Is there a better way of solving this first order linear differential equation?

[alevis]

3y'+2y-2sin(3x)+2e^(-3x)+x³+4=0

Variables
x - independent
y - dependent

Attempt at a solution
I rewrote the equation in form dy/dx+P(x)y=Q(x) and used an integrating factor of \mu(x)=ke^(2/3)x
with P(x) = 2/3 and Q(x) = 2sin(3x)-2e^(-3x)-x³-4
Since y(x) = 1/ke^(2/3)x∫e^(2x/3)(2sin(3x)-2e^(-3x)-x³-4)dx

This integral lead to two integrals that required integration by parts to be solved which was a tedious process for me. Is there a better way solve this differential equation?

 

[LCKurtz]

An alternative approach would be to solve the constant coefficient homogeneous equation and use undetermined coefficients to find a particular solution of the non-homogeneous equation. I'm guessing it might be a bit less work.

 

[JJacquelin]

3y'+2y-2sin(3x)+2exp(-3x)+x^3+4=0
Solutions of 3Y'+2Y=0 are Y(x)=C*exp(-2x/3)
A particular solution of 3(y1)'+2(y1)+4=0 obviously is (y1)=-2
A particular solution of 3(y2)'+2(y2)+x^3=0 is (y2)=a(x^3)+b(x^2)+cx+d ; compute a,b, c, d.
A particular solution of 3(y3)'+2(y3)+2exp(-3x)=0 is (y3)=a*exp(-3x) ; compute a.
A particular solution of 3(y4)'+2(y4)-2sin(3x)=0 is (y4)=a*cos(3x)+b*sin(3x) : compute a, b;
The solutions of the complete EDO are : y(x)=Y(x)+(y1)+(y2)+(y3)+(y4)