2.46 problem Problem 18(k)

Internal problem ID [12266]
Internal file name [OUTPUT/10919_Thursday_September_28_2023_01_08_46_AM_24811125/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 18(k).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact linear second order ode", "second_order_integrable_as_is", "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2", "second_order_ode_non_constant_coeff_transformation_on_B"

Maple gives the following as the ode type

[[_2nd_order, _linear, _nonhomogeneous]]

\[ \boxed {y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y=\cos \left (x \right )} \]

2.46.1 Solving as second order change of variable on x method 2 ode

This is second order non-homogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y = 0 \] In normal form the ode \begin {align*} y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y&=0 \tag {1} \end {align*}

Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}

Where \begin {align*} p \left (x \right )&=\cot \left (x \right )\\ q \left (x \right )&=-\csc \left (x \right )^{2} \end {align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}

Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}

Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}

This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \cot \left (x \right )d x \right )}d x\\ &= \int e^{-\ln \left (\sin \left (x \right )\right )} \,dx\\ &= \int \csc \left (x \right )d x\\ &= -\ln \left (\csc \left (x \right )+\cot \left (x \right )\right )\tag {6} \end {align*}

Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {-\csc \left (x \right )^{2}}{\csc \left (x \right )^{2}}\\ &= -1\tag {7} \end {align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )-y \left (\tau \right )&=0 \end {align*}

The above ode is now solved for \(y \left (\tau \right )\).This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A y''(\tau ) + B y'(\tau ) + C y(\tau ) = 0 \] Where in the above \(A=1, B=0, C=-1\). Let the solution be \(y \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda \tau }-{\mathrm e}^{\lambda \tau } = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives \[ \lambda ^{2}-1 = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=-1\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-1\right )}\\ &= \pm 1 \end {align*}

Hence \begin{align*} \lambda _1 &= + 1 \\ \lambda _2 &= - 1 \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= 1 \\ \lambda _2 &= -1 \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} y \left (\tau \right ) &= c_{1} e^{\lambda _1 \tau } + c_{2} e^{\lambda _2 \tau } \\ y \left (\tau \right ) &= c_{1} e^{\left (1\right )\tau } +c_{2} e^{\left (-1\right )\tau } \\ \end{align*} Or \[ y \left (\tau \right ) =c_{1} {\mathrm e}^{\tau }+c_{2} {\mathrm e}^{-\tau } \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= -\left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \csc \left (x \right ) \end {align*}

Therefore the homogeneous solution \(y_h\) is \[ y_h = -\left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \csc \left (x \right ) \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= -\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ y_2 &= \csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} -\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) & \csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ \frac {d}{dx}\left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) & \frac {d}{dx}\left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} -\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) & \csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ \csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right ) & -\csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )-\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right ) \end {vmatrix} \] Therefore \[ W = \left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right )\left (-\csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )-\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right )\right ) - \left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right )\left (\csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right )\right ) \] Which simplifies to \[ W = -2 \csc \left (x \right )^{2} \sin \left (x \right ) \] Which simplifies to \[ W = -2 \csc \left (x \right ) \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \cos \left (x \right )}{-2 \csc \left (x \right )}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {\left (1+\cos \left (x \right )\right ) \cos \left (x \right )}{2}d x \] Hence \[ u_1 = \frac {\cos \left (x \right ) \sin \left (x \right )}{4}+\frac {x}{4}+\frac {\sin \left (x \right )}{2} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \cos \left (x \right )}{-2 \csc \left (x \right )}\,dx \] Which simplifies to \[ u_2 = \int \frac {\left (\cos \left (x \right )-1\right ) \cos \left (x \right )}{2}d x \] Hence \[ u_2 = \frac {\cos \left (x \right ) \sin \left (x \right )}{4}+\frac {x}{4}-\frac {\sin \left (x \right )}{2} \] Which simplifies to \begin{align*} u_1 &= \frac {\sin \left (x \right ) \left (2+\cos \left (x \right )\right )}{4}+\frac {x}{4} \\ u_2 &= \frac {\sin \left (x \right ) \left (-2+\cos \left (x \right )\right )}{4}+\frac {x}{4} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \left (\frac {\sin \left (x \right ) \left (2+\cos \left (x \right )\right )}{4}+\frac {x}{4}\right ) \left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right )+\left (\frac {\sin \left (x \right ) \left (-2+\cos \left (x \right )\right )}{4}+\frac {x}{4}\right ) \left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \] Which simplifies to \[ y_p(x) = -\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (-\left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \csc \left (x \right )\right ) + \left (-\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \csc \left (x \right )-\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2} \\ \end{align*}

