Internal problem ID [285]
Internal file name [OUTPUT/285_Sunday_June_05_2022_01_38_15_AM_78402842/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 10.
ODE order: 4.
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]]
\[ \boxed {5 y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }=0} \] The characteristic equation is \[ 5 \lambda ^{4}+3 \lambda ^{3} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -{\frac {3}{5}}\\ \lambda _2 &= 0\\ \lambda _3 &= 0\\ \lambda _4 &= 0 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)=c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-\frac {3 x}{5}} c_{4} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= x\\ y_3 &= x^{2}\\ y_4 &= {\mathrm e}^{-\frac {3 x}{5}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-\frac {3 x}{5}} c_{4} \\ \end{align*}
Verification of solutions
\[ y = c_{3} x^{2}+c_{2} x +c_{1} +{\mathrm e}^{-\frac {3 x}{5}} c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 5 y^{\prime \prime \prime \prime }+3 y^{\prime \prime \prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=-\frac {3 y^{\prime \prime \prime }}{5} \\ \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 \prime }+\frac {3 y^{\prime \prime \prime }}{5}=0 \\ \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 {Define new variable}\hspace {3pt} y_{4}\left (x \right ) \\ {} & {} & y_{4}\left (x \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (x \right )=-\frac {3 y_{4}\left (x \right )}{5} \\ & {} & \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_{4}\left (x \right )=y_{3}^{\prime }\left (x \right ), y_{4}^{\prime }\left (x \right )=-\frac {3 y_{4}\left (x \right )}{5}\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 ) \\ y_{4}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (x \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -\frac {3}{5} \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -\frac {3}{5} \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 [-\frac {3}{5}, \left [\begin {array}{c} -\frac {125}{27} \\ \frac {25}{9} \\ -\frac {5}{3} \\ 1 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-\frac {3}{5}, \left [\begin {array}{c} -\frac {125}{27} \\ \frac {25}{9} \\ -\frac {5}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-\frac {3 x}{5}}\cdot \left [\begin {array}{c} -\frac {125}{27} \\ \frac {25}{9} \\ -\frac {5}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 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 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 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 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}=\left [\begin {array}{c} 0 \\ 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}+c_{4} {\moverset {\rightarrow }{y}}_{4} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-\frac {3 x}{5}}\cdot \left [\begin {array}{c} -\frac {125}{27} \\ \frac {25}{9} \\ -\frac {5}{3} \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{2} \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y=-\frac {125 c_{1} {\mathrm e}^{-\frac {3 x}{5}}}{27}+c_{2} \end {array} \]
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: 0.0 (sec). Leaf size: 20
dsolve(5*diff(y(x),x$4)+3*diff(y(x),x$3)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} +c_{2} x +c_{3} x^{2}+c_{4} {\mathrm e}^{-\frac {3 x}{5}} \]
✓ Solution by Mathematica
Time used: 0.037 (sec). Leaf size: 30
DSolve[5*y''''[x]+3*y'''[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\frac {125}{27} c_1 e^{-3 x/5}+x (c_4 x+c_3)+c_2 \]