problem 1109

Internal problem ID [9438]
Internal ﬁle name [OUTPUT/8378_Monday_June_06_2022_02_53_00_AM_94158363/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1109.
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"

\[ \boxed {x y^{\prime \prime }-x y^{\prime }-y=x \left (x +1\right ) {\mathrm e}^{x}} \]

3.105.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y^{\prime \prime }-x y^{\prime }-y\right )d x &= \int x \left (x +1\right ) {\mathrm e}^{x}d x\\ \left (-x -1\right ) y+x y^{\prime } = \left (x^{2}-x +1\right ) {\mathrm e}^{x} + 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 {x +1}{x}\\ q(x) &=\frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }-\frac {\left (x +1\right ) y}{x} = \frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x} \end {align*}

The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {x +1}{x}d x} \\ &= {\mathrm e}^{-x -\ln \left (x \right )} \\ \end{align*} Which simpliﬁes to \[ \mu = \frac {{\mathrm e}^{-x}}{x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (\frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {{\mathrm e}^{-x} y}{x}\right ) &= \left (\frac {{\mathrm e}^{-x}}{x}\right ) \left (\frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x}\right )\\ \mathrm {d} \left (\frac {{\mathrm e}^{-x} y}{x}\right ) &= \left (\frac {x^{2}-x +1+c_{1} {\mathrm e}^{-x}}{x^{2}}\right )\, \mathrm {d} x \end {align*}

Integrating gives \begin {align*} \frac {{\mathrm e}^{-x} y}{x} &= \int {\frac {x^{2}-x +1+c_{1} {\mathrm e}^{-x}}{x^{2}}\,\mathrm {d} x}\\ \frac {{\mathrm e}^{-x} y}{x} &= x -\frac {1}{x}-\ln \left (-x \right )-c_{1} \left (\frac {{\mathrm e}^{-x}}{x}-\operatorname {expIntegral}_{1}\left (x \right )\right ) + c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu =\frac {{\mathrm e}^{-x}}{x}\) results in \begin {align*} y &= x \,{\mathrm e}^{x} \left (x -\frac {1}{x}-\ln \left (-x \right )-c_{1} \left (\frac {{\mathrm e}^{-x}}{x}-\operatorname {expIntegral}_{1}\left (x \right )\right )\right )+c_{2} x \,{\mathrm e}^{x} \end {align*}

which simpliﬁes to \begin {align*} y &= -\ln \left (-x \right ) x \,{\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) c_{1} x \,{\mathrm e}^{x}+\left (c_{2} x +x^{2}-1\right ) {\mathrm e}^{x}-c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\ln \left (-x \right ) x \,{\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) c_{1} x \,{\mathrm e}^{x}+\left (c_{2} x +x^{2}-1\right ) {\mathrm e}^{x}-c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\ln \left (-x \right ) x \,{\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) c_{1} x \,{\mathrm e}^{x}+\left (c_{2} x +x^{2}-1\right ) {\mathrm e}^{x}-c_{1} \] Veriﬁed OK.

3.105.2 Solving as type second_order_integrable_as_is (not using ABC version)

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

Hence the ode is \begin {align*} y^{\prime }-\frac {\left (x +1\right ) y}{x} = \frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x} \end {align*}

Summary

Veriﬁcation of solutions

\[ y = -\ln \left (-x \right ) x \,{\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) c_{1} x \,{\mathrm e}^{x}+\left (c_{2} x +x^{2}-1\right ) {\mathrm e}^{x}-c_{1} \] Veriﬁed OK.

3.105.3 Solving using Kovacic algorithm

Writing the ode as \begin {align*} x y^{\prime \prime }-x y^{\prime }-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 \\ B &= -x\tag {3} \\ C &= -1 \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*}

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 {4+x}{4 x}\tag {6} \end {align*}

Comparing the above to (5) shows that \begin {align*} s &= 4+x\\ t &= 4 x \end {align*}

Therefore eq. (4) becomes \begin {align*} z''(x) &= \left ( \frac {4+x}{4 x}\right ) z(x)\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	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 265: Necessary conditions for each Kovacic case

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) \\ &= 1 - 1 \\ &= 0 \end {align*}

The poles of \(r\) in eq. (7) and the order of each pole are determined by solving for the roots of \(t=4 x\). There is a pole at \(x=0\) of order \(1\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(0\) then the necessary conditions for case one are met. Therefore \begin {align*} L &= [1] \end {align*}

Looking at poles of order 1. For the pole at \(x = 0\) of order 1 then \begin {align*} [\sqrt r]_c &= 0 \\ \alpha _c^+ &= 1 \\ \alpha _c^- &= 1 \end {align*}

