problem 27

Internal problem ID [14459]
Internal ﬁle name [OUTPUT/14140_Monday_March_25_2024_09_50_24_PM_76676701/index.tex]

Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.1, page 141
Problem number: 27.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "reduction_of_order", "second_order_linear_constant_coeﬀ", "linear_second_order_ode_solved_by_an_integrating_factor"

\[ \boxed {y^{\prime \prime }-4 y^{\prime }+4 y=0} \] Given that one solution of the ode is \begin {align*} y_1 &= {\mathrm e}^{2 t} \end {align*}

With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 1] \end {align*}

9.16.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(t)y^{\prime } + q(t) y &= F \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-4 y^{\prime }+4 y = 0 \end {align*}

Given one basis solution \(y_{1}\left (t \right )\), then the second basis solution is given by \[ y_{2}\left (t \right ) = y_{1} \left (\int \frac {{\mathrm e}^{-\left (\int p d t \right )}}{y_{1}^{2}}d t \right ) \] Where \(p(x)\) is the coeﬃcient of \(y^{\prime }\) when the ode is written in the normal form \[ y^{\prime \prime }+p \left (t \right ) y^{\prime }+q \left (t \right ) y = f \left (t \right ) \] Looking at the ode to solve shows that \[ p \left (t \right ) = -4 \] Therefore \begin{align*} y_{2}\left (t \right ) &= {\mathrm e}^{2 t} \left (\int {\mathrm e}^{-\left (\int \left (-4\right )d t \right )} {\mathrm e}^{-4 t}d t \right ) \\ y_{2}\left (t \right ) &= {\mathrm e}^{2 t} \int \frac {{\mathrm e}^{4 t}}{{\mathrm e}^{4 t}} , dt \\ y_{2}\left (t \right ) &= {\mathrm e}^{2 t} \left (\int 1d t \right ) \\ y_{2}\left (t \right ) &= t \,{\mathrm e}^{2 t} \\ \end{align*} Hence the solution is \begin{align*} y &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{2 t} t \\ \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}^{2 t}+c_{2} {\mathrm e}^{2 t} t \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 \(t = 0\) in the above gives \begin {align*} 0 = c_{1}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = 2 c_{1} {\mathrm e}^{2 t}+2 c_{2} {\mathrm e}^{2 t} t +c_{2} {\mathrm e}^{2 t} \end {align*}

substituting \(y^{\prime } = 1\) and \(t = 0\) in the above gives \begin {align*} 1 = 2 c_{1} +c_{2}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=0\\ c_{2}&=1 \end {align*}

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

Summary

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

Veriﬁcation of solutions

\[ y = t \,{\mathrm e}^{2 t} \] Veriﬁed OK.

9.16.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-4 y^{\prime }+4 y=0, y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}-4 r +4=0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & \left (r -2\right )^{2}=0 \\ \bullet & {} & \textrm {Root of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =2 \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )={\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \,{\mathrm e}^{2 t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y=c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y=c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{2 t} t \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=c_{1} {\mathrm e}^{2 t}+c_{2} {\mathrm e}^{2 t} t \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0=c_{1} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=2 c_{1} {\mathrm e}^{2 t}+2 c_{2} {\mathrm e}^{2 t} t +c_{2} {\mathrm e}^{2 t} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=1 \\ {} & {} & 1=2 c_{1} +c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=t \,{\mathrm e}^{2 t} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=t \,{\mathrm e}^{2 t} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

9.16 problem 27

9.16.1 Existence and uniqueness analysis

9.16.2 Maple step by step solution