Internal problem ID [298]
Internal file name [OUTPUT/298_Sunday_June_05_2022_01_38_23_AM_15154783/index.tex]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 5.3, Higher-Order Linear Differential Equations. Homogeneous Equations with
Constant Coefficients. Page 300
Problem number: problem 26.
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_x]]
\[ \boxed {y^{\prime \prime \prime }+10 y^{\prime \prime }+25 y^{\prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 3, y^{\prime }\left (0\right ) = 4, y^{\prime \prime }\left (0\right ) = 5] \end {align*}
The characteristic equation is \[ \lambda ^{3}+10 \lambda ^{2}+25 \lambda = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 0\\ \lambda _2 &= -5\\ \lambda _3 &= -5 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{1} +{\mathrm e}^{-5 x} c_{2} +x \,{\mathrm e}^{-5 x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= {\mathrm e}^{-5 x}\\ y_3 &= x \,{\mathrm e}^{-5 x} \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}^{-5 x} c_{2} +x \,{\mathrm e}^{-5 x} c_{3} \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 \(y = 3\) and \(x = 0\) in the above gives \begin {align*} 3 = c_{1} +c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -5 \,{\mathrm e}^{-5 x} c_{2} +{\mathrm e}^{-5 x} c_{3} -5 x \,{\mathrm e}^{-5 x} c_{3} \end {align*}
substituting \(y^{\prime } = 4\) and \(x = 0\) in the above gives \begin {align*} 4 = -5 c_{2} +c_{3}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = 25 \,{\mathrm e}^{-5 x} c_{2} -10 \,{\mathrm e}^{-5 x} c_{3} +25 x \,{\mathrm e}^{-5 x} c_{3} \end {align*}
substituting \(y^{\prime \prime } = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = 25 c_{2} -10 c_{3}\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 {24}{5}}\\ c_{2}&=-{\frac {9}{5}}\\ c_{3}&=-5 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = \frac {24}{5}-\frac {9 \,{\mathrm e}^{-5 x}}{5}-5 x \,{\mathrm e}^{-5 x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {24}{5}-\frac {9 \,{\mathrm e}^{-5 x}}{5}-5 x \,{\mathrm e}^{-5 x} \\ \end{align*}
Verification of solutions
\[ y = \frac {24}{5}-\frac {9 \,{\mathrm e}^{-5 x}}{5}-5 x \,{\mathrm e}^{-5 x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime }+10 y^{\prime \prime }+25 y^{\prime }=0, y \left (0\right )=3, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=4, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5\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 )=-10 y_{3}\left (x \right )-25 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 )=-10 y_{3}\left (x \right )-25 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 & -25 & -10 \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 & -25 & -10 \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 [-5, \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]\right ], \left [-5, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [-5, \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} -5 \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}\left (x \right )={\mathrm e}^{-5 x}\cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =-5\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =-5 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{2}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} -5 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -25 & -10 \end {array}\right ]-\left (-5\right )\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{125} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} -5 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{-5 x}\cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{125} \\ 0 \\ 0 \end {array}\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}}_{3}=\left [\begin {array}{c} 1 \\ 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}\left (x \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+c_{3} {\moverset {\rightarrow }{y}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-5 x}\cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]+{\mathrm e}^{-5 x} c_{2} \cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]+\left [\begin {array}{c} \frac {1}{125} \\ 0 \\ 0 \end {array}\right ]\right )+\left [\begin {array}{c} c_{3} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=\frac {\left (\left (5 x +1\right ) c_{2} +5 c_{1} \right ) {\mathrm e}^{-5 x}}{125}+c_{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=3 \\ {} & {} & 3=\frac {c_{2}}{125}+\frac {c_{1}}{25}+c_{3} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {{\mathrm e}^{-5 x} c_{2}}{25}-\frac {\left (\left (5 x +1\right ) c_{2} +5 c_{1} \right ) {\mathrm e}^{-5 x}}{25} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=4 \\ {} & {} & 4=-\frac {c_{1}}{5} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {2 \,{\mathrm e}^{-5 x} c_{2}}{5}+\frac {\left (\left (5 x +1\right ) c_{2} +5 c_{1} \right ) {\mathrm e}^{-5 x}}{5} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=-\frac {c_{2}}{5}+c_{1} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{c_{1} =-20, c_{2} =-125, c_{3} =\frac {24}{5}\right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=\frac {24}{5}-\frac {9 \,{\mathrm e}^{-5 x}}{5}-5 x \,{\mathrm e}^{-5 x} \end {array} \]
Maple trace
`Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 19
dsolve([diff(y(x),x$3)+10*diff(y(x),x$2)+25*diff(y(x),x)=0,y(0) = 3, D(y)(0) = 4, (D@@2)(y)(0) = 5],y(x), singsol=all)
\[ y \left (x \right ) = \frac {24}{5}-\frac {9 \,{\mathrm e}^{-5 x}}{5}-5 \,{\mathrm e}^{-5 x} x \]
✓ Solution by Mathematica
Time used: 0.045 (sec). Leaf size: 26
DSolve[{y'''[x]+10*y''[x]+25*y'[x]==0,{y[0]==3,y'[0]==4,y''[0]==5}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{5} e^{-5 x} \left (-25 x+24 e^{5 x}-9\right ) \]