Verification of solutions

\[ y = -\left (\left (c_{1} -c_{2} \right ) \cos \left (x \right )-c_{1} -c_{2} \right ) \csc \left (x \right )-\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2} \] Verified OK.

2.46.2 Solving as second order change of variable on x method 1 ode

This is second order non-homogeneous ODE. In standard form the ODE is \[ A y''(x) + B y'(x) + C y(x) = f(x) \] Where \(A=1, B=\cot \left (x \right ), C=-\csc \left (x \right )^{2}, f(x)=\cos \left (x \right )\). Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from \[ y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y = 0 \] In normal form the ode \begin {align*} y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y&=0 \tag {1} \end {align*}

Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}

Where \begin {align*} p \left (x \right )&=\cot \left (x \right )\\ q \left (x \right )&=-\csc \left (x \right )^{2} \end {align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}

Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}

Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {-\csc \left (x \right )^{2}}}{c}\tag {6} \\ \tau '' &= \frac {\csc \left (x \right )^{2} \cot \left (x \right )}{c \sqrt {-\csc \left (x \right )^{2}}} \end {align*}

Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {\frac {\csc \left (x \right )^{2} \cot \left (x \right )}{c \sqrt {-\csc \left (x \right )^{2}}}+\cot \left (x \right )\frac {\sqrt {-\csc \left (x \right )^{2}}}{c}}{\left (\frac {\sqrt {-\csc \left (x \right )^{2}}}{c}\right )^2} \\ &=0 \end {align*}

Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}

The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= c_{1} \cos \left (c \tau \right )+c_{2} \sin \left (c \tau \right ) \end {align*}

Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {-\csc \left (x \right )^{2}}d x}{c}\\ &= \frac {\sqrt {-\csc \left (x \right )^{2}}\, \ln \left (-\cot \left (x \right )+\csc \left (x \right )\right ) \sin \left (x \right )}{c} \end {align*}

