+0  
 
0
379
1
avatar
y''+y'-6y=e^(-3x)?
Guest Feb 18, 2012
 #1
avatar+3077 
0
isaac3211:

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)

[input]diff(diff(y(x),x),x)+diff(y(x),x)-6*y(x)=0[/input]

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
[input]diff( x*a1*e^(-3x), x)[/input]

yp(x) d^2/dx^2 = a1*( -6*e^(-3x)+9*e^(-3x)*x )
[input]diff(diff( x*a1*e^(-3x), x),x)[/input]

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
admin  Feb 18, 2012

16 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.