Internal problem ID [11254]
Internal file name [OUTPUT/10240_Sunday_December_11_2022_01_27_06_AM_79243032/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter VII, Linear differential equations with constant coefficients. Article 47.
Particular integral. Page 100
Problem number: Ex 3.
ODE order: 3.
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
[[_3rd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime \prime }+3 y^{\prime \prime }+3 y^{\prime }+y=2 \,{\mathrm e}^{-x}-x^{2} {\mathrm e}^{-x}} \] This is higher order nonhomogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to \[ y^{\prime \prime \prime }+3 y^{\prime \prime }+3 y^{\prime }+y = 0 \] The characteristic equation is \[ \lambda ^{3}+3 \lambda ^{2}+3 \lambda +1 = 0 \] The roots of the above equation are \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -1\\ \lambda _3 &= -1 \end {align*}
Therefore the homogeneous solution is \[ y_h(x)={\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +x^{2} {\mathrm e}^{-x} c_{3} \] The fundamental set of solutions for the homogeneous solution are the following \begin{align*} y_1 &= {\mathrm e}^{-x} \\ y_2 &= x \,{\mathrm e}^{-x} \\ y_3 &= x^{2} {\mathrm e}^{-x} \\ \end{align*} Now the particular solution to the given ODE is found \[ y^{\prime \prime \prime }+3 y^{\prime \prime }+3 y^{\prime }+y = 2 \,{\mathrm e}^{-x}-x^{2} {\mathrm e}^{-x} \] The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is \[ 2 \,{\mathrm e}^{-x}-x^{2} {\mathrm e}^{-x} \] Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is \[ [\{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x}\}] \] While the set of the basis functions for the homogeneous solution found earlier is \[ \{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}, {\mathrm e}^{-x}\} \] Since \({\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x \,{\mathrm e}^{-x}, x^{2} {\mathrm e}^{-x}, x^{3} {\mathrm e}^{-x}\}] \] Since \(x \,{\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{2} {\mathrm e}^{-x}, x^{3} {\mathrm e}^{-x}, x^{4} {\mathrm e}^{-x}\}] \] Since \(x^{2} {\mathrm e}^{-x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes \[ [\{x^{3} {\mathrm e}^{-x}, x^{4} {\mathrm e}^{-x}, x^{5} {\mathrm e}^{-x}\}] \] Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set. \[ y_p = A_{1} x^{3} {\mathrm e}^{-x}+A_{2} x^{4} {\mathrm e}^{-x}+A_{3} x^{5} {\mathrm e}^{-x} \] The unknowns \(\{A_{1}, A_{2}, A_{3}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives \[ 24 A_{2} x \,{\mathrm e}^{-x}+60 A_{3} x^{2} {\mathrm e}^{-x}+6 A_{1} {\mathrm e}^{-x} = 2 \,{\mathrm e}^{-x}-x^{2} {\mathrm e}^{-x} \] Solving for the unknowns by comparing coefficients results in \[ \left [A_{1} = {\frac {1}{3}}, A_{2} = 0, A_{3} = -{\frac {1}{60}}\right ] \] Substituting the above back in the above trial solution \(y_p\), gives the particular solution \[ y_p = \frac {x^{3} {\mathrm e}^{-x}}{3}-\frac {x^{5} {\mathrm e}^{-x}}{60} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_{1} +x \,{\mathrm e}^{-x} c_{2} +x^{2} {\mathrm e}^{-x} c_{3}\right ) + \left (\frac {x^{3} {\mathrm e}^{-x}}{3}-\frac {x^{5} {\mathrm e}^{-x}}{60}\right ) \\ \end{align*} Which simplifies to \[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{3} {\mathrm e}^{-x}}{3}-\frac {x^{5} {\mathrm e}^{-x}}{60} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{3} {\mathrm e}^{-x}}{3}-\frac {x^{5} {\mathrm e}^{-x}}{60} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{-x} \left (c_{3} x^{2}+c_{2} x +c_{1} \right )+\frac {x^{3} {\mathrm e}^{-x}}{3}-\frac {x^{5} {\mathrm e}^{-x}}{60} \] 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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 32
dsolve(diff(y(x),x$3)+3*diff(y(x),x$2)+3*diff(y(x),x)+y(x)=2*exp(-x)-x^2*exp(-x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {{\mathrm e}^{-x} \left (x^{5}-60 c_{2} x^{2}-20 x^{3}-60 c_{3} x -60 c_{1} \right )}{60} \]
✓ Solution by Mathematica
Time used: 0.017 (sec). Leaf size: 41
DSolve[y'''[x]+3*y''[x]+3*y'[x]+y[x]==2*Exp[-x]-x^2*Exp[-x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{60} e^{-x} \left (-x^5+20 x^3+60 c_3 x^2+60 c_2 x+60 c_1\right ) \]