Internal problem ID [9558]
Internal file name [OUTPUT/8498_Monday_June_06_2022_03_14_33_AM_39556210/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1229.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "kovacic", "exact linear second order ode", "second_order_integrable_as_is", "second_order_change_of_variable_on_y_method_1", "linear_second_order_ode_solved_by_an_integrating_factor"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _nonhomogeneous]]
\[ \boxed {\left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y=2 \cos \left (x \right )-2 x} \]
The ode satisfies this form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+\frac {\left (p \left (x \right )^{2}+p^{\prime }\left (x \right )\right ) y}{2} = f \left (x \right ) \] Where \( p(x) = \frac {4 x}{x^{2}+1}\). Therefore, there is an integrating factor given by \begin {align*} M(x) &= e^{\frac {1}{2} \int p \, dx} \\ &= e^{ \int \frac {4 x}{x^{2}+1} \, dx} \\ &= x^{2}+1 \end {align*}
Multiplying both sides of the ODE by the integrating factor \(M(x)\) makes the left side of the ODE a complete differential \begin{align*} \left ( M(x) y \right )'' &= 2 \cos \left (x \right )-2 x \\ \left ( \left (x^{2}+1\right ) y \right )'' &= 2 \cos \left (x \right )-2 x \\ \end{align*} Integrating once gives \[ \left ( \left (x^{2}+1\right ) y \right )' = -x^{2}+2 \sin \left (x \right )+c_{1} \] Integrating again gives \[ \left ( \left (x^{2}+1\right ) y \right ) = c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2} \] Hence the solution is \begin{align*} y &= \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}}{x^{2}+1} \\ \end{align*} Or \[ y = -\frac {x^{3}}{3 \left (x^{2}+1\right )}+\frac {c_{1} x}{x^{2}+1}+\frac {c_{2}}{x^{2}+1}-\frac {2 \cos \left (x \right )}{x^{2}+1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{3}}{3 \left (x^{2}+1\right )}+\frac {c_{1} x}{x^{2}+1}+\frac {c_{2}}{x^{2}+1}-\frac {2 \cos \left (x \right )}{x^{2}+1} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{3}}{3 \left (x^{2}+1\right )}+\frac {c_{1} x}{x^{2}+1}+\frac {c_{2}}{x^{2}+1}-\frac {2 \cos \left (x \right )}{x^{2}+1} \] Verified OK.
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 \[ \left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y = 0 \] In normal form the given ode is written as \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 )&=\frac {4 x}{x^{2}+1}\\ q \left (x \right )&=\frac {2}{x^{2}+1} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= \frac {2}{x^{2}+1} - \frac {\left (\frac {4 x}{x^{2}+1}\right )'}{2}- \frac {\left (\frac {4 x}{x^{2}+1}\right )^2}{4} \\ &= \frac {2}{x^{2}+1} - \frac {\left (\frac {4}{x^{2}+1}-\frac {8 x^{2}}{\left (x^{2}+1\right )^{2}}\right )}{2}- \frac {\left (\frac {16 x^{2}}{\left (x^{2}+1\right )^{2}}\right )}{4} \\ &= \frac {2}{x^{2}+1} - \left (\frac {2}{x^{2}+1}-\frac {4 x^{2}}{\left (x^{2}+1\right )^{2}}\right )-\frac {4 x^{2}}{\left (x^{2}+1\right )^{2}}\\ &= 0 \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {\frac {4 x}{x^{2}+1}}{2} }\\ &= \frac {1}{x^{2}+1}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = \frac {v \left (x \right )}{x^{2}+1}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} v^{\prime \prime }\left (x \right ) = 2 \cos \left (x \right )-2 x \end {align*}
Which is now solved for \(v \left (x \right )\) Integrating once gives \[ v^{\prime }\left (x \right )= -x^{2}+2 \sin \left (x \right ) + c_{1} \] Integrating again gives \[ v \left (x \right )= -\frac {x^{3}}{3}-2 \cos \left (x \right ) + c_{1} x + c_{2} \] Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= \frac {1}{x^{2}+1} \end {align*}
Hence (7) becomes \begin {align*} y = \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}}{x^{2}+1} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}}{x^{2}+1} \] 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 &= \frac {x}{x^{2}+1} \\ y_2 &= \frac {1}{x^{2}+1} \\ \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 \[ W = \begin {vmatrix} \frac {x}{x^{2}+1} & \frac {1}{x^{2}+1} \\ \frac {d}{dx}\left (\frac {x}{x^{2}+1}\right ) & \frac {d}{dx}\left (\frac {1}{x^{2}+1}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {x}{x^{2}+1} & \frac {1}{x^{2}+1} \\ \frac {1}{x^{2}+1}-\frac {2 x^{2}}{\left (x^{2}+1\right )^{2}} & -\frac {2 x}{\left (x^{2}+1\right )^{2}} \end {vmatrix} \] Therefore \[ W = \left (\frac {x}{x^{2}+1}\right )\left (-\frac {2 x}{\left (x^{2}+1\right )^{2}}\right ) - \left (\frac {1}{x^{2}+1}\right )\left (\frac {1}{x^{2}+1}-\frac {2 x^{2}}{\left (x^{2}+1\right )^{2}}\right ) \] Which simplifies to \[ W = -\frac {1}{\left (x^{2}+1\right )^{2}} \] Which simplifies to \[ W = -\frac {1}{\left (x^{2}+1\right )^{2}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {2 \cos \left (x \right )-2 x}{x^{2}+1}}{-\frac {1}{x^{2}+1}}\,dx \] Which simplifies to \[ u_1 = - \int \left (-2 \cos \left (x \right )+2 x \right )d x \] Hence \[ u_1 = -x^{2}+2 \sin \left (x \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {x \left (2 \cos \left (x \right )-2 x \right )}{x^{2}+1}}{-\frac {1}{x^{2}+1}}\,dx \] Which simplifies to \[ u_2 = \int -2 x \left (\cos \left (x \right )-x \right )d x \] Hence \[ u_2 = -2 \cos \left (x \right )-2 \sin \left (x \right ) x +\frac {2 x^{3}}{3} \] Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {\left (-x^{2}+2 \sin \left (x \right )\right ) x}{x^{2}+1}+\frac {-2 \cos \left (x \right )-2 \sin \left (x \right ) x +\frac {2 x^{3}}{3}}{x^{2}+1} \] Which simplifies to \[ y_p(x) = \frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}}{x^{2}+1}\right ) + \left (\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}}{x^{2}+1}+\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )+c_{2}}{x^{2}+1}+\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \] Verified OK.
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y\right )d x &= \int \left (2 \cos \left (x \right )-2 x \right )d x\\ 2 y x +\left (x^{2}+1\right ) y^{\prime } = -x^{2}+2 \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) &=\frac {2 x}{x^{2}+1}\\ q(x) &=\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {2 y x}{x^{2}+1} = \frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2 x}{x^{2}+1}d x} \\ &= x^{2}+1 \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (x^{2}+1\right ) y\right ) &= \left (x^{2}+1\right ) \left (\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1}\right )\\ \mathrm {d} \left (\left (x^{2}+1\right ) y\right ) &= \left (-x^{2}+2 \sin \left (x \right )+c_{1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \left (x^{2}+1\right ) y &= \int {-x^{2}+2 \sin \left (x \right )+c_{1}\,\mathrm {d} x}\\ \left (x^{2}+1\right ) y &= c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right ) + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =x^{2}+1\) results in \begin {align*} y &= \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )}{x^{2}+1}+\frac {c_{2}}{x^{2}+1} \end {align*}
which simplifies to \begin {align*} y &= \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \\ \end{align*}
Verification of solutions
\[ y = \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \] Verified OK.
Writing the ode as \[ \left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y = 2 \cos \left (x \right )-2 x \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y\right )d x &= \int \left (2 \cos \left (x \right )-2 x \right )d x\\ y^{\prime }+y^{\prime } x^{2}+2 y x = -x^{2}+2 \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) &=\frac {2 x}{x^{2}+1}\\ q(x) &=\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {2 y x}{x^{2}+1} = \frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2 x}{x^{2}+1}d x} \\ &= x^{2}+1 \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (x^{2}+1\right ) y\right ) &= \left (x^{2}+1\right ) \left (\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1}\right )\\ \mathrm {d} \left (\left (x^{2}+1\right ) y\right ) &= \left (-x^{2}+2 \sin \left (x \right )+c_{1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \left (x^{2}+1\right ) y &= \int {-x^{2}+2 \sin \left (x \right )+c_{1}\,\mathrm {d} x}\\ \left (x^{2}+1\right ) y &= c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right ) + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =x^{2}+1\) results in \begin {align*} y &= \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )}{x^{2}+1}+\frac {c_{2}}{x^{2}+1} \end {align*}
which simplifies to \begin {align*} y &= \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \\ \end{align*}
Verification of solutions
\[ y = \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \] Verified OK.
