#### djeitnstine

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**1. The problem statement, all variables and given/known data**

[tex]y''+y=tan(x)+e^{3x}-1[/tex]

**2. Relevant equations**

homogeneous solution:

[tex]y_{hom..}=C_{1}cos(x)+C_{2}sin(x)[/tex]

particular solution:

[tex]y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)[/tex]

**3. The attempt at a solution**

[tex]v_{1}' cos(x)+v_{2}' sin(x)=0[/tex] (1)

[tex]-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 [/tex] (2)

(1)* by sin (2)* by cos

and solve for v2

[tex]sin(x)+e^{3x}cos(x)-cos(x)=v_{2}'[/tex]

[tex]-cos(x)+ \frac{3}{10}e^{3x}cos(x)-\frac{1}{10}sin(x)-sin(x)=v_{2}[/tex]

Now finding v1 is the problem here. if I reverse the multiplication and do (1)* by cos (2)* by sin it seems even worse =(. I saw the complete answer and if I remember correctly its [tex]y_{p}=\frac{1}{2}cos(x)ln(sin(x)+1)-\frac{1}{2}cos(x)ln(sin(x)-1)-\frac{1}{10}e^{3x}+1[/tex]

I really just need to know what to multiply by to get going