Substituting the above into the solution obtained gives \[ y = -i \cot \left (x \right ) c_{2} +c_{1} \cosh \left (\ln \left (-\cot \left (x \right )+\csc \left (x \right )\right )\right ) \] Now the particular solution to this ODE is found \[ y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y = \cos \left (x \right ) \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= -\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ y_2 &= \csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} -\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) & \csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ \frac {d}{dx}\left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) & \frac {d}{dx}\left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} -\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) & \csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \\ \csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right ) & -\csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )-\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right ) \end {vmatrix} \] Therefore \[ W = \left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right )\left (-\csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )-\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right )\right ) - \left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right )\left (\csc \left (x \right ) \cot \left (x \right ) \cos \left (x \right )+\csc \left (x \right ) \sin \left (x \right )-\csc \left (x \right ) \cot \left (x \right )\right ) \] Which simplifies to \[ W = -2 \csc \left (x \right )^{2} \sin \left (x \right ) \] Which simplifies to \[ W = -2 \csc \left (x \right ) \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \cos \left (x \right )}{-2 \csc \left (x \right )}\,dx \] Which simplifies to \[ u_1 = - \int -\frac {\left (1+\cos \left (x \right )\right ) \cos \left (x \right )}{2}d x \] Hence \[ u_1 = \frac {\cos \left (x \right ) \sin \left (x \right )}{4}+\frac {x}{4}+\frac {\sin \left (x \right )}{2} \] And Eq. (3) becomes \[ u_2 = \int \frac {\left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \cos \left (x \right )}{-2 \csc \left (x \right )}\,dx \] Which simplifies to \[ u_2 = \int \frac {\left (\cos \left (x \right )-1\right ) \cos \left (x \right )}{2}d x \] Hence \[ u_2 = \frac {\cos \left (x \right ) \sin \left (x \right )}{4}+\frac {x}{4}-\frac {\sin \left (x \right )}{2} \] Which simplifies to \begin{align*} u_1 &= \frac {\sin \left (x \right ) \left (2+\cos \left (x \right )\right )}{4}+\frac {x}{4} \\ u_2 &= \frac {\sin \left (x \right ) \left (-2+\cos \left (x \right )\right )}{4}+\frac {x}{4} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \left (\frac {\sin \left (x \right ) \left (2+\cos \left (x \right )\right )}{4}+\frac {x}{4}\right ) \left (-\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right )+\left (\frac {\sin \left (x \right ) \left (-2+\cos \left (x \right )\right )}{4}+\frac {x}{4}\right ) \left (\csc \left (x \right ) \cos \left (x \right )+\csc \left (x \right )\right ) \] Which simplifies to \[ y_p(x) = -\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (-i \cot \left (x \right ) c_{2} +c_{1} \cosh \left (\ln \left (-\cot \left (x \right )+\csc \left (x \right )\right )\right )\right ) + \left (-\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2}\right ) \\ &= -\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2}-i \cot \left (x \right ) c_{2} +c_{1} \cosh \left (\ln \left (-\cot \left (x \right )+\csc \left (x \right )\right )\right ) \\ \end{align*} Which simplifies to \[ y = \frac {\left (2 c_{1} +x \right ) \csc \left (x \right )}{2}-i \cot \left (x \right ) c_{2} -\frac {\cos \left (x \right )}{2} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{1} +x \right ) \csc \left (x \right )}{2}-i \cot \left (x \right ) c_{2} -\frac {\cos \left (x \right )}{2} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (2 c_{1} +x \right ) \csc \left (x \right )}{2}-i \cot \left (x \right ) c_{2} -\frac {\cos \left (x \right )}{2} \] Verified OK.

2.46.3 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y\right )d x &= \int \cos \left (x \right )d x\\ \cot \left (x \right ) y+y^{\prime } = \sin \left (x \right ) + c_{1} \end {align*}

Which is now solved for \(y\).

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=\cot \left (x \right )\\ q(x) &=\sin \left (x \right )+c_{1} \end {align*}

Hence the ode is \begin {align*} \cot \left (x \right ) y+y^{\prime } = \sin \left (x \right )+c_{1} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \cot \left (x \right )d x} \\ &= \sin \left (x \right ) \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\sin \left (x \right )+c_{1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\sin \left (x \right ) y\right ) &= \left (\sin \left (x \right )\right ) \left (\sin \left (x \right )+c_{1}\right )\\ \mathrm {d} \left (\sin \left (x \right ) y\right ) &= \left (\left (\sin \left (x \right )+c_{1} \right ) \sin \left (x \right )\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} \sin \left (x \right ) y &= \int {\left (\sin \left (x \right )+c_{1} \right ) \sin \left (x \right )\,\mathrm {d} x}\\ \sin \left (x \right ) y &= -\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}-\cos \left (x \right ) c_{1} + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu =\sin \left (x \right )\) results in \begin {align*} y &= \csc \left (x \right ) \left (-\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}-\cos \left (x \right ) c_{1} \right )+\csc \left (x \right ) c_{2} \end {align*}

which simplifies to \begin {align*} y &= \frac {\left (x +2 c_{2} \right ) \csc \left (x \right )}{2}-\cot \left (x \right ) c_{1} -\frac {\cos \left (x \right )}{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (x +2 c_{2} \right ) \csc \left (x \right )}{2}-\cot \left (x \right ) c_{1} -\frac {\cos \left (x \right )}{2} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (x +2 c_{2} \right ) \csc \left (x \right )}{2}-\cot \left (x \right ) c_{1} -\frac {\cos \left (x \right )}{2} \] Verified OK.

