Ansatz of particular solution to a second order ODE

Question

I have the following second order differential equation

$$y''+y'-6y=2(x+1)e^{2x}+2e^{3x}$$

Please can you help me to choose the particular solution of this equation.

Answer 1

A particular solution of $$\tag{1} y''+y'-6y=2e^{3x} $$ is of the form $$\tag{2} y=Ae^{3x}. $$ Substituting $(2)$ into $(1)$ we get $$ (9A+3A-6A)e^{3x}=6Ae^{2x}=2e^{3x}, $$ i.e. $A=\dfrac13$.

A particular solution of $$\tag{3} y''+y'-6y=2(x+1)e^{2x} $$ is of the form $$\tag{4} y=(ax^2+bx+c)e^{2x} $$ Since \begin{eqnarray} y'&=&2(ax^2+bx+c)e^{2x}+(2ax+b)e^{2x}=[2ax^2+(2a+2b)x+b+2c]e^{2x}\\ y''&=&2[2ax^2+2(a+b)x+b+2c]e^{2x}+[4ax+2(a+b)]e^{2x}\\ &=&[4ax^2+(8a+4b)x+2a+3b+4c]e^{2x} \end{eqnarray} substituting $(4)$ into $(3)$ we get $$ 2(x+1)e^{2x}=2(5ax+a+2b)e^{2x} $$ i.e. $$ a=\dfrac15,\quad b=\dfrac25 $$ Aparticular solution of the original DE is then $$ y_p=\dfrac15(x^2+2x)e^{2x}+\dfrac13e^{3x} $$

Answer 2

Your equation is $$ (D+3)(D-2)y = 2(x+1)e^{2x}+2e^{3x} $$ The term $e^{3x}$ is annihilated by $(D-3)$, and $2(x+1)e^{2x}$ is annihilated by $(D-2)^2$. Therefore $y$ satisfies $$ (D-3)(D-2)^2(D+3)(D-2)y = 0, \\ (D-3)(D+3)(D-2)^3y = 0. $$ The general solution of the above is $$ y = Ae^{3x}+Be^{-3x}+Ce^{2x}+Exe^{2x}+Fx^2e^{2x}. $$ For this $y$, \begin{align} (D+3)(D-2)y & = (D+3)(D-2)(Ae^{3x}+Be^{-3x}+Ce^{2x}+Exe^{2x}+Fx^2e^{2x}) \\ & = (D+3)(D-2)(Ae^{3x}+Exe^{2x}+Fx^2e^{2x}) \\ & = 6Ae^{3x}+(D+3)(D-2)(Exe^{2x}+Fx^2e^{2x}) \\ & = 6Ae^{3x}+\{ (D-2)^2+5(D-2) \}(Exe^{2x}+Fx^2e^{2x}) \\ & = 6Ae^{3x}+e^{2x}\{ D^2+5D \}(Ex+Fx^2) \\ & = 6Ae^{3x}+e^{2x}\{ 2F+5E+10Fx \} \end{align} (The reduction from $D-2$ to $D$ occurs because $(D-2)f=e^{2x}D(e^{-2x}f)$.) Comparing this to the first equation gives $$ (2+2x)e^{2x}+2e^{3x} = \{(2F+5E)+10Fx\}e^{2x}+6Ae^{3x} \\ \implies 2=2F+5E,\;\;2=10F,\;\; 2=6A \\ \implies A=\frac{1}{3},\;\; F=\frac{1}{5},\;\; E=\frac{8}{25} $$ The general solution is $$ y = \frac{1}{3}e^{3x}+\frac{8}{25}xe^{2x}+\frac{1}{5}x^2e^{2x}+Be^{-3x}+Ce^{2x}. $$