2.42 problem 42

Internal problem ID [8131]
Book : Second order enumerated odes
Section : section 2
Problem number : 42
Date solved : Monday, October 21, 2024 at 04:54:10 PM
CAS classification : [[_high_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3 y^{\prime \prime }+5 y^{\prime }-2 y&=x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{-2 x} \end{align*}

2.42.1 Solved as higher order constant coeff ode

Time used: 0.167 (sec)

The characteristic equation is

\[ \lambda ^{4}-\lambda ^{3}-3 \lambda ^{2}+5 \lambda -2 = 0 \]

The roots of the above equation are

\begin{align*} \lambda _1 &= -2\\ \lambda _2 &= 1\\ \lambda _3 &= 1\\ \lambda _4 &= 1 \end{align*}

Therefore the homogeneous solution is

\[ y_h(x)={\mathrm e}^{-2 x} c_1 +{\mathrm e}^{x} c_2 +x \,{\mathrm e}^{x} c_3 +x^{2} {\mathrm e}^{x} c_4 \]

The fundamental set of solutions for the homogeneous solution are the following

\begin{align*} y_1 &= {\mathrm e}^{-2 x}\\ y_2 &= {\mathrm e}^{x}\\ y_3 &= x \,{\mathrm e}^{x}\\ y_4 &= x^{2} {\mathrm e}^{x} \end{align*}

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

\[ y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3 y^{\prime \prime }+5 y^{\prime }-2 y = 0 \]

Now the particular solution to the given ODE is found

\[ y^{\prime \prime \prime \prime }-y^{\prime \prime \prime }-3 y^{\prime \prime }+5 y^{\prime }-2 y = \left (x \,{\mathrm e}^{3 x}+3\right ) {\mathrm e}^{-2 x} \]

The particular solution is now found using the method of undetermined coefficients.

Looking at the RHS of the ode, which is

\[ \left (x \,{\mathrm e}^{3 x}+3\right ) {\mathrm e}^{-2 x} \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{{\mathrm e}^{-2 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}, {\mathrm e}^{x}, {\mathrm e}^{-2 x}\} \]

Since \({\mathrm e}^{-2 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x \,{\mathrm e}^{-2 x}\}, \{x \,{\mathrm e}^{x}, {\mathrm e}^{x}\}] \]

Since \({\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x \,{\mathrm e}^{-2 x}\}, \{x \,{\mathrm e}^{x}, x^{2} {\mathrm e}^{x}\}] \]

Since \(x \,{\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x \,{\mathrm e}^{-2 x}\}, \{x^{2} {\mathrm e}^{x}, x^{3} {\mathrm e}^{x}\}] \]

Since \(x^{2} {\mathrm e}^{x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x \,{\mathrm e}^{-2 x}\}, \{x^{3} {\mathrm e}^{x}, x^{4} {\mathrm e}^{x}\}] \]

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.

\[ y_p = A_{1} x \,{\mathrm e}^{-2 x}+A_{2} x^{3} {\mathrm e}^{x}+A_{3} x^{4} {\mathrm e}^{x} \]

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

\[ 72 A_{3} x \,{\mathrm e}^{x}+18 A_{2} {\mathrm e}^{x}+24 A_{3} {\mathrm e}^{x}-27 A_{1} {\mathrm e}^{-2 x} = x \,{\mathrm e}^{x}+3 \,{\mathrm e}^{-2 x} \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = -{\frac {1}{9}}, A_{2} = -{\frac {1}{54}}, A_{3} = {\frac {1}{72}}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = -\frac {x \,{\mathrm e}^{-2 x}}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {x^{4} {\mathrm e}^{x}}{72} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-2 x} c_1 +{\mathrm e}^{x} c_2 +x \,{\mathrm e}^{x} c_3 +x^{2} {\mathrm e}^{x} c_4\right ) + \left (-\frac {x \,{\mathrm e}^{-2 x}}{9}-\frac {x^{3} {\mathrm e}^{x}}{54}+\frac {x^{4} {\mathrm e}^{x}}{72}\right ) \\ \end{align*}

2.42.2 Maple step by step solution

2.42.3 Maple trace

Methods for high order ODEs:
 

2.42.4 Maple dsolve solution

Solving time : 0.006 (sec)
Leaf size : 52

dsolve(diff(diff(diff(diff(y(x),x),x),x),x)-diff(diff(diff(y(x),x),x),x)-3*diff(diff(y(x),x),x)+5*diff(y(x),x)-2*y(x) = x*exp(x)+3*exp(-2*x), 
       y(x),singsol=all)
 

\[ y = \frac {\left (\left (x^{4}-\frac {4 x^{3}}{3}+\left (72 c_4 +\frac {4}{3}\right ) x^{2}+\left (72 c_3 -\frac {8}{9}\right ) x +72 c_1 +\frac {8}{27}\right ) {\mathrm e}^{3 x}-8 x +72 c_2 -8\right ) {\mathrm e}^{-2 x}}{72} \]

2.42.5 Mathematica DSolve solution

Solving time : 0.408 (sec)
Leaf size : 64

DSolve[{D[y[x],{x,4}]-D[y[x],{x,3}]-3*D[y[x],{x,2}]+5*D[y[x],x]-2*y[x]==x*Exp[x]+3*Exp[-2*x],{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to e^x \left (\frac {x^4}{72}-\frac {x^3}{54}+\left (\frac {1}{54}+c_4\right ) x^2+\left (-\frac {1}{81}+c_3\right ) x+\frac {1}{243}+c_2\right )-\frac {1}{9} e^{-2 x} (x+1-9 c_1) \]