problem 661

Internal problem ID [15425]
Internal ﬁle name [OUTPUT/15426_Wednesday_May_08_2024_03_58_33_PM_17635758/index.tex]

Book: A book of problems in ordinary diﬀerential equations. M.L. KRASNOV, A.L. KISELYOV, G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 15.5 Linear equations with variable coeﬃcients. The Lagrange method. Exercises page 148
Problem number: 661.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime }+y^{\prime \prime }=\frac {x -1}{x^{3}}} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime }+y^{\prime \prime } = 0 \] The characteristic equation is \[ \lambda ^{3}+\lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= 1 \\ y_3 &= x \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+y^{\prime \prime } = \frac {x -1}{x^{3}} \] Let the particular solution be \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Where \(y_i\) are the basis solutions found above for the homogeneous solution \(y_h\) and \(U_i(x)\) are functions to be determined as follows \[ U_i = (-1)^{n-i} \int { \frac {F(x) W_i(x) }{a W(x)} \, dx} \] Where \(W(x)\) is the Wronskian and \(W_i(x)\) is the Wronskian that results after deleting the last row and the \(i\)-th column of the determinant and \(n\) is the order of the ODE or equivalently, the number of basis solutions, and \(a\) is the coeﬃcient of the leading derivative in the ODE, and \(F(x)\) is the RHS of the ODE. Therefore, the ﬁrst step is to ﬁnd the Wronskian \(W \left (x \right )\). This is given by \begin {equation*} W(x) = \begin {vmatrix} y_1&y_2&y_3\\ y_1'&y_2'&y_3'\\ y_1''&y_2''&y_3''\\ \end {vmatrix} \end {equation*} Substituting the fundamental set of solutions \(y_i\) found above in the Wronskian gives \begin {align*} W &= \left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & x \\ -{\mathrm e}^{-x} & 0 & 1 \\ {\mathrm e}^{-x} & 0 & 0 \end {array}\right ] \\ |W| &= {\mathrm e}^{-x} \end {align*}

The determinant simpliﬁes to \begin {align*} |W| &= {\mathrm e}^{-x} \end {align*}

Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{cc} 1 & x \\ 0 & 1 \end {array}\right ] \\ &= 1 \end {align*}

\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} {\mathrm e}^{-x} & x \\ -{\mathrm e}^{-x} & 1 \end {array}\right ] \\ &= {\mathrm e}^{-x} \left (x +1\right ) \end {align*}

\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} {\mathrm e}^{-x} & 1 \\ -{\mathrm e}^{-x} & 0 \end {array}\right ] \\ &= {\mathrm e}^{-x} \end {align*}

Now we are ready to evaluate each \(U_i(x)\). \begin {align*} U_1 &= (-1)^{3-1} \int { \frac {F(x) W_1(x) }{a W(x)} \, dx}\\ &= (-1)^{2} \int { \frac { \left (\frac {x -1}{x^{3}}\right ) \left (1\right )}{\left (1\right ) \left ({\mathrm e}^{-x}\right )} \, dx} \\ &= \int { \frac {\frac {x -1}{x^{3}}}{{\mathrm e}^{-x}} \, dx}\\ &= \int {\left (\frac {\left (x -1\right ) {\mathrm e}^{x}}{x^{3}}\right ) \, dx} \\ &= -\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{2}+\frac {{\mathrm e}^{x}}{2 x^{2}} \end {align*}

\begin {align*} U_2 &= (-1)^{3-2} \int { \frac {F(x) W_2(x) }{a W(x)} \, dx}\\ &= (-1)^{1} \int { \frac { \left (\frac {x -1}{x^{3}}\right ) \left ({\mathrm e}^{-x} \left (x +1\right )\right )}{\left (1\right ) \left ({\mathrm e}^{-x}\right )} \, dx} \\ &= - \int { \frac {\frac {\left (x -1\right ) {\mathrm e}^{-x} \left (x +1\right )}{x^{3}}}{{\mathrm e}^{-x}} \, dx}\\ &= - \int {\left (\frac {x^{2}-1}{x^{3}}\right ) \, dx} \\ &= -\ln \left (x \right )-\frac {1}{2 x^{2}} \end {align*}

\begin {align*} U_3 &= (-1)^{3-3} \int { \frac {F(x) W_3(x) }{a W(x)} \, dx}\\ &= (-1)^{0} \int { \frac { \left (\frac {x -1}{x^{3}}\right ) \left ({\mathrm e}^{-x}\right )}{\left (1\right ) \left ({\mathrm e}^{-x}\right )} \, dx} \\ &= \int { \frac {\frac {\left (x -1\right ) {\mathrm e}^{-x}}{x^{3}}}{{\mathrm e}^{-x}} \, dx}\\ &= \int {\left (\frac {x -1}{x^{3}}\right ) \, dx}\\ &= \frac {1}{2 x^{2}}-\frac {1}{x} \end {align*}

Now that all the \(U_i\) functions have been determined, the particular solution is found from \[ y_p = U_1 y_1+U_2 y_2+U_3 y_3 \] Hence \begin {equation*} \begin {split} y_p &= \left (-\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{2}+\frac {{\mathrm e}^{x}}{2 x^{2}}\right ) \left ({\mathrm e}^{-x}\right ) \\ &+\left (-\ln \left (x \right )-\frac {1}{2 x^{2}}\right ) \\ &+\left (\frac {1}{2 x^{2}}-\frac {1}{x}\right ) \left (x\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = -\frac {{\mathrm e}^{-x} \operatorname {expIntegral}_{1}\left (-x \right )}{2}-\ln \left (x \right )-1 \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x\right ) + \left (-\frac {{\mathrm e}^{-x} \operatorname {expIntegral}_{1}\left (-x \right )}{2}-\ln \left (x \right )-1\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x -\frac {{\mathrm e}^{-x} \operatorname {expIntegral}_{1}\left (-x \right )}{2}-\ln \left (x \right )-1 \\ \end{align*}

Veriﬁcation of solutions

\[ y = c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} x -\frac {{\mathrm e}^{-x} \operatorname {expIntegral}_{1}\left (-x \right )}{2}-\ln \left (x \right )-1 \] Veriﬁed OK.

20.22.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime \prime }=\frac {x -1}{x^{3}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{3}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{3}^{\prime }\left (x \right )=-\frac {y_{3}\left (x \right ) x^{3}-x +1}{x^{3}} \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=y_{1}^{\prime }\left (x \right ), y_{3}\left (x \right )=y_{2}^{\prime }\left (x \right ), y_{3}^{\prime }\left (x \right )=-\frac {y_{3}\left (x \right ) x^{3}-x +1}{x^{3}}\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & -1 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & -1 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{-x}+c_{2} \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(_b(_a)*_a^3-_a+1)/_a^3, _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful`

\[ y \left (x \right ) = -\frac {\left (\int \int \frac {{\mathrm e}^{-x} \operatorname {expIntegral}_{1}\left (-x \right ) x^{2}-2 \,{\mathrm e}^{-x} c_{1} x^{2}+x -1}{x^{2}}d x d x \right )}{2}+c_{2} x +c_{3} \]

\[ y(x)\to \frac {e^{-x} \operatorname {ExpIntegralEi}(x)}{2}-\log (x)+c_1 e^{-x}+c_3 x+c_2 \]

20.22 problem 661

20.22.1 Maple step by step solution