lambda x +\mu \right )^{k} a \,c^{2}+b n \,x^{n -1}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (c_{1} -i c_{2} \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )\right ) \sqrt {\sin \left (\lambda x +\mu \right )}\, \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{-\frac {k}{2}} {\mathrm e}^{-\frac {\left (\int \left (2 a \left (b \,x^{n}+c \right ) \sec \left (\lambda x +\mu \right ) \cos \left (\lambda x +\mu \right )^{k +1}+\lambda \left (k \tan \left (\lambda x +\mu \right )+\cot \left (\lambda x +\mu \right )\right )\right )d x \right )}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = \sqrt {\sin \left (\lambda x +\mu \right )}\, \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{-1-\frac {k}{2}} \sec \left (\lambda x +\mu \right ) {\mathrm e}^{-\frac {\left (\int \left (2 a \left (b \,x^{n}+c \right ) \sec \left (\lambda x +\mu \right ) \cos \left (\lambda x +\mu \right )^{k +1}+\lambda \left (k \tan \left (\lambda x +\mu \right )+\cot \left (\lambda x +\mu \right )\right )\right )d x \right )}{2}} \left (\left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right ) \left (i \left (b \,x^{n}+c \right ) a c_{2} \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )-x^{n} c_{1} a b -i \lambda c_{2} -a c c_{1} \right ) \cos \left (\lambda x +\mu \right )^{k +1}+\frac {k \lambda \left (\left (-\sin \left (\lambda x \right ) \cos \left (\mu \right )-\cos \left (\lambda x \right ) \sin \left (\mu \right )\right ) \cos \left (\lambda x +\mu \right )+\sin \left (\lambda x +\mu \right ) \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )\right ) \left (i c_{2} \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )-c_{1} \right )}{2}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )^{-1-\frac {k}{2}} \sec \left (\lambda x +\mu \right ) \left (\left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right ) \left (i \left (b \,x^{n}+c \right ) a c_{2} \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )-x^{n} c_{1} a b -i \lambda c_{2} -a c c_{1} \right ) \cos \left (\lambda x +\mu \right )^{k +1}+\frac {k \lambda \left (\left (-\sin \left (\lambda x \right ) \cos \left (\mu \right )-\cos \left (\lambda x \right ) \sin \left (\mu \right )\right ) \cos \left (\lambda x +\mu \right )+\sin \left (\lambda x +\mu \right ) \left (\cos \left (\mu \right ) \cos \left (\lambda x \right )-\sin \left (\lambda x \right ) \sin \left (\mu \right )\right )\right ) \left (i c_{2} \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )-c_{1} \right )}{2}\right ) \cos \left (\lambda x +\mu \right )^{-k} \cos \left (\lambda x +\mu \right )^{\frac {k}{2}}}{a \left (c_{1} -i c_{2} \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (b \,x^{n}+c \right ) a \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )+i x^{n} c_{3} a b -\lambda +i a c c_{3}}{a \left (\lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )+i c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (b \,x^{n}+c \right ) a \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )+i x^{n} c_{3} a b -\lambda +i a c c_{3}}{a \left (\lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )+i c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (b \,x^{n}+c \right ) a \lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )+i x^{n} c_{3} a b -\lambda +i a c c_{3}}{a \left (\lambda \left (\int \cos \left (\lambda x +\mu \right )^{k}d x \right )+i c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \cos \left (\lambda x +\mu \right )^{k} \left (y-b \,x^{n}-c \right )^{2}=b n \,x^{n -1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \cos \left (\lambda x +\mu \right )^{k} \left (y-b \,x^{n}-c \right )^{2}+b n \,x^{n -1} \end {array} \]
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 case Kamke (d) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 38
dsolve(diff(y(x),x)=a*cos(lambda*x+mu)^k*(y(x)-b*x^n-c)^2+b*n*x^(n-1),y(x), singsol=all)
\[ y \left (x \right ) = b \,x^{n}+c +\frac {1}{c_{1} -a \left (\int \left (\cos \left (x \lambda \right ) \cos \left (\mu \right )-\sin \left (x \lambda \right ) \sin \left (\mu \right )\right )^{k}d x \right )} \]
✓ Solution by Mathematica
Time used: 6.016 (sec). Leaf size: 92
DSolve[y'[x]==a*Cos[\[Lambda]*x+\[Mu]]^k*(y[x]-b*x^n-c)^2+b*n*x^(n-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{\frac {a \sqrt {\sin ^2(\mu +\lambda x)} \csc (\mu +\lambda x) \cos ^{k+1}(\mu +\lambda x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {k+1}{2},\frac {k+3}{2},\cos ^2(x \lambda +\mu )\right )}{(k+1) \lambda }+c_1}+b x^n+c \\ y(x)\to b x^n+c \\ \end{align*}