Since the order of \(r\) at \(\infty \) is \(O_r(\infty ) = 0\) then \begin {alignat*} {3} v &= \frac {-O_r(\infty )}{2} &&= \frac {0}{2} &&= 0 \end {alignat*}

\([\sqrt r]_\infty \) is the sum of terms involving \(x^i\) for \(0\leq i \leq v\) in the Laurent series for \(\sqrt r\) at \(\infty \). Therefore \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{v} a_i x^i \\ &= \sum _{i=0}^{0} a_i x^i \tag {8} \end {align*}

Let \(a\) be the coeﬃcient of \(x^v=x^0\) in the above sum. The Laurent series of \(\sqrt r\) at \(\infty \) is \[ \sqrt r \approx \frac {1}{2}+\frac {1}{x}-\frac {1}{x^{2}}+\frac {2}{x^{3}}-\frac {5}{x^{4}}+\frac {14}{x^{5}}-\frac {42}{x^{6}}+\frac {132}{x^{7}} + \dots \tag {9} \] Comparing Eq. (9) with Eq. (8) shows that \[ a = {\frac {1}{2}} \] From Eq. (9) the sum up to \(v=0\) gives \begin {align*} [\sqrt r]_\infty &= \sum _{i=0}^{0} a_i x^i \\ &= {\frac {1}{2}} \tag {10} \end {align*}

Now we need to ﬁnd \(b\), where \(b\) be the coeﬃcient of \(x^{v-1} = x^{-1}=\frac {1}{x}\) in \(r\) minus the coeﬃcient of same term but in \(\left ( [\sqrt r]_\infty \right )^2 \) where \([\sqrt r]_\infty \) was found above in Eq (10). Hence \[ \left ( [\sqrt r]_\infty \right )^2 = {\frac {1}{4}} \] This shows that the coeﬃcient of \(\frac {1}{x}\) in the above is \(0\). Now we need to ﬁnd the coeﬃcient of \(\frac {1}{x}\) in \(r\). How this is done depends on if \(v=0\) or not. Since \(v=0\) then starting from \(r=\frac {s}{t}\) and doing long division in the form \[ r = Q + \frac {R}{t} \] Where \(Q\) is the quotient and \(R\) is the remainder. Then the coeﬃcient of \(\frac {1}{x}\) in \(r\) will be the coeﬃcient in \(R\) of the term in \(x\) of degree of \(t\) minus one, divided by the leading coeﬃcient in \(t\). Doing long division gives \begin {align*} r &= \frac {s}{t} \\ &= \frac {4+x}{4 x} \\ &= Q + \frac {R}{4 x}\\ &= \left ({\frac {1}{4}}\right ) + \left ( \frac {1}{x}\right ) \\ &= \frac {1}{4}+\frac {1}{x} \end {align*}

Since the degree of \(t\) is \(1\), then we see that the coeﬃcient of the term \(1\) in the remainder \(R\) is \(4\). Dividing this by leading coeﬃcient in \(t\) which is \(4\) gives \(1\). Now \(b\) can be found. \begin {align*} b &= \left (1\right )-\left (0\right )\\ &= 1 \end {align*}

Hence \begin {alignat*} {3} [\sqrt r]_\infty &= {\frac {1}{2}}\\ \alpha _{\infty }^{+} &= \frac {1}{2} \left ( \frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( \frac {1}{{\frac {1}{2}}} - 0 \right ) &&= 1\\ \alpha _{\infty }^{-} &= \frac {1}{2} \left ( -\frac {b}{a} - v \right ) &&= \frac {1}{2} \left ( -\frac {1}{{\frac {1}{2}}} - 0 \right ) &&= -1 \end {alignat*}

The following table summarizes the ﬁndings so far for poles and for the order of \(r\) at \(\infty \) where \(r\) is \[ r=\frac {4+x}{4 x} \]

Now that the all \([\sqrt r]_c\) and its associated \(\alpha _c^{\pm }\) have been determined for all the poles in the set \(\Gamma \) and \([\sqrt r]_\infty \) and its associated \(\alpha _\infty ^{\pm }\) have also been found, the next step is to determine possible non negative integer \(d\) from these using \begin {align*} d &= \alpha _\infty ^{s(\infty )} - \sum _{c \in \Gamma } \alpha _c^{s(c)} \end {align*}

Where \(s(c)\) is either \(+\) or \(-\) and \(s(\infty )\) is the sign of \(\alpha _\infty ^{\pm }\). This is done by trial over all set of families \(s=(s(c))_{c \in \Gamma \cup {\infty }}\) until such \(d\) is found to work in ﬁnding candidate \(\omega \). Trying \(\alpha _\infty ^{+} = 1\) then \begin {align*} d &= \alpha _\infty ^{+} - \left ( \alpha _{c_1}^{-} \right ) \\ &= 1 - \left ( 1 \right ) \\ &= 0 \end {align*}