Writing the ode as \begin {align*} \left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end {align*}
Comparing (1) and (2) shows that \begin {align*} A &= x^{2}+1 \\ B &= 4 x\tag {3} \\ C &= 2 \end {align*}
Applying the Liouville transformation on the dependent variable gives \begin {align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end {align*}
Then (2) becomes \begin {align*} z''(x) = r z(x)\tag {4} \end {align*}
Where \(r\) is given by \begin {align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end {align*}
Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives \begin {align*} r &= \frac {0}{1}\tag {6} \end {align*}
Comparing the above to (5) shows that \begin {align*} s &= 0\\ t &= 1 \end {align*}
Therefore eq. (4) becomes \begin {align*} z''(x) &= 0 \tag {7} \end {align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation \begin {align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \end {align*}
The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.
| Case |
Allowed pole order for \(r\) |
Allowed value for \(\mathcal {O}(\infty )\) |
| 1 |
\(\left \{ 0,1,2,4,6,8,\cdots \right \} \) |
\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \) |
|
2 |
Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\). |
no condition |
| 3 |
\(\left \{ 1,2\right \} \) |
\(\left \{ 2,3,4,5,6,7,\cdots \right \} \) |
The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore \begin {align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end {align*}
There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore \begin {align*} L &= [1] \end {align*}
Since \(r = 0\) is not a function of \(x\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is \[ z_1(x) = 1 \] Using the above, the solution for the original ode can now be found. The first solution to the original ode in \(y\) is found from \begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {4 x}{x^{2}+1} \,dx} \\ &= z_1 e^{-\ln \left (x^{2}+1\right )} \\ &= z_1 \left (\frac {1}{x^{2}+1}\right ) \\ \end{align*} Which simplifies to \[ y_1 = \frac {1}{x^{2}+1} \] The second solution \(y_2\) to the original ode is found using reduction of order \[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \] Substituting gives \begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {4 x}{x^{2}+1} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{-2 \ln \left (x^{2}+1\right )}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (x\right ) \\ \end{align*} Therefore the solution is
\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (\frac {1}{x^{2}+1}\right ) + c_{2} \left (\frac {1}{x^{2}+1}\left (x\right )\right ) \\ \end{align*} This is second order nonhomogeneous 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 nonhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ \left (x^{2}+1\right ) y^{\prime \prime }+4 y^{\prime } x +2 y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = \frac {c_{1}}{x^{2}+1}+\frac {c_{2} x}{x^{2}+1} \] 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 &= \frac {1}{x^{2}+1} \\ y_2 &= \frac {x}{x^{2}+1} \\ \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 \[ W = \begin {vmatrix} \frac {1}{x^{2}+1} & \frac {x}{x^{2}+1} \\ \frac {d}{dx}\left (\frac {1}{x^{2}+1}\right ) & \frac {d}{dx}\left (\frac {x}{x^{2}+1}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {1}{x^{2}+1} & \frac {x}{x^{2}+1} \\ -\frac {2 x}{\left (x^{2}+1\right )^{2}} & \frac {1}{x^{2}+1}-\frac {2 x^{2}}{\left (x^{2}+1\right )^{2}} \end {vmatrix} \] Therefore \[ W = \left (\frac {1}{x^{2}+1}\right )\left (\frac {1}{x^{2}+1}-\frac {2 x^{2}}{\left (x^{2}+1\right )^{2}}\right ) - \left (\frac {x}{x^{2}+1}\right )\left (-\frac {2 x}{\left (x^{2}+1\right )^{2}}\right ) \] Which