Internal problem ID [3312]
Internal file name [OUTPUT/2804_Sunday_June_05_2022_08_40_47_AM_2643711/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 2
Problem number: 48.
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]
\[ \boxed {y^{\prime }+\left (\sin \left (x \right )-y\right ) y=\cos \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -y \sin \left (x \right )+y^{2}+\cos \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y \sin \left (x \right )+y^{2}+\cos \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\cos \left (x \right )\), \(f_1(x)=-\sin \left (x \right )\) 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 (x \right )\\ f_2^2 f_0 &=\cos \left (x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\sin \left (x \right ) u^{\prime }\left (x \right )+\cos \left (x \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (c_{1} \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+c_{2} \right ) {\mathrm e}^{\cos \left (x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = -\sin \left (x \right ) \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right ) {\mathrm e}^{\cos \left (x \right )} c_{1} -\sin \left (x \right ) {\mathrm e}^{\cos \left (x \right )} c_{2} +c_{1} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-\sin \left (x \right ) \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right ) {\mathrm e}^{\cos \left (x \right )} c_{1} -\sin \left (x \right ) {\mathrm e}^{\cos \left (x \right )} c_{2} +c_{1} \right ) {\mathrm e}^{-\cos \left (x \right )}}{c_{1} \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {c_{3} \sin \left (x \right ) \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+\sin \left (x \right )-c_{3} {\mathrm e}^{-\cos \left (x \right )}}{c_{3} \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{3} \sin \left (x \right ) \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+\sin \left (x \right )-c_{3} {\mathrm e}^{-\cos \left (x \right )}}{c_{3} \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+1} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{3} \sin \left (x \right ) \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+\sin \left (x \right )-c_{3} {\mathrm e}^{-\cos \left (x \right )}}{c_{3} \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\left (\sin \left (x \right )-y\right ) y=\cos \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\cos \left (x \right )-\left (\sin \left (x \right )-y\right ) y \end {array} \]
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 with symmetry of the form [0, exp(-Int(f,x))/P*(y*P-f)^2] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 38
dsolve(diff(y(x),x) = cos(x)-(sin(x)-y(x))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sin \left (x \right ) \left (\int {\mathrm e}^{-\cos \left (x \right )}d x \right )+c_{1} \sin \left (x \right )-{\mathrm e}^{-\cos \left (x \right )}}{c_{1} +\int {\mathrm e}^{-\cos \left (x \right )}d x} \]
✓ Solution by Mathematica
Time used: 42.807 (sec). Leaf size: 158
DSolve[y'[x]==Cos[x]-(Sin[x]-y[x])y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {c_1 \sin (x) \int _1^xe^{-\cos (K[1])}dK[1]+\sin (x)+c_1 \left (-e^{-\cos (x)}\right )}{1+c_1 \int _1^xe^{-\cos (K[1])}dK[1]} \\ y(x)\to \sin (x) \\ y(x)\to \frac {\sin ^3(x) e^{\cos (x)} \int _1^{\cos (x)}\frac {e^{-K[1]} K[1]}{\left (1-K[1]^2\right )^{3/2}}dK[1]}{\sin ^2(x) e^{\cos (x)} \int _1^{\cos (x)}\frac {e^{-K[1]} K[1]}{\left (1-K[1]^2\right )^{3/2}}dK[1]-\sqrt {\sin ^2(x)}} \\ \end{align*}