2.46.4 Solving as second order ode non constant coeff transformation on B ode

Given an ode of the form \begin {align*} A y^{\prime \prime } + B y^{\prime } + C y &= F(x) \end {align*}

This method reduces the order ode the ODE by one by applying the transformation \begin {align*} y&= B v \end {align*}

This results in \begin {align*} y' &=B' v+ v' B \\ y'' &=B'' v+ B' v' +v'' B + v' B' \\ &=v'' B+2 v'+ B'+B'' v \end {align*}

And now the original ode becomes \begin {align*} A\left ( v'' B+2v'B'+ B'' v\right )+B\left ( B'v+ v' B\right ) +CBv & =0\\ ABv'' +\left ( 2AB'+B^{2}\right ) v'+\left (AB''+BB'+CB\right ) v & =0 \tag {1} \end {align*}

If the term \(AB''+BB'+CB\) is zero, then this method works and can be used to solve \[ ABv''+\left ( 2AB' +B^{2}\right ) v'=0 \] By Using \(u=v'\) which reduces the order of the above ode to one. The new ode is \[ ABu'+\left ( 2AB'+B^{2}\right ) u=0 \] The above ode is first order ode which is solved for \(u\). Now a new ode \(v'=u\) is solved for \(v\) as first order ode. Then the final solution is obtain from \(y=Bv\).

This method works only if the term \(AB''+BB'+CB\) is zero. The given ODE shows that \begin {align*} A &= 1\\ B &= \cot \left (x \right )\\ C &= -\csc \left (x \right )^{2}\\ F &= \cos \left (x \right ) \end {align*}

The above shows that for this ode \begin {align*} AB''+BB'+CB &= \left (1\right ) \left (-2 \cot \left (x \right ) \left (-1-\cot \left (x \right )^{2}\right )\right ) + \left (\cot \left (x \right )\right ) \left (-1-\cot \left (x \right )^{2}\right ) + \left (-\csc \left (x \right )^{2}\right ) \left (\cot \left (x \right )\right ) \\ &=-\cot \left (x \right ) \left (-1-\cot \left (x \right )^{2}\right )-\csc \left (x \right )^{2} \cot \left (x \right )\\ &=0 \end {align*}

Hence the ode in \(v\) given in (1) now simplifies to \begin {align*} \cot \left (x \right ) v'' +\left ( -\cot \left (x \right )^{2}-2\right ) v' & =0 \end {align*}

Now by applying \(v'=u\) the above becomes \begin {align*} -u \left (x \right ) \cot \left (x \right )^{2}+\cot \left (x \right ) u^{\prime }\left (x \right )-2 u \left (x \right ) = 0 \end {align*}

Which is now solved for \(u\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (\cot \left (x \right )^{2}+2\right )}{\cot \left (x \right )} \end {align*}

Where \(f(x)=\frac {\cot \left (x \right )^{2}+2}{\cot \left (x \right )}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {\cot \left (x \right )^{2}+2}{\cot \left (x \right )} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {\cot \left (x \right )^{2}+2}{\cot \left (x \right )} \,d x}\\ \ln \left (u \right )&=\ln \left (\sin \left (x \right )\right )-2 \ln \left (\cos \left (x \right )\right )+c_{1}\\ u&={\mathrm e}^{\ln \left (\sin \left (x \right )\right )-2 \ln \left (\cos \left (x \right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (\sin \left (x \right )\right )-2 \ln \left (\cos \left (x \right )\right )} \end {align*}

Which simplifies to \[ u \left (x \right ) = \frac {c_{1} \sin \left (x \right )}{\cos \left (x \right )^{2}} \] The ode for \(v\) now becomes \begin {align*} v' &= u\\ &=\frac {c_{1} \sin \left (x \right )}{\cos \left (x \right )^{2}} \end {align*}

