problem 13.6 (e)

Internal problem ID [13510]
Internal ﬁle name [OUTPUT/12682_Friday_February_16_2024_12_10_52_AM_74301330/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 13. Higher order equations: Extending ﬁrst order concepts. Additional exercises page 259
Problem number: 13.6 (e).
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_linear_constant_coeﬃcients_ODE"

\[ \boxed {y^{\prime \prime \prime }-y^{\prime \prime }=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 10, y^{\prime }\left (0\right ) = 5, y^{\prime \prime }\left (0\right ) = 2] \end {align*}

The characteristic equation is \[ \lambda ^{3}-\lambda ^{2} = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= 1\\ \lambda _2 &= 0\\ \lambda _3 &= 0 \end {align*}

Therefore the homogeneous solution is \[ y_h(x)=c_{2} x +c_{1} +{\mathrm e}^{x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= 1\\ y_2 &= x\\ y_3 &= {\mathrm e}^{x} \end {align*}

Looking at the above solution \begin {align*} y = c_{2} x +c_{1} +{\mathrm e}^{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 \(y = 10\) and \(x = 0\) in the above gives \begin {align*} 10 = c_{1} +c_{3}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = c_{2} +{\mathrm e}^{x} c_{3} \end {align*}

substituting \(y^{\prime } = 5\) and \(x = 0\) in the above gives \begin {align*} 5 = c_{2} +c_{3}\tag {2A} \end {align*}

Taking two derivatives of the solution gives \begin {align*} y^{\prime \prime } = {\mathrm e}^{x} c_{3} \end {align*}

substituting \(y^{\prime \prime } = 2\) and \(x = 0\) in the above gives \begin {align*} 2 = 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}&=8\\ c_{2}&=3\\ c_{3}&=2 \end {align*}

Substituting these values back in above solution results in \begin {align*} y = 8+2 \,{\mathrm e}^{x}+3 x \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= 8+2 \,{\mathrm e}^{x}+3 x \\ \end{align*}

Veriﬁcation of solutions

\[ y = 8+2 \,{\mathrm e}^{x}+3 x \] Veriﬁed OK.

8.39.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y^{\prime \prime }-\frac {d}{d x}y^{\prime }=0, y \left (0\right )=10, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5, \left (\frac {d}{d x}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\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 )=\frac {d}{d x}y^{\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 )=y_{3}\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 )=y_{3}\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 & 0 & 1 \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 & 0 & 1 \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 [0, \left [\begin {array}{c} 0 \\ 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 [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 eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 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} 0 \\ 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 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} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}={\mathrm e}^{x} c_{3} \cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{x} c_{3} +c_{1} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (0\right )=10 \\ {} & {} & 10=c_{1} +c_{3} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{x} c_{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=5 \\ {} & {} & 5=c_{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }={\mathrm e}^{x} c_{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d x}y^{\prime }\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=2 \\ {} & {} & 2=c_{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ \bullet & {} & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \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`

8.39 problem 13.6 (e)

8.39.1 Maple step by step solution