Internal problem ID [14668]
Internal file name [OUTPUT/14348_Wednesday_April_03_2024_02_17_26_PM_55540013/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.5, page 175
Problem number: 50.
ODE order: 6.
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
[[_high_order, _missing_x]]
\[ \boxed {y^{\left (6\right )}-3 y^{\prime \prime \prime \prime }+3 y^{\prime \prime }-y=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 16, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime \prime }\left (0\right ) = 0, y^{\left (5\right )}\left (0\right ) = 0] \end {align*}
The characteristic equation is \[ \lambda ^{6}-3 \lambda ^{4}+3 \lambda ^{2}-1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= -1\\ \lambda _3 &= 1\\ \lambda _4 &= -1\\ \lambda _5 &= 1\\ \lambda _6 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(t)=c_{1} {\mathrm e}^{-t}+c_{2} t \,{\mathrm e}^{-t}+t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+t \,{\mathrm e}^{t} c_{5} +t^{2} {\mathrm e}^{t} c_{6} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= {\mathrm e}^{-t}\\ y_2 &= t \,{\mathrm e}^{-t}\\ y_3 &= t^{2} {\mathrm e}^{-t}\\ y_4 &= {\mathrm e}^{t}\\ y_5 &= t \,{\mathrm e}^{t}\\ y_6 &= {\mathrm e}^{t} t^{2} \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}^{-t}+c_{2} t \,{\mathrm e}^{-t}+t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+t \,{\mathrm e}^{t} c_{5} +t^{2} {\mathrm e}^{t} c_{6} \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 \(y = 16\) and \(t = 0\) in the above gives \begin {align*} 16 = c_{1} +c_{4}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} {\mathrm e}^{-t}+c_{2} {\mathrm e}^{-t}-c_{2} t \,{\mathrm e}^{-t}+2 t \,{\mathrm e}^{-t} c_{3} -t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+{\mathrm e}^{t} c_{5} +t \,{\mathrm e}^{t} c_{5} +2 t \,{\mathrm e}^{t} c_{6} +t^{2} {\mathrm e}^{t} c_{6} \end {align*}
substituting \(y^{\prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = -c_{1} +c_{2} +c_{4} +c_{5}\tag {2A} \end {align*}
Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = c_{1} {\mathrm e}^{-t}-2 c_{2} {\mathrm e}^{-t}+c_{2} t \,{\mathrm e}^{-t}+2 c_{3} {\mathrm e}^{-t}-4 t \,{\mathrm e}^{-t} c_{3} +t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+2 \,{\mathrm e}^{t} c_{5} +t \,{\mathrm e}^{t} c_{5} +2 \,{\mathrm e}^{t} c_{6} +4 t \,{\mathrm e}^{t} c_{6} +t^{2} {\mathrm e}^{t} c_{6} \end {align*}
substituting \(y^{\prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{1} -2 c_{2} +2 c_{3} +c_{4} +2 c_{5} +2 c_{6}\tag {3A} \end {align*}
Taking three derivatives of the solution gives \begin {align*} y^{\prime \prime \prime } = -c_{1} {\mathrm e}^{-t}+3 c_{2} {\mathrm e}^{-t}-c_{2} t \,{\mathrm e}^{-t}-6 c_{3} {\mathrm e}^{-t}+6 t \,{\mathrm e}^{-t} c_{3} -t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+3 \,{\mathrm e}^{t} c_{5} +t \,{\mathrm e}^{t} c_{5} +6 \,{\mathrm e}^{t} c_{6} +6 t \,{\mathrm e}^{t} c_{6} +t^{2} {\mathrm e}^{t} c_{6} \end {align*}
