problem 14.3 (e)

Internal problem ID [13553]
Internal ﬁle name [OUTPUT/12725_Saturday_February_17_2024_08_44_13_AM_59055855/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 14. Higher order equations and the reduction of order method. Additional exercises page 277
Problem number: 14.3 (e).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "reduction_of_order", "exact linear second order ode", "second_order_integrable_as_is", "second_order_change_of_variable_on_y_method_1", "second_order_change_of_variable_on_y_method_2"

\[ \boxed {x y^{\prime \prime }+\left (2+2 x \right ) y^{\prime }+2 y=8 \,{\mathrm e}^{2 x}} \] Given that one solution of the ode is \begin {align*} y_1 &= \frac {1}{x} \end {align*}

This is second order nonhomogeneous ODE. In standard form the ODE is \[ A y''(x) + B y'(x) + C y(x) = f(x) \] Where \(A=x, B=2+2 x, C=2, f(x)=8 \,{\mathrm e}^{2 x}\). 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 inhomogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ x y^{\prime \prime }+\left (2+2 x \right ) y^{\prime }+2 y = 0 \] Given one basis solution \(y_{1}\left (x \right )\), then the second basis solution is given by \[ y_{2}\left (x \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d x \right )}}{y_{1}^{2}}d x \right ) \] Where \(p(x)\) is the coeﬃcient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y = f \left (x \right ) \] Looking at the ode to solve shows that \[ p \left (x \right ) = \frac {2+2 x}{x} \] Therefore \begin{align*} y_{2}\left (x \right ) &= \frac {\int {\mathrm e}^{-\left (\int \frac {2+2 x}{x}d x \right )} x^{2}d x}{x} \\ y_{2}\left (x \right ) &= \frac {1}{x} \int \frac {{\mathrm e}^{-2 x -2 \ln \left (x \right )}}{\frac {1}{x^{2}}} , dx \\ y_{2}\left (x \right ) &= \frac {\int {\mathrm e}^{-2 x}d x}{x} \\ y_{2}\left (x \right ) &= -\frac {{\mathrm e}^{-2 x}}{2 x} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= \frac {c_{1}}{x}-\frac {c_{2} {\mathrm e}^{-2 x}}{2 x} \\ \end{align*} Therefore the homogeneous solution \(y_h\) is \[ y_h = \frac {c_{1}}{x}-\frac {c_{2} {\mathrm e}^{-2 x}}{2 x} \] 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} \\ y_2 &= -\frac {{\mathrm e}^{-2 x}}{2 x} \\ \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 \[ W = \begin {vmatrix} \frac {1}{x} & -\frac {{\mathrm e}^{-2 x}}{2 x} \\ \frac {d}{dx}\left (\frac {1}{x}\right ) & \frac {d}{dx}\left (-\frac {{\mathrm e}^{-2 x}}{2 x}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \frac {1}{x} & -\frac {{\mathrm e}^{-2 x}}{2 x} \\ -\frac {1}{x^{2}} & \frac {{\mathrm e}^{-2 x}}{2 x^{2}}+\frac {{\mathrm e}^{-2 x}}{x} \end {vmatrix} \] Therefore \[ W = \left (\frac {1}{x}\right )\left (\frac {{\mathrm e}^{-2 x}}{2 x^{2}}+\frac {{\mathrm e}^{-2 x}}{x}\right ) - \left (-\frac {{\mathrm e}^{-2 x}}{2 x}\right )\left (-\frac {1}{x^{2}}\right ) \] Which simpliﬁes to \[ W = \frac {{\mathrm e}^{-2 x}}{x^{2}} \] Which simpliﬁes to \[ W = \frac {{\mathrm e}^{-2 x}}{x^{2}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {-\frac {4 \,{\mathrm e}^{-2 x} {\mathrm e}^{2 x}}{x}}{\frac {{\mathrm e}^{-2 x}}{x}}\,dx \] Which simpliﬁes to \[ u_1 = - \int -4 \,{\mathrm e}^{2 x}d x \] Hence \[ u_1 = 2 \,{\mathrm e}^{2 x} \] And Eq. (3) becomes \[ u_2 = \int \frac {\frac {8 \,{\mathrm e}^{2 x}}{x}}{\frac {{\mathrm e}^{-2 x}}{x}}\,dx \] Which simpliﬁes to \[ u_2 = \int 8 \,{\mathrm e}^{4 x}d x \] Hence \[ u_2 = 2 \,{\mathrm e}^{4 x} \] Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {2 \,{\mathrm e}^{2 x}}{x}-\frac {{\mathrm e}^{4 x} {\mathrm e}^{-2 x}}{x} \] Which simpliﬁes to \[ y_p(x) = \frac {{\mathrm e}^{2 x}}{x} \] Therefore the general solution is \begin {align*} y &= y_h + y_p \\ &= \left (\frac {c_{1}}{x}-\frac {c_{2} {\mathrm e}^{-2 x}}{2 x}\right ) + \left (\frac {{\mathrm e}^{2 x}}{x}\right ) \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x}-\frac {c_{2} {\mathrm e}^{-2 x}}{2 x}+\frac {{\mathrm e}^{2 x}}{x} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {c_{1}}{x}-\frac {c_{2} {\mathrm e}^{-2 x}}{2 x}+\frac {{\mathrm e}^{2 x}}{x} \] 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`

\[ y \left (x \right ) = \frac {{\mathrm e}^{-2 x} c_{2} +{\mathrm e}^{2 x}+c_{1}}{x} \]

9.29 problem 14.3 (e)