Since \(d\) an integer and \(d \geq 0\) then it can be used to ﬁnd \(\omega \) using \begin {align*} \omega &= \sum _{c \in \Gamma } \left ( s(c) [\sqrt r]_c + \frac {\alpha _c^{s(c)}}{x-c} \right ) + s(\infty ) [\sqrt r]_\infty \end {align*}

Substituting the above values in the above results in \begin {align*} \omega &= \left ( (-)[\sqrt r]_{c_1} + \frac { \alpha _{c_1}^{-} }{x- c_1}\right ) + (+) [\sqrt r]_\infty \\ &= \frac {1}{x} + \left ( {\frac {1}{2}} \right ) \\ &= \frac {1}{2}+\frac {1}{x}\\ &= \frac {1}{2}+\frac {1}{x} \end {align*}

Now that \(\omega \) is determined, the next step is ﬁnd a corresponding minimal polynomial \(p(x)\) of degree \(d=0\) to solve the ode. The polynomial \(p(x)\) needs to satisfy the equation \begin {align*} p'' + 2 \omega p' + \left ( \omega ' +\omega ^2 -r\right ) p = 0 \tag {1A} \end {align*}

Substituting the above in eq. (1A) gives \begin {align*} \left (0\right ) + 2 \left (\frac {1}{2}+\frac {1}{x}\right ) \left (0\right ) + \left ( \left (-\frac {1}{x^{2}}\right ) + \left (\frac {1}{2}+\frac {1}{x}\right )^2 - \left (\frac {4+x}{4 x}\right ) \right ) &= 0\\ 0 = 0 \end {align*}

The equation is satisﬁed since both sides are zero. Therefore the ﬁrst solution to the ode \(z'' = r z\) is \begin {align*} z_1(x) &= p e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \left (\frac {1}{2}+\frac {1}{x}\right )d x}\\ &= x \,{\mathrm e}^{\frac {x}{2}} \end {align*}

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 {-x}{x} \,dx} \\ &= z_1 e^{\frac {x}{2}} \\ &= z_1 \left ({\mathrm e}^{\frac {x}{2}}\right ) \\ \end{align*} Which simpliﬁes to \[ y_1 = x \,{\mathrm e}^{x} \] 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 {-x}{x} \,dx}}{\left (y_1\right )^2} \,dx \\ &= y_1 \int \frac { e^{x}}{\left (y_1\right )^2} \,dx \\ &= y_1 \left (\frac {\operatorname {expIntegral}_{1}\left (x \right ) x -{\mathrm e}^{-x}}{x}\right ) \\ \end{align*} Therefore the solution is

