problem Problem 24.37

Internal problem ID [5213]
Internal ﬁle name [OUTPUT/4706_Sunday_June_05_2022_03_03_45_PM_1165393/index.tex]

Book: Schaums Outline Diﬀerential Equations, 4th edition. Bronson and Costa. McGraw Hill 2014
Section: Chapter 24. Solutions of linear DE by Laplace transforms. Supplementary Problems. page 248
Problem number: Problem 24.37.
ODE order: 3.
ODE degree: 1.

\[ \boxed {y^{\prime \prime \prime }-3 y^{\prime \prime }+3 y^{\prime }-y=x^{2} {\mathrm e}^{x}} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 2, y^{\prime \prime }\left (0\right ) = 3] \end {align*}

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right ) \end {align*}

The given ode becomes an algebraic equation in the Laplace domain \[ s^{3} Y \left (s \right )-y^{\prime \prime }\left (0\right )-s y^{\prime }\left (0\right )-s^{2} y \left (0\right )-3 s^{2} Y \left (s \right )+3 y^{\prime }\left (0\right )+3 s y \left (0\right )+3 s Y \left (s \right )-3 y \left (0\right )-Y \left (s \right ) = \frac {2}{\left (s -1\right )^{3}}\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=1\\ y^{\prime }\left (0\right )&=2\\ y^{\prime \prime }\left (0\right )&=3 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \[ s^{3} Y \left (s \right )+s -s^{2}-3 s^{2} Y \left (s \right )+3 s Y \left (s \right )-Y \left (s \right ) = \frac {2}{\left (s -1\right )^{3}} \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {s^{5}-4 s^{4}+6 s^{3}-4 s^{2}+s +2}{\left (s -1\right )^{3} \left (s^{3}-3 s^{2}+3 s -1\right )} \] Applying partial fractions decomposition results in \[ Y(s)= \frac {2}{\left (s -1\right )^{6}}+\frac {1}{\left (s -1\right )^{2}}+\frac {1}{s -1} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {2}{\left (s -1\right )^{6}}\right ) &= \frac {x^{5} {\mathrm e}^{x}}{60}\\ \mathcal {L}^{-1}\left (\frac {1}{\left (s -1\right )^{2}}\right ) &= x \,{\mathrm e}^{x}\\ \mathcal {L}^{-1}\left (\frac {1}{s -1}\right ) &= {\mathrm e}^{x} \end {align*}

Adding the above results and simplifying gives \[ y=\frac {{\mathrm e}^{x} \left (x^{5}+60 x +60\right )}{60} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{x} \left (x^{5}+60 x +60\right )}{60} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {{\mathrm e}^{x} \left (x^{5}+60 x +60\right )}{60} \] Veriﬁed OK.

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

\[ y \left (x \right ) = \frac {{\mathrm e}^{x} \left (x^{5}+60 x +60\right )}{60} \]

5.14 problem Problem 24.37