How to Solve a Differential Equation Using Laplace Transformations?

[gonch76]

Homework Statement

Solve df/dx (x)+ f(x)= 3 under the condition f(0)= -2 using the Laplace transformations

Homework Equations

The Attempt at a Solution

Not even sure where to start on this one. Dont have any examples that point me in the right direction. Any ideas anyone??

 

[losiu99]

Do you know what is Laplace transform of f'(x) (for general f)? That's what makes Laplace transform a tool suitable to solving differential equation.

 

[gomunkul51]

Did you studied Laplace transforms?
if you did it should be pretty obvious :)

Tell us if it rings a bell:

http://www.math.oregonstate.edu/hom...QuestStudyGuides/ode/laplace/solve/solve.html

*do you know/have a Laplace transform table? :)

 

[gonch76]

Manged to final get to this answer:

df/dx (x)+ f(x)= 3
ℓ ( df/dx+ dx) = l 3
ℓ ( df/dx )+ l (dx) = l 3
ℓ ( df/dx )= f^' (x)= sF(s)- f(0)
ℓ (dx) = F(s)
∴ sF(s)- f(0) + F(s) = l 3
∴ sF(s) – f(0) + F(s) = 3/s
F(s)(s + 1) – f(0) = 3/s
F(s)(s + 1) – (-2) = 3/s
F(s)(s + 1) +2 = 3/s
F(s)(s + 1) = 3/s – 2
F(s) = 3/(s(s+1)) - 2/(s+1)
3/(s(s+1)) = 3/s- 3/(s+1)
∴ F(s) = 3/s - 3/(s+1) - 2/(s+1)
∴ F(s) = 3/s - 5/(s+1)
l^(-1) (3/s ) = 3
l^(-1) (5/(s+1) )= 〖5e〗^(-t)

∴ F(s) = 3 - 5e^(-t)


think its pretty good but would like some feedback. Cheers.

 

[vela]

Looks good though there are some typos. The last line should be f(t)=..., not F(s)=...

 

[gonch76]

Thanks Vela. If that's all i got wrong then i am quite happy!