y''+y'-6y=e^(-3x)?

isaac3211 wrote:y''+y'-6y=e^(-3x)?

first the complementary solution for: y''+y'-6y=0
y(x)=e^(λ*x)

e^(λ*x) d^2/dx^2 + e^(λ*x) d/dx - 6*e^(λ*x) = 0

substitute:
e^(λ*x) d^2/dx^2 with λ^2*e^(λ*x)
e^(λ*x) d/dx with λ*e^(λ*x)

λ^2*e^(λ*x)+λ*e^(λ*x)-6*e^(λ*x) = 0

(λ^2+λ-6)*e^(λ*x) = 0
solve:
λ^2+λ-6=0
λ=-3 and λ=2

so:
y1(x) = c1*e^(-3*x)
y2(x) = c2*e^(2*x)
(c1,c2 = constant)
=> the complementary solution y(x):
y(x) = y1(x)+y2(x) = c1*e^(-3*x) + c2*e^(2*x)



ok, looks good. now to the particular solution: y''+y'-6y=e^(-3x)
yp(x) = x*a1*e^(-3x)

yp(x) d/dx = a1*e^(-3x)*-3*a1*e^(-3x)*x


yp(x) d^2/dx^2 = a1*( -6*e^(-3x)+9*e^(-3x)*x )


now insert both diff'd yp(x) and yp(x) into y''+y'-6y=e^(-3x)
a1*( -6*e^(-3x)+9*e^(-3x)*x ) + a1*e^(-3x)*-3*a1*e^(-3x)*x - 6*x*a1*e^(-3x) = e^(-3x)
-5*a1*e^(-3x) = e^(-3x)
a1 = -1/5
yp(x) = -(1/5)*e^(-3x)*x

y(x) = c1*e^(-3*x) + c2*e^(2*x) + y(p)
y(x) = c1*e^(-3x) + c2*e^(2x) - (1/5)*e^(-3x)*x