Which is now solved for \(v\). Integrating both sides gives \begin {align*} v \left (x \right ) &= \int { \frac {c_{1} \sin \left (x \right )}{\cos \left (x \right )^{2}}\,\mathop {\mathrm {d}x}}\\ &= \frac {c_{1}}{\cos \left (x \right )}+c_{2} \end {align*}

Therefore the homogeneous solution is \begin {align*} y_h(x) &= B v\\ &= \left (\cot \left (x \right )\right ) \left (\frac {c_{1}}{\cos \left (x \right )}+c_{2}\right ) \\ &= \cot \left (x \right ) c_{2} +\csc \left (x \right ) c_{1} \end {align*}

And now the particular solution \(y_p(x)\) will be found. The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= \cot \left (x \right ) \\ y_2 &= \csc \left (x \right ) \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} \cot \left (x \right ) & \csc \left (x \right ) \\ \frac {d}{dx}\left (\cot \left (x \right )\right ) & \frac {d}{dx}\left (\csc \left (x \right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \cot \left (x \right ) & \csc \left (x \right ) \\ -1-\cot \left (x \right )^{2} & -\csc \left (x \right ) \cot \left (x \right ) \end {vmatrix} \] Therefore \[ W = \left (\cot \left (x \right )\right )\left (-\csc \left (x \right ) \cot \left (x \right )\right ) - \left (\csc \left (x \right )\right )\left (-1-\cot \left (x \right )^{2}\right ) \] Which simplifies to \[ W = \csc \left (x \right ) \] Which simplifies to \[ W = \csc \left (x \right ) \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\csc \left (x \right ) \cos \left (x \right )}{\csc \left (x \right )}\,dx \] Which simplifies to \[ u_1 = - \int \cos \left (x \right )d x \] Hence \[ u_1 = -\sin \left (x \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {\cos \left (x \right ) \cot \left (x \right )}{\csc \left (x \right )}\,dx \] Which simplifies to \[ u_2 = \int \cos \left (x \right )^{2}d x \] Hence \[ u_2 = \frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2} \] Which simplifies to \begin{align*} u_1 &= -\sin \left (x \right ) \\ u_2 &= \frac {\sin \left (2 x \right )}{4}+\frac {x}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\cot \left (x \right ) \sin \left (x \right )+\left (\frac {\sin \left (2 x \right )}{4}+\frac {x}{2}\right ) \csc \left (x \right ) \] Which simplifies to \[ y_p(x) = -\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2} \] Hence the complete solution is \begin {align*} y(x) &= y_h + y_p \\ &= \left (\cot \left (x \right ) c_{2} +\csc \left (x \right ) c_{1}\right ) + \left (-\frac {\cos \left (x \right )}{2}+\frac {\csc \left (x \right ) x}{2}\right )\\ &= \frac {\left (2 c_{1} +x \right ) \csc \left (x \right )}{2}+\cot \left (x \right ) c_{2} -\frac {\cos \left (x \right )}{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (2 c_{1} +x \right ) \csc \left (x \right )}{2}+\cot \left (x \right ) c_{2} -\frac {\cos \left (x \right )}{2} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (2 c_{1} +x \right ) \csc \left (x \right )}{2}+\cot \left (x \right ) c_{2} -\frac {\cos \left (x \right )}{2} \] Verified OK.

2.46.5 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y = \cos \left (x \right ) \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y\right )d x &= \int \cos \left (x \right )d x\\ \int \left (y^{\prime \prime }+\cot \left (x \right ) y^{\prime }-\csc \left (x \right )^{2} y\right )d x = \sin \left (x \right ) +c_{1} \end {align*}

Which is now solved for \(y\).