\begin{align*} y &= c_{1} y_1 + c_{2} y_2 \\ &= c_{1} \left (x \,{\mathrm e}^{x}\right ) + c_{2} \left (x \,{\mathrm e}^{x}\left (\frac {\operatorname {expIntegral}_{1}\left (x \right ) x -{\mathrm e}^{-x}}{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 \[ x y^{\prime \prime }-x y^{\prime }-y = 0 \] The homogeneous solution is found using the Kovacic algorithm which results in \[ y_h = {\mathrm e}^{x} c_{1} x +c_{2} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) \] 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 &= x \,{\mathrm e}^{x} \\ y_2 &= \operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-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 \[ W = \begin {vmatrix} x \,{\mathrm e}^{x} & \operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1 \\ \frac {d}{dx}\left (x \,{\mathrm e}^{x}\right ) & \frac {d}{dx}\left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} x \,{\mathrm e}^{x} & \operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1 \\ x \,{\mathrm e}^{x}+{\mathrm e}^{x} & -{\mathrm e}^{x} {\mathrm e}^{-x}+\operatorname {expIntegral}_{1}\left (x \right ) {\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x} \end {vmatrix} \] Therefore \[ W = \left (x \,{\mathrm e}^{x}\right )\left (-{\mathrm e}^{x} {\mathrm e}^{-x}+\operatorname {expIntegral}_{1}\left (x \right ) {\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}\right ) - \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right )\left (x \,{\mathrm e}^{x}+{\mathrm e}^{x}\right ) \] Which simpliﬁes to \[ W = -{\mathrm e}^{2 x} {\mathrm e}^{-x} x +x \,{\mathrm e}^{x}+{\mathrm e}^{x} \] Which simpliﬁes to \[ W = {\mathrm e}^{x} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) x \left (x +1\right ) {\mathrm e}^{x}}{x \,{\mathrm e}^{x}}\,dx \] Which simpliﬁes to \[ u_1 = - \int \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) \left (x +1\right )d x \] Hence \[ u_1 = -\left (\int _{0}^{x}\left (\operatorname {expIntegral}_{1}\left (\alpha \right ) \alpha \,{\mathrm e}^{\alpha }-1\right ) \left (\alpha +1\right )d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {x^{2} {\mathrm e}^{2 x} \left (x +1\right )}{x \,{\mathrm e}^{x}}\,dx \] Which simpliﬁes to \[ u_2 = \int x \left (x +1\right ) {\mathrm e}^{x}d x \] Hence \[ u_2 = \left (x^{2}-x +1\right ) {\mathrm e}^{x} \] Therefore the particular solution, from equation (1) is \[ y_p(x) = -\left (\int _{0}^{x}\left (\operatorname {expIntegral}_{1}\left (\alpha \right ) \alpha \,{\mathrm e}^{\alpha }-1\right ) \left (\alpha +1\right )d \alpha \right ) x \,{\mathrm e}^{x}+\left (x^{2}-x +1\right ) {\mathrm e}^{x} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{x} c_{1} x +c_{2} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right )\right ) + \left (-\left (\int _{0}^{x}\left (\operatorname {expIntegral}_{1}\left (\alpha \right ) \alpha \,{\mathrm e}^{\alpha }-1\right ) \left (\alpha +1\right )d \alpha \right ) x \,{\mathrm e}^{x}+\left (x^{2}-x +1\right ) {\mathrm e}^{x} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{x} c_{1} x +c_{2} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right )-\left (\int _{0}^{x}\left (\operatorname {expIntegral}_{1}\left (\alpha \right ) \alpha \,{\mathrm e}^{\alpha }-1\right ) \left (\alpha +1\right )d \alpha \right ) x \,{\mathrm e}^{x}+\left (x^{2}-x +1\right ) {\mathrm e}^{x} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{x} c_{1} x +c_{2} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right )-\left (\int _{0}^{x}\left (\operatorname {expIntegral}_{1}\left (\alpha \right ) \alpha \,{\mathrm e}^{\alpha }-1\right ) \left (\alpha +1\right )d \alpha \right ) x \,{\mathrm e}^{x}+\left (x^{2}-x +1\right ) {\mathrm e}^{x} \left (\operatorname {expIntegral}_{1}\left (x \right ) x \,{\mathrm e}^{x}-1\right ) \] Veriﬁed OK.

3.105.4 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*}

For the given ode we have \begin {align*} p(x) &= x\\ q(x) &= -x\\ r(x) &= -1\\ s(x) &= x \left (x +1\right ) {\mathrm e}^{x} \end {align*}

Therefore (1) becomes \begin {align*} 0- \left (-1\right ) + \left (-1\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*} \left (-x -1\right ) y+x y^{\prime }&=\int {x \left (x +1\right ) {\mathrm e}^{x}\, dx} \end {align*}

We now have a ﬁrst order ode to solve which is \begin {align*} \left (-x -1\right ) y+x y^{\prime } = \left (x^{2}-x +1\right ) {\mathrm e}^{x}+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 {x +1}{x}\\ q(x) &=\frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x} \end {align*}

Hence the ode is \begin {align*} y^{\prime }-\frac {\left (x +1\right ) y}{x} = \frac {{\mathrm e}^{x} x^{2}-x \,{\mathrm e}^{x}+{\mathrm e}^{x}+c_{1}}{x} \end {align*}

Summary

Veriﬁcation of solutions

\[ y = -\ln \left (-x \right ) x \,{\mathrm e}^{x}+\operatorname {expIntegral}_{1}\left (x \right ) c_{1} x \,{\mathrm e}^{x}+\left (c_{2} x +x^{2}-1\right ) {\mathrm e}^{x}-c_{1} \] 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`

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

\[ y(x)\to -c_2 \left (e^x x \operatorname {ExpIntegralEi}(-x)+1\right )+e^x \left (x^2+x-x \log (-x)-1\right )+c_1 e^x x \]


Order of \(r\) at \(\infty \)	\([\sqrt r]_\infty \)	\(\alpha _\infty ^{+}\)	\(\alpha _\infty ^{-}\)

\(0\)	\(\frac {1}{2}\)	\(1\)	\(-1\)


pole \(c\) location	pole order	\([\sqrt r]_c\)	\(\alpha _c^{+}\)	\(\alpha _c^{-}\)

\(0\)	\(1\)	\(0\)	\(0\)	\(1\)

3.105 problem 1109

3.105.1 Solving as second order integrable as is ode

3.105.2 Solving as type second_order_integrable_as_is (not using ABC version)

3.105.3 Solving using Kovacic algorithm

3.105.4 Solving as exact linear second order ode ode