substituting \(y^{\prime \prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = -c_{1} +3 c_{2} -6 c_{3} +c_{4} +3 c_{5} +6 c_{6}\tag {4A} \end {align*}
Taking four derivatives of the solution gives \begin {align*} y^{\prime \prime \prime \prime } = c_{1} {\mathrm e}^{-t}-4 c_{2} {\mathrm e}^{-t}+c_{2} t \,{\mathrm e}^{-t}+12 c_{3} {\mathrm e}^{-t}-8 t \,{\mathrm e}^{-t} c_{3} +t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+4 \,{\mathrm e}^{t} c_{5} +t \,{\mathrm e}^{t} c_{5} +12 \,{\mathrm e}^{t} c_{6} +8 t \,{\mathrm e}^{t} c_{6} +t^{2} {\mathrm e}^{t} c_{6} \end {align*}
substituting \(y^{\prime \prime \prime \prime } = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = c_{1} -4 c_{2} +12 c_{3} +c_{4} +4 c_{5} +12 c_{6}\tag {5A} \end {align*}
Taking five derivatives of the solution gives \begin {align*} y^{\left (5\right )} = -c_{1} {\mathrm e}^{-t}+5 c_{2} {\mathrm e}^{-t}-c_{2} t \,{\mathrm e}^{-t}-20 c_{3} {\mathrm e}^{-t}+10 t \,{\mathrm e}^{-t} c_{3} -t^{2} {\mathrm e}^{-t} c_{3} +c_{4} {\mathrm e}^{t}+5 \,{\mathrm e}^{t} c_{5} +t \,{\mathrm e}^{t} c_{5} +20 \,{\mathrm e}^{t} c_{6} +10 t \,{\mathrm e}^{t} c_{6} +t^{2} {\mathrm e}^{t} c_{6} \end {align*}
substituting \(y^{\left (5\right )} = 0\) and \(t = 0\) in the above gives \begin {align*} 0 = -c_{1} +5 c_{2} -20 c_{3} +c_{4} +5 c_{5} +20 c_{6}\tag {6A} \end {align*}
Equations {1A,2A,3A,4A,5A,6A} are now solved for \(\{c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}\}\). Solving for the constants gives \begin {align*} c_{1}&=8\\ c_{2}&=5\\ c_{3}&=1\\ c_{4}&=8\\ c_{5}&=-5\\ c_{6}&=1 \end {align*}
Substituting these values back in above solution results in \begin {align*} y = t^{2} {\mathrm e}^{-t}+{\mathrm e}^{t} t^{2}+5 t \,{\mathrm e}^{-t}-5 t \,{\mathrm e}^{t}+8 \,{\mathrm e}^{-t}+8 \,{\mathrm e}^{t} \end {align*}
Which simplifies to \[ y = \left (t^{2}+5 t +8\right ) {\mathrm e}^{-t}+{\mathrm e}^{t} \left (t^{2}-5 t +8\right ) \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (t^{2}+5 t +8\right ) {\mathrm e}^{-t}+{\mathrm e}^{t} \left (t^{2}-5 t +8\right ) \\ \end{align*}
Verification of solutions
\[ y = \left (t^{2}+5 t +8\right ) {\mathrm e}^{-t}+{\mathrm e}^{t} \left (t^{2}-5 t +8\right ) \] Verified OK.
Maple trace
`Methods for high order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 29
dsolve([diff(y(t),t$6)-3*diff(y(t),t$4)+3*diff(y(t),t$2)-y(t)=0,y(0) = 16, D(y)(0) = 0, (D@@2)(y)(0) = 0, (D@@3)(y)(0) = 0, (D@@4)(y)(0) = 0, (D@@5)(y)(0) = 0],y(t), singsol=all)
\[ y \left (t \right ) = \left (t^{2}+5 t +8\right ) {\mathrm e}^{-t}+{\mathrm e}^{t} \left (t^{2}-5 t +8\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 33
DSolve[{y''''''[t]-3*y''''[t]+3*y''[t]-y[t]==0,{y[0]==16,y'[0]==0,y''[0]==0,y'''[0]==0,y''''[0]==0,y'''''[0]==0}},y[t],t,IncludeSingularSolutions -> True]
\[ y(t)\to e^{-t} \left (t^2+e^{2 t} \left (t^2-5 t+8\right )+5 t+8\right ) \]