2.46.6 Solving as exact linear second order ode ode

An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}

is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}

For the given ode we have \begin {align*} p(x) &= 1\\ q(x) &= \cot \left (x \right )\\ r(x) &= -\csc \left (x \right )^{2}\\ s(x) &= \cos \left (x \right ) \end {align*}

Hence \begin {align*} p''(x) &= 0\\ q'(x) &= -1-\cot \left (x \right )^{2} \end {align*}

Therefore (1) becomes \begin {align*} 0- \left (-1-\cot \left (x \right )^{2}\right ) + \left (-\csc \left (x \right )^{2}\right )&=0 \end {align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}

Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}

Substituting the above values for \(p,q,r,s\) gives \begin {align*} \cot \left (x \right ) y+y^{\prime }&=\int {\cos \left (x \right )\, dx} \end {align*}

We now have a first order ode to solve which is \begin {align*} \cot \left (x \right ) y+y^{\prime } = \sin \left (x \right )+c_{1} \end {align*}

Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}

Where here \begin {align*} p(x) &=\cot \left (x \right )\\ q(x) &=\sin \left (x \right )+c_{1} \end {align*}

Hence the ode is \begin {align*} \cot \left (x \right ) y+y^{\prime } = \sin \left (x \right )+c_{1} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \cot \left (x \right )d x} \\ &= \sin \left (x \right ) \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\sin \left (x \right )+c_{1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\sin \left (x \right ) y\right ) &= \left (\sin \left (x \right )\right ) \left (\sin \left (x \right )+c_{1}\right )\\ \mathrm {d} \left (\sin \left (x \right ) y\right ) &= \left (\left (\sin \left (x \right )+c_{1} \right ) \sin \left (x \right )\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} \sin \left (x \right ) y &= \int {\left (\sin \left (x \right )+c_{1} \right ) \sin \left (x \right )\,\mathrm {d} x}\\ \sin \left (x \right ) y &= -\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}-\cos \left (x \right ) c_{1} + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu =\sin \left (x \right )\) results in \begin {align*} y &= \csc \left (x \right ) \left (-\frac {\cos \left (x \right ) \sin \left (x \right )}{2}+\frac {x}{2}-\cos \left (x \right ) c_{1} \right )+\csc \left (x \right ) c_{2} \end {align*}

which simplifies to \begin {align*} y &= \frac {\left (x +2 c_{2} \right ) \csc \left (x \right )}{2}-\cot \left (x \right ) c_{1} -\frac {\cos \left (x \right )}{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (x +2 c_{2} \right ) \csc \left (x \right )}{2}-\cot \left (x \right ) c_{1} -\frac {\cos \left (x \right )}{2} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (x +2 c_{2} \right ) \csc \left (x \right )}{2}-\cot \left (x \right ) c_{1} -\frac {\cos \left (x \right )}{2} \] Verified OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
trying symmetries linear in x and y(x) 
-> Try solving first the homogeneous part of the ODE 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   <- linear_1 successful 
<- solving first the homogeneous part of the ODE successful`
 

✓ Solution by Maple

Time used: 0.015 (sec). Leaf size: 37

dsolve(diff(y(x),x$2)+cot(x)*diff(y(x),x)-csc(x)^2*y(x)=cos(x),y(x), singsol=all)
 

\[ y \left (x \right ) = -\cos \left (\frac {x}{2}\right )^{2}+\frac {1}{2}+\frac {\sec \left (\frac {x}{2}\right ) \csc \left (\frac {x}{2}\right ) x}{4}+\cot \left (\frac {x}{2}\right ) c_{2} +\tan \left (\frac {x}{2}\right ) c_{1} \]

✓ Solution by Mathematica

Time used: 0.349 (sec). Leaf size: 45

DSolve[y''[x]+Cot[x]*y'[x]-Csc[x]^2*y[x]==Cos[x],y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {1}{2} \left (x \csc (x)+\frac {2 c_1}{\sqrt {\sin ^2(x)}}+\cos (x) \left (-1-\frac {2 i c_2}{\sqrt {\sin ^2(x)}}\right )\right ) \]