Internal problem ID [12273]
Internal file name [OUTPUT/10926_Thursday_September_28_2023_01_09_07_AM_72583590/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Differential Equations. Problems page
221
Problem number: Problem 19(f).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "exact nonlinear second order ode"
Maple gives the following as the ode type
[[_2nd_order, _exact, _nonlinear], [_2nd_order, _reducible, _mu_y_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {\left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime \prime }-{y^{\prime }}^{2} \left (2 \sin \left (y\right )+y \cos \left (y\right )\right )=\sin \left (x \right )} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime \prime }+\left (-y \cos \left (y\right ) y^{\prime }-2 y^{\prime } \sin \left (y\right )\right ) y^{\prime }\right )d x &= \int \sin \left (x \right )d x\\ \left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime } = -\cos \left (x \right ) + c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-\cos \left (x \right )+c_{1}}{\cos \left (y \right )-y \sin \left (y \right )} \end {align*}
Where \(f(x)=-\cos \left (x \right )+c_{1}\) and \(g(y)=\frac {1}{\cos \left (y \right )-y \sin \left (y \right )}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{\cos \left (y \right )-y \sin \left (y \right )}} \,dy &= -\cos \left (x \right )+c_{1} \,d x \\ \int { \frac {1}{\frac {1}{\cos \left (y \right )-y \sin \left (y \right )}} \,dy} &= \int {-\cos \left (x \right )+c_{1} \,d x} \\ y \cos \left (y \right )&=c_{1} x -\sin \left (x \right )+c_{2} \\ \end{align*} The solution is \[ y \cos \left (y\right )-c_{1} x +\sin \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} y \cos \left (y\right )-c_{1} x +\sin \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ y \cos \left (y\right )-c_{1} x +\sin \left (x \right )-c_{2} = 0 \] Verified OK.
Writing the ode as \[ \left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime \prime }+\left (-y \cos \left (y\right ) y^{\prime }-2 y^{\prime } \sin \left (y\right )\right ) y^{\prime } = \sin \left (x \right ) \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime \prime }+\left (-y \cos \left (y\right ) y^{\prime }-2 y^{\prime } \sin \left (y\right )\right ) y^{\prime }\right )d x &= \int \sin \left (x \right )d x\\ \left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime } = -\cos \left (x \right ) +c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {-\cos \left (x \right )+c_{1}}{\cos \left (y \right )-y \sin \left (y \right )} \end {align*}
Where \(f(x)=-\cos \left (x \right )+c_{1}\) and \(g(y)=\frac {1}{\cos \left (y \right )-y \sin \left (y \right )}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{\cos \left (y \right )-y \sin \left (y \right )}} \,dy &= -\cos \left (x \right )+c_{1} \,d x \\ \int { \frac {1}{\frac {1}{\cos \left (y \right )-y \sin \left (y \right )}} \,dy} &= \int {-\cos \left (x \right )+c_{1} \,d x} \\ y \cos \left (y \right )&=c_{1} x -\sin \left (x \right )+c_{2} \\ \end{align*} The solution is \[ y \cos \left (y\right )-c_{1} x +\sin \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} y \cos \left (y\right )-c_{1} x +\sin \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ y \cos \left (y\right )-c_{1} x +\sin \left (x \right )-c_{2} = 0 \] Verified OK.
An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}
Where the following conditions are satisfied \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}
Looking at the the ode given we see that \begin {align*} a_2 &= \cos \left (y\right )-y \sin \left (y\right )\\ a_1 &= -y \cos \left (y\right ) y^{\prime }-2 y^{\prime } \sin \left (y\right )\\ a_0 &= -\sin \left (x \right ) \end {align*}
Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to first order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {\cos \left (y\right )-y \sin \left (y\right )\,d y'} + \int {-y \cos \left (y\right ) y^{\prime }-2 y^{\prime } \sin \left (y\right )\,d y} + \int {-\sin \left (x \right )\,d x} &= c_{1} \end {align*}
Which results in \begin {align*} \left (\cos \left (y\right )-y \sin \left (y\right )\right ) y^{\prime }+\left (\cos \left (y\right )-y \sin \left (y\right )+1\right ) y^{\prime }+\cos \left (x \right ) = c_{1} \end {align*}
Which is now solved In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {\cos \left (x \right )-c_{1}}{2 y \sin \left (y \right )-2 \cos \left (y \right )-1} \end {align*}
Where \(f(x)=\cos \left (x \right )-c_{1}\) and \(g(y)=\frac {1}{2 y \sin \left (y \right )-2 \cos \left (y \right )-1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{2 y \sin \left (y \right )-2 \cos \left (y \right )-1}} \,dy &= \cos \left (x \right )-c_{1} \,d x \\ \int { \frac {1}{\frac {1}{2 y \sin \left (y \right )-2 \cos \left (y \right )-1}} \,dy} &= \int {\cos \left (x \right )-c_{1} \,d x} \\ -y -2 y \cos \left (y \right )&=-c_{1} x +\sin \left (x \right )+c_{2} \\ \end{align*} The solution is \[ -y-2 y \cos \left (y\right )+c_{1} x -\sin \left (x \right )-c_{2} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} -y-2 y \cos \left (y\right )+c_{1} x -\sin \left (x \right )-c_{2} &= 0 \\ \end{align*}
Verification of solutions
\[ -y-2 y \cos \left (y\right )+c_{1} x -\sin \left (x \right )-c_{2} = 0 \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying a quadrature <- quadrature successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.157 (sec). Leaf size: 20
dsolve((cos(y(x))-y(x)*sin(y(x)))*diff(y(x),x$2)- diff(y(x),x)^2* (2*sin(y(x))+y(x)*cos(y(x))) =sin(x),y(x), singsol=all)
\[ -y \left (x \right ) \cos \left (y \left (x \right )\right )-c_{1} x -\sin \left (x \right )+c_{2} = 0 \]
✓ Solution by Mathematica
Time used: 0.531 (sec). Leaf size: 28
DSolve[(Cos[y[x]]-y[x]*Sin[y[x]])*y''[x]- y'[x]^2* (2*Sin[y[x]]+y[x]*Cos[y[x]])==Sin[x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {y(x) \cos (y(x))}{x}+\frac {\sin (x)}{x}+\frac {c_1}{x}=c_2,y(x)\right ] \]