First order linear differential equations help?

divide by dx

dy/dx + 3y = e^(-3x) -----------(1)

Equation of the form dy/dx +yp(x)=q(x)

p(x)=3

q(x)=e^(-3x)

Compute the integrating factor e^∫p(x) dx = e^∫3 dx = e^(3x)

Multiply both sides of (1) by the integraing factor e^(3x)

e^(3x) dy/dx + 3y e^(3x) = e^(-3x)e^(3x)

e^(3x) dy/dx + 3y e^(3x) = 1 --------(2)

The left side of (2) is the derivative of ye^(3x) or [ye^(3x)]'

(2) may be written as

[ye^(3x)] ' = 1

Integrate both sides

∫[ye^(3x)] ' = ∫ 1 dx

y e^(3x) = x + C

divide both sides by e^(3x)

y = x e^(-3x) + C e^(-3x)

Lol so easy

jst divide by dy

u wil get

1 + 3y (dx/dy) =e^(-3x) (dx/dy)

nw after arranging

(e^(-3x) - 3y ) (dx/dy) = 1

nw

dx/dy = (1 / (whole thing) )

this is simplified and i hope u can solve next steps

2(dy/dx) = 5 - 6y This is actually seperable: dy/(5 - 6y) = dx/x Integrating both sides: -1/6*ln|5 - 6y| = ln|x| + C solve for y.