Internal problem ID [15372]
Internal file name [OUTPUT/15373_Wednesday_May_08_2024_03_57_19_PM_36055126/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 15.3 Nonhomogeneous linear equations with
constant coefficients. Initial value problem. Exercises page 140
Problem number: 604.
ODE order: 3.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_linear_constant_coefficients_ODE"
Maple gives the following as the ode type
[[_3rd_order, _missing_y]]
\[ \boxed {y^{\prime \prime \prime }-y^{\prime }=-2 x} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1, y^{\prime \prime }\left (0\right ) = 2] \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 \[ y^{\prime \prime \prime }-y^{\prime } = 0 \] The characteristic equation is \[ \lambda ^{3}-\lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= 1\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} {\mathrm e}^{x} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= 1 \\ y_3 &= {\mathrm e}^{x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }-y^{\prime } = -2 x \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ x \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{1, x\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{1, {\mathrm e}^{x}, {\mathrm e}^{-x}\} \] Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x, x^{2}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{2} x^{2}+A_{1} x \] The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ -2 x A_{2}-A_{1} = -2 x \] Solving for the unknowns by comparing coefficients results in \[ [A_{1} = 0, A_{2} = 1] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = x^{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} {\mathrm e}^{x}\right ) + \left (x^{2}\right ) \\ \end{align*} Initial conditions are used to solve for the constants of integration.
Looking at the above solution \begin {align*} y = c_{1} {\mathrm e}^{-x}+c_{2} +c_{3} {\mathrm e}^{x}+x^{2} \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 \(x = 0\) in the above gives \begin {align*} 0 = c_{1} +c_{2} +c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} {\mathrm e}^{-x}+c_{3} {\mathrm e}^{x}+2 x \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 0\) in the above gives \begin {align*} 1 = -c_{1} +c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{-x}+c_{3} {\mathrm e}^{x}+2 \end {align*}
substituting \(y^{\prime \prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = c_{1} +c_{3} +2\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}&=-{\frac {1}{2}}\\ c_{2}&=0\\ c_{3}&={\frac {1}{2}} \end {align*}
Substituting these values back in above solution results in \begin {align*} y = -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2}+x^{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2}+x^{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2}+x^{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }-y^{\prime }=-2 x , y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2\right ] \\ \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 )=-2 x +y_{2}\left (x \right ) \\ & {} & \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 )=-2 x +y_{2}\left (x \right )\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 )+\left [\begin {array}{c} 0 \\ 0 \\ -2 x \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ -2 x \end {array}\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 )+{\moverset {\rightarrow }{f}} \\ \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 [1, \left [\begin {array}{c} 1 \\ 1 \\ 1 \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 [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}}_{3}={\mathrm e}^{x}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & {\mathrm e}^{x} \\ -{\mathrm e}^{-x} & 0 & {\mathrm e}^{x} \\ {\mathrm e}^{-x} & 0 & {\mathrm e}^{x} \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & 1 & {\mathrm e}^{x} \\ -{\mathrm e}^{-x} & 0 & {\mathrm e}^{x} \\ {\mathrm e}^{-x} & 0 & {\mathrm e}^{x} \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{ccc} 1 & 1 & 1 \\ -1 & 0 & 1 \\ 1 & 0 & 1 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} 1 & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {{\mathrm e}^{-x}}{2}-1+\frac {{\mathrm e}^{x}}{2} \\ 0 & \frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} \\ 0 & -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} & \frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (x \right )=\Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (x \right )=\frac {1}{\Phi \left (x \right )}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} x^{2}-{\mathrm e}^{-x}+2-{\mathrm e}^{x} \\ 2 x +{\mathrm e}^{-x}-{\mathrm e}^{x} \\ -{\mathrm e}^{-x}+2-{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+\left [\begin {array}{c} x^{2}-{\mathrm e}^{-x}+2-{\mathrm e}^{x} \\ 2 x +{\mathrm e}^{-x}-{\mathrm e}^{x} \\ -{\mathrm e}^{-x}+2-{\mathrm e}^{x} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\left (c_{1} -1\right ) {\mathrm e}^{-x}+\left (c_{3} -1\right ) {\mathrm e}^{x}+x^{2}+c_{2} +2 \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} +c_{2} +c_{3} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (c_{1} -1\right ) {\mathrm e}^{-x}+\left (c_{3} -1\right ) {\mathrm e}^{x}+2 x \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=-c_{1} +c_{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\left (c_{1} -1\right ) {\mathrm e}^{-x}+2+\left (c_{3} -1\right ) {\mathrm e}^{x} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=c_{1} +c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =\frac {1}{2}, c_{2} =-2, c_{3} =\frac {3}{2}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2}+x^{2} \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)-2*_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: 0.016 (sec). Leaf size: 18
dsolve([diff(y(x),x$3)-diff(y(x),x)=-2*x,y(0) = 0, D(y)(0) = 1, (D@@2)(y)(0) = 2],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {{\mathrm e}^{-x}}{2}+\frac {{\mathrm e}^{x}}{2}+x^{2} \]
✓ Solution by Mathematica
Time used: 0.024 (sec). Leaf size: 25
DSolve[{y'''[x]-y'[x]==-2*x,{y[0]==0,y'[0]==1,y''[0]==2}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^2-\frac {e^{-x}}{2}+\frac {e^x}{2} \]