simplifies to \[ W = \frac {1}{\left (x^{2}+1\right )^{2}} \] Which simplifies to \[ W = \frac {1}{\left (x^{2}+1\right )^{2}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\frac {x \left (2 \cos \left (x \right )-2 x \right )}{x^{2}+1}}{\frac {1}{x^{2}+1}}\,dx \] Which simplifies to \[ u_1 = - \int x \left (2 \cos \left (x \right )-2 x \right )d x \] Hence \[ u_1 = -2 \cos \left (x \right )-2 \sin \left (x \right ) x +\frac {2 x^{3}}{3} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {2 \cos \left (x \right )-2 x}{x^{2}+1}}{\frac {1}{x^{2}+1}}\,dx \] Which simplifies to \[ u_2 = \int \left (2 \cos \left (x \right )-2 x \right )d x \] Hence \[ u_2 = -x^{2}+2 \sin \left (x \right ) \] Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {\left (-x^{2}+2 \sin \left (x \right )\right ) x}{x^{2}+1}+\frac {-2 \cos \left (x \right )-2 \sin \left (x \right ) x +\frac {2 x^{3}}{3}}{x^{2}+1} \] Which simplifies to \[ y_p(x) = \frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {c_{1}}{x^{2}+1}+\frac {c_{2} x}{x^{2}+1}\right ) + \left (\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3}\right ) \\ \end{align*} Which simplifies to \[ y = \frac {c_{2} x +c_{1}}{x^{2}+1}+\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{2} x +c_{1}}{x^{2}+1}+\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{2} x +c_{1}}{x^{2}+1}+\frac {-x^{3}-6 \cos \left (x \right )}{3 x^{2}+3} \] Verified OK.
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) &= x^{2}+1\\ q(x) &= 4 x\\ r(x) &= 2\\ s(x) &= 2 \cos \left (x \right )-2 x \end {align*}
Hence \begin {align*} p''(x) &= 2\\ q'(x) &= 4 \end {align*}
Therefore (1) becomes \begin {align*} 2- \left (4\right ) + \left (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*} 2 y x +\left (x^{2}+1\right ) y^{\prime }&=\int {2 \cos \left (x \right )-2 x\, dx} \end {align*}
We now have a first order ode to solve which is \begin {align*} 2 y x +\left (x^{2}+1\right ) y^{\prime } = -x^{2}+2 \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) &=\frac {2 x}{x^{2}+1}\\ q(x) &=\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1} \end {align*}
Hence the ode is \begin {align*} y^{\prime }+\frac {2 y x}{x^{2}+1} = \frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2 x}{x^{2}+1}d x} \\ &= x^{2}+1 \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\left (x^{2}+1\right ) y\right ) &= \left (x^{2}+1\right ) \left (\frac {-x^{2}+2 \sin \left (x \right )+c_{1}}{x^{2}+1}\right )\\ \mathrm {d} \left (\left (x^{2}+1\right ) y\right ) &= \left (-x^{2}+2 \sin \left (x \right )+c_{1}\right )\, \mathrm {d} x \end {align*}
Integrating gives \begin {align*} \left (x^{2}+1\right ) y &= \int {-x^{2}+2 \sin \left (x \right )+c_{1}\,\mathrm {d} x}\\ \left (x^{2}+1\right ) y &= c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right ) + c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =x^{2}+1\) results in \begin {align*} y &= \frac {c_{1} x -\frac {x^{3}}{3}-2 \cos \left (x \right )}{x^{2}+1}+\frac {c_{2}}{x^{2}+1} \end {align*}
which simplifies to \begin {align*} y &= \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \\ \end{align*}
Verification of solutions
\[ y = \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable <- high order exact linear fully integrable successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 31
dsolve((x^2+1)*diff(diff(y(x),x),x)+4*x*diff(y(x),x)+2*y(x)-2*cos(x)+2*x=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-x^{3}+3 c_{1} x -6 \cos \left (x \right )+3 c_{2}}{3 x^{2}+3} \]
✓ Solution by Mathematica
Time used: 0.085 (sec). Leaf size: 33
DSolve[2*x - 2*Cos[x] + 2*y[x] + 4*x*y'[x] + (1 + x^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {x^3+6 \cos (x)-3 c_2 x-3 c_1}{3 x^2+3} \]