Internal problem ID [12744]
Internal file name [OUTPUT/11397_Friday_November_03_2023_06_32_22_AM_11601804/index.tex]
Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 4. N-th Order Linear Differential Equations. Exercises 4.1, page 186
Problem number: 2.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_missing_y"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {x y^{\prime \prime \prime }+y^{\prime } x=4} \] With initial conditions \begin {align*} [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = 1, y^{\prime \prime }\left (1\right ) = -1] \end {align*}
Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} v^{\prime \prime }\left (x \right ) x +v \left (x \right ) x = 0 \end {align*}
This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A v''(x) + B v'(x) + C v(x) = 0 \] Where in the above \(A=1, B=0, C=1\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda x}+{\mathrm e}^{\lambda x} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) 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 i \end {align*}
Hence \begin {align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end {align*}
Which simplifies to \begin{align*} \lambda _1 &= i \\ \lambda _2 &= -i \\ \end{align*} Since roots are complex conjugate of each others, then let the roots be \[ \lambda _{1,2} = \alpha \pm i \beta \] Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution, when using Euler relation, can be written as \[ v \left (x \right ) = e^{\alpha x} \left ( c_{1} \cos (\beta x) + c_{2} \sin (\beta x) \right ) \] Which becomes \[ v \left (x \right ) = e^{0}\left (\cos \left (x \right ) c_{1}+c_{2} \sin \left (x \right )\right ) \] Or \[ v \left (x \right ) = \cos \left (x \right ) c_{1}+c_{2} \sin \left (x \right ) \] But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \cos \left (x \right ) c_{1} +c_{2} \sin \left (x \right )\). Integrating both sides gives \begin {align*} y &= \int { \cos \left (x \right ) c_{1} +c_{2} \sin \left (x \right )\,\mathop {\mathrm {d}x}}\\ &= -\cos \left (x \right ) c_{2} +c_{1} \sin \left (x \right )+c_{3} \end {align*}
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 \[ x y^{\prime \prime \prime }+y^{\prime } x = 0 \] 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} 1 & \cos \left (x \right ) & \sin \left (x \right ) \\ 0 & -\sin \left (x \right ) & \cos \left (x \right ) \\ 0 & -\cos \left (x \right ) & -\sin \left (x \right ) \end {array}\right ] \\ |W| &= \sin \left (x \right )^{2}+\cos \left (x \right )^{2} \end {align*}
The determinant simplifies to \begin {align*} |W| &= 1 \end {align*}
Now we determine \(W_i\) for each \(U_i\). \begin {align*} W_1(x) &= \det \,\left [\begin {array}{cc} \cos \left (x \right ) & \sin \left (x \right ) \\ -\sin \left (x \right ) & \cos \left (x \right ) \end {array}\right ] \\ &= 1 \end {align*}
\begin {align*} W_2(x) &= \det \,\left [\begin {array}{cc} 1 & \sin \left (x \right ) \\ 0 & \cos \left (x \right ) \end {array}\right ] \\ &= \cos \left (x \right ) \end {align*}
\begin {align*} W_3(x) &= \det \,\left [\begin {array}{cc} 1 & \cos \left (x \right ) \\ 0 & -\sin \left (x \right ) \end {array}\right ] \\ &= -\sin \left (x \right ) \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 (4\right ) \left (1\right )}{\left (x\right ) \left (1\right )} \, dx} \\ &= \int { \frac {4}{x} \, dx}\\ &= \int {\left (\frac {4}{x}\right ) \, dx}\\ &= 4 \ln \left (x \right ) \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 (4\right ) \left (\cos \left (x \right )\right )}{\left (x\right ) \left (1\right )} \, dx} \\ &= - \int { \frac {4 \cos \left (x \right )}{x} \, dx}\\ &= - \int {\left (\frac {4 \cos \left (x \right )}{x}\right ) \, dx}\\ &= -4 \,\operatorname {Ci}\left (x \right ) \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 (4\right ) \left (-\sin \left (x \right )\right )}{\left (x\right ) \left (1\right )} \, dx} \\ &= \int { \frac {-4 \sin \left (x \right )}{x} \, dx}\\ &= \int {\left (-\frac {4 \sin \left (x \right )}{x}\right ) \, dx}\\ &= -4 \,\operatorname {Si}\left (x \right ) \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 (4 \ln \left (x \right )\right ) \\ &+\left (-4 \,\operatorname {Ci}\left (x \right )\right ) \left (\cos \left (x \right )\right ) \\ &+\left (-4 \,\operatorname {Si}\left (x \right )\right ) \left (\sin \left (x \right )\right ) \end {split} \end {equation*} Therefore the particular solution is \[ y_p = 4 \ln \left (x \right )-4 \,\operatorname {Ci}\left (x \right ) \cos \left (x \right )-4 \,\operatorname {Si}\left (x \right ) \sin \left (x \right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (y &= -\cos \left (x \right ) c_{2} +c_{1} \sin \left (x \right )+c_{3}\right ) + \left (4 \ln \left (x \right )-4 \,\operatorname {Ci}\left (x \right ) \cos \left (x \right )-4 \,\operatorname {Si}\left (x \right ) \sin \left (x \right )\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = -\cos \left (x \right ) c_{2} +c_{1} \sin \left (x \right )+c_{3} +4 \ln \left (x \right )-4 \,\operatorname {Ci}\left (x \right ) \cos \left (x \right )-4 \,\operatorname {Si}\left (x \right ) \sin \left (x \right ) \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = \left (-c_{2} -4 \,\operatorname {Ci}\left (1\right )\right ) \cos \left (1\right )+\left (c_{1} -4 \,\operatorname {Si}\left (1\right )\right ) \sin \left (1\right )+c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} \sin \left (x \right )+\cos \left (x \right ) c_{1} +\frac {4}{x}-\frac {4 \cos \left (x \right )^{2}}{x}+4 \,\operatorname {Ci}\left (x \right ) \sin \left (x \right )-\frac {4 \sin \left (x \right )^{2}}{x}-4 \,\operatorname {Si}\left (x \right ) \cos \left (x \right ) \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = \left (c_{1} -4 \,\operatorname {Si}\left (1\right )\right ) \cos \left (1\right )+\sin \left (1\right ) \left (c_{2} +4 \,\operatorname {Ci}\left (1\right )\right )\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = \cos \left (x \right ) c_{2} -c_{1} \sin \left (x \right )-\frac {4}{x^{2}}+\frac {4 \cos \left (x \right )^{2}}{x^{2}}+4 \,\operatorname {Ci}\left (x \right ) \cos \left (x \right )+\frac {4 \sin \left (x \right )^{2}}{x^{2}}+4 \,\operatorname {Si}\left (x \right ) \sin \left (x \right ) \end {align*}
substituting \(y^{\prime \prime } = -1\) and \(x = 1\) in the above gives \begin {align*} -1 = \left (c_{2} +4 \,\operatorname {Ci}\left (1\right )\right ) \cos \left (1\right )-\left (c_{1} -4 \,\operatorname {Si}\left (1\right )\right ) \sin \left (1\right )\tag {3A} \end {align*}
Equations {1A,2A,3A} are now solved for \(\{c_{1}, c_{2}, c_{3}\}\). Solving for the constants gives \begin {align*} c_{1}&=\cos \left (1\right )+\sin \left (1\right )+4 \,\operatorname {Si}\left (1\right )\\ c_{2}&=-\cos \left (1\right )+\sin \left (1\right )-4 \,\operatorname {Ci}\left (1\right )\\ c_{3}&=-1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -1+4 \cos \left (x \right ) \operatorname {Ci}\left (1\right )+4 \sin \left (x \right ) \operatorname {Si}\left (1\right )+4 \ln \left (x \right )-4 \,\operatorname {Ci}\left (x \right ) \cos \left (x \right )-4 \,\operatorname {Si}\left (x \right ) \sin \left (x \right )+\cos \left (x \right ) \cos \left (1\right )-\cos \left (x \right ) \sin \left (1\right )+\sin \left (x \right ) \cos \left (1\right )+\sin \left (x \right ) \sin \left (1\right ) \end {align*}
Which simplifies to \[ y = \left (4 \,\operatorname {Ci}\left (1\right )-4 \,\operatorname {Ci}\left (x \right )+\cos \left (1\right )-\sin \left (1\right )\right ) \cos \left (x \right )+\left (4 \,\operatorname {Si}\left (1\right )-4 \,\operatorname {Si}\left (x \right )+\cos \left (1\right )+\sin \left (1\right )\right ) \sin \left (x \right )+4 \ln \left (x \right )-1 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (4 \,\operatorname {Ci}\left (1\right )-4 \,\operatorname {Ci}\left (x \right )+\cos \left (1\right )-\sin \left (1\right )\right ) \cos \left (x \right )+\left (4 \,\operatorname {Si}\left (1\right )-4 \,\operatorname {Si}\left (x \right )+\cos \left (1\right )+\sin \left (1\right )\right ) \sin \left (x \right )+4 \ln \left (x \right )-1 \\ \end{align*}
Verification of solutions
\[ y = \left (4 \,\operatorname {Ci}\left (1\right )-4 \,\operatorname {Ci}\left (x \right )+\cos \left (1\right )-\sin \left (1\right )\right ) \cos \left (x \right )+\left (4 \,\operatorname {Si}\left (1\right )-4 \,\operatorname {Si}\left (x \right )+\cos \left (1\right )+\sin \left (1\right )\right ) \sin \left (x \right )+4 \ln \left (x \right )-1 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x y^{\prime \prime \prime }+y^{\prime } x =4, y \left (1\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-\frac {y^{\prime } x -4}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }+y^{\prime }=\frac {4}{x} \\ \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_{2}\left (x \right ) x -4}{x} \\ & {} & \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_{2}\left (x \right ) x -4}{x}\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 & -1 & 0 \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 & -1 & 0 \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ], \left [\mathrm {-I}, \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ], \left [\mathrm {I}, \left [\begin {array}{c} -1 \\ \mathrm {-I} \\ 1 \end {array}\right ]\right ]\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}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [\mathrm {-I}, \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\mathrm {-I} x}\cdot \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right )\cdot \left [\begin {array}{c} -1 \\ \mathrm {I} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\cos \left (x \right )+\mathrm {I} \sin \left (x \right ) \\ \mathrm {I} \left (-\mathrm {I} \sin \left (x \right )+\cos \left (x \right )\right ) \\ -\mathrm {I} \sin \left (x \right )+\cos \left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (x \right )=\left [\begin {array}{c} -\cos \left (x \right ) \\ \sin \left (x \right ) \\ \cos \left (x \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (x \right )=\left [\begin {array}{c} \sin \left (x \right ) \\ \cos \left (x \right ) \\ -\sin \left (x \right ) \end {array}\right ]\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}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (x \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} c_{3} \sin \left (x \right )-\cos \left (x \right ) c_{2} +c_{1} \\ c_{3} \cos \left (x \right )+c_{2} \sin \left (x \right ) \\ -c_{3} \sin \left (x \right )+\cos \left (x \right ) c_{2} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=c_{3} \sin \left (x \right )-\cos \left (x \right ) c_{2} +c_{1} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=c_{3} \sin \left (1\right )-\cos \left (1\right ) c_{2} +c_{1} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=c_{3} \cos \left (x \right )+c_{2} \sin \left (x \right ) \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & 1=c_{3} \cos \left (1\right )+c_{2} \sin \left (1\right ) \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-c_{3} \sin \left (x \right )+\cos \left (x \right ) c_{2} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=-1 \\ {} & {} & -1=-c_{3} \sin \left (1\right )+\cos \left (1\right ) c_{2} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-1, c_{2} =-\frac {-\sin \left (1\right )+\cos \left (1\right )}{\cos \left (1\right )^{2}+\sin \left (1\right )^{2}}, c_{3} =\frac {\cos \left (1\right )+\sin \left (1\right )}{\cos \left (1\right )^{2}+\sin \left (1\right )^{2}}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-1+\left (-\sin \left (1\right )+\cos \left (1\right )\right ) \cos \left (x \right )+\left (\cos \left (1\right )+\sin \left (1\right )\right ) \sin \left (x \right ) \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 3; linear nonhomogeneous with symmetry [0,1] -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a) = -(_b(_a)*_a-4)/_a, _b(_a)` *** Sublevel 2 *** 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)] -> Try solving first the homogeneous part of the ODE checking if the LODE has constant coefficients <- constant coefficients successful <- solving first the homogeneous part of the ODE successful <- differential order: 3; linear nonhomogeneous with symmetry [0,1] successful`
✓ Solution by Maple
Time used: 0.171 (sec). Leaf size: 49
dsolve([x*diff(y(x),x$3)+x*diff(y(x),x)=4,y(1) = 0, D(y)(1) = 1, (D@@2)(y)(1) = -1],y(x), singsol=all)
\[ y \left (x \right ) = \left (4 \,\operatorname {Ci}\left (1\right )-4 \,\operatorname {Ci}\left (x \right )+\cos \left (1\right )-\sin \left (1\right )\right ) \cos \left (x \right )+\left (4 \,\operatorname {Si}\left (1\right )-4 \,\operatorname {Si}\left (x \right )+\cos \left (1\right )+\sin \left (1\right )\right ) \sin \left (x \right )+4 \ln \left (x \right )-1 \]
✓ Solution by Mathematica
Time used: 0.184 (sec). Leaf size: 85
DSolve[{x*y'''[x]+x*y'[x]==4,{y[1]==0,y'[1]==1,y''[1]==-1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -4 \operatorname {CosIntegral}(x) \cos (x)+4 \operatorname {CosIntegral}(1) \cos (x)-2 \text {sinc}(1) \cos (2-x)-6 \text {sinc}(1) \cos (x)+8 \text {sinc}(1) \cos (1)-4 \text {Si}(x) \sin (x)+4 \text {Si}(1) \sin (x)+4 \log (x)+\sin (1-x)+\sin (3-x)+3 \sin (x+1)+\cos (1-x)-1-4